Python | Pandas DataFrame.reset_index ()

Python Methods and Functions | reset_index | reset_index

Pandas reset_index() — it is a method to reset the index of a data frame. The .reset_index () method sets a list of integers ranging from 0 to the length of the data as the index.

Syntax:
DataFrame..reset_index (level = None, drop = False, inplace = False, col_level = 0, col_fill = ”)

Parameters:
level: int, string or a list to select and remove passed column from index.
drop: Boolean value, Adds the replaced index column to the data if False.
inplace: Boolean value, make changes in the original data frame itself if True.
col_level: Select in which column level to insert the labels.
col_fill: Object, to determine how the other levels are named.

Return type: DataFrame

To load the CSV file you are using, press here.

Example # 1: Resetting the index
In this example, to reset the index, the Name column was first set as the column index, and then a new index was created using the reset index.

# import pandas package

import pandas as pd

 
# create data frame from CSV file

data = pd.read_csv ( "employees.csv" )

 
# set name as index column

data.set_index ([ "First Name" ], inplace = True ,

  append = True , drop = True )

 
# reset index

data. .reset_index (inplace = True )

 
# display
data.head ( )

Output:
As shown in the output images, a new index mark was created called level_0.

Before reset —

After reset —

Example # 2: Working with a multilevel index
In this example, 2 columns (Name and Gender) are added to the index column and then one level is removed using the .reset_index () method.

# import pandas package

import pandas as pd

 
# create data frame from CSV file

data = pd.read_csv ( "employees.csv" )

 
# set name as index column

data.set_index ([ "First Name" , "Gender" ], inplace = True ,

  append = True , drop = True )

  
# reset index

data..reset_index (level = 2 , inplace = True , col_level = 1 )

 
# display
d ata.head ()

Output:
As shown in the output image, the floor column in the index column was replaced because its level was 2.

Before reset —

After reset —





Python | Pandas DataFrame.reset_index (): StackOverflow Questions

How to reset index in a pandas dataframe?

I have a dataframe from which I remove some rows. As a result, I get a dataframe in which index is something like that: [1,5,6,10,11] and I would like to reset it to [0,1,2,3,4]. How can I do it?


The following seems to work:

df = df.reset_index()
del df["index"]

The following does not work:

df = df.reindex()

Answer #1

Quick Answer:

The simplest way to get row counts per group is by calling .size(), which returns a Series:

df.groupby(["col1","col2"]).size()


Usually you want this result as a DataFrame (instead of a Series) so you can do:

df.groupby(["col1", "col2"]).size().reset_index(name="counts")


If you want to find out how to calculate the row counts and other statistics for each group continue reading below.


Detailed example:

Consider the following example dataframe:

In [2]: df
Out[2]: 
  col1 col2  col3  col4  col5  col6
0    A    B  0.20 -0.61 -0.49  1.49
1    A    B -1.53 -1.01 -0.39  1.82
2    A    B -0.44  0.27  0.72  0.11
3    A    B  0.28 -1.32  0.38  0.18
4    C    D  0.12  0.59  0.81  0.66
5    C    D -0.13 -1.65 -1.64  0.50
6    C    D -1.42 -0.11 -0.18 -0.44
7    E    F -0.00  1.42 -0.26  1.17
8    E    F  0.91 -0.47  1.35 -0.34
9    G    H  1.48 -0.63 -1.14  0.17

First let"s use .size() to get the row counts:

In [3]: df.groupby(["col1", "col2"]).size()
Out[3]: 
col1  col2
A     B       4
C     D       3
E     F       2
G     H       1
dtype: int64

Then let"s use .size().reset_index(name="counts") to get the row counts:

In [4]: df.groupby(["col1", "col2"]).size().reset_index(name="counts")
Out[4]: 
  col1 col2  counts
0    A    B       4
1    C    D       3
2    E    F       2
3    G    H       1


Including results for more statistics

When you want to calculate statistics on grouped data, it usually looks like this:

In [5]: (df
   ...: .groupby(["col1", "col2"])
   ...: .agg({
   ...:     "col3": ["mean", "count"], 
   ...:     "col4": ["median", "min", "count"]
   ...: }))
Out[5]: 
            col4                  col3      
          median   min count      mean count
col1 col2                                   
A    B    -0.810 -1.32     4 -0.372500     4
C    D    -0.110 -1.65     3 -0.476667     3
E    F     0.475 -0.47     2  0.455000     2
G    H    -0.630 -0.63     1  1.480000     1

The result above is a little annoying to deal with because of the nested column labels, and also because row counts are on a per column basis.

To gain more control over the output I usually split the statistics into individual aggregations that I then combine using join. It looks like this:

In [6]: gb = df.groupby(["col1", "col2"])
   ...: counts = gb.size().to_frame(name="counts")
   ...: (counts
   ...:  .join(gb.agg({"col3": "mean"}).rename(columns={"col3": "col3_mean"}))
   ...:  .join(gb.agg({"col4": "median"}).rename(columns={"col4": "col4_median"}))
   ...:  .join(gb.agg({"col4": "min"}).rename(columns={"col4": "col4_min"}))
   ...:  .reset_index()
   ...: )
   ...: 
Out[6]: 
  col1 col2  counts  col3_mean  col4_median  col4_min
0    A    B       4  -0.372500       -0.810     -1.32
1    C    D       3  -0.476667       -0.110     -1.65
2    E    F       2   0.455000        0.475     -0.47
3    G    H       1   1.480000       -0.630     -0.63



Footnotes

The code used to generate the test data is shown below:

In [1]: import numpy as np
   ...: import pandas as pd 
   ...: 
   ...: keys = np.array([
   ...:         ["A", "B"],
   ...:         ["A", "B"],
   ...:         ["A", "B"],
   ...:         ["A", "B"],
   ...:         ["C", "D"],
   ...:         ["C", "D"],
   ...:         ["C", "D"],
   ...:         ["E", "F"],
   ...:         ["E", "F"],
   ...:         ["G", "H"] 
   ...:         ])
   ...: 
   ...: df = pd.DataFrame(
   ...:     np.hstack([keys,np.random.randn(10,4).round(2)]), 
   ...:     columns = ["col1", "col2", "col3", "col4", "col5", "col6"]
   ...: )
   ...: 
   ...: df[["col3", "col4", "col5", "col6"]] = 
   ...:     df[["col3", "col4", "col5", "col6"]].astype(float)
   ...: 


Disclaimer:

If some of the columns that you are aggregating have null values, then you really want to be looking at the group row counts as an independent aggregation for each column. Otherwise you may be misled as to how many records are actually being used to calculate things like the mean because pandas will drop NaN entries in the mean calculation without telling you about it.

Answer #2

The idiomatic way to do this with Pandas is to use the .sample method of your dataframe to sample all rows without replacement:

df.sample(frac=1)

The frac keyword argument specifies the fraction of rows to return in the random sample, so frac=1 means return all rows (in random order).


Note: If you wish to shuffle your dataframe in-place and reset the index, you could do e.g.

df = df.sample(frac=1).reset_index(drop=True)

Here, specifying drop=True prevents .reset_index from creating a column containing the old index entries.

Follow-up note: Although it may not look like the above operation is in-place, python/pandas is smart enough not to do another malloc for the shuffled object. That is, even though the reference object has changed (by which I mean id(df_old) is not the same as id(df_new)), the underlying C object is still the same. To show that this is indeed the case, you could run a simple memory profiler:

$ python3 -m memory_profiler .	est.py
Filename: .	est.py

Line #    Mem usage    Increment   Line Contents
================================================
     5     68.5 MiB     68.5 MiB   @profile
     6                             def shuffle():
     7    847.8 MiB    779.3 MiB       df = pd.DataFrame(np.random.randn(100, 1000000))
     8    847.9 MiB      0.1 MiB       df = df.sample(frac=1).reset_index(drop=True)

Answer #3

I would suggest using the duplicated method on the Pandas Index itself:

df3 = df3[~df3.index.duplicated(keep="first")]

While all the other methods work, .drop_duplicates is by far the least performant for the provided example. Furthermore, while the groupby method is only slightly less performant, I find the duplicated method to be more readable.

Using the sample data provided:

>>> %timeit df3.reset_index().drop_duplicates(subset="index", keep="first").set_index("index")
1000 loops, best of 3: 1.54 ms per loop

>>> %timeit df3.groupby(df3.index).first()
1000 loops, best of 3: 580 µs per loop

>>> %timeit df3[~df3.index.duplicated(keep="first")]
1000 loops, best of 3: 307 µs per loop

Note that you can keep the last element by changing the keep argument to "last".

It should also be noted that this method works with MultiIndex as well (using df1 as specified in Paul"s example):

>>> %timeit df1.groupby(level=df1.index.names).last()
1000 loops, best of 3: 771 µs per loop

>>> %timeit df1[~df1.index.duplicated(keep="last")]
1000 loops, best of 3: 365 µs per loop

Answer #4

df.to_numpy() is better than df.values, here"s why.*

It"s time to deprecate your usage of values and as_matrix().

pandas v0.24.0 introduced two new methods for obtaining NumPy arrays from pandas objects:

  1. to_numpy(), which is defined on Index, Series, and DataFrame objects, and
  2. array, which is defined on Index and Series objects only.

If you visit the v0.24 docs for .values, you will see a big red warning that says:

Warning: We recommend using DataFrame.to_numpy() instead.

See this section of the v0.24.0 release notes, and this answer for more information.

* - to_numpy() is my recommended method for any production code that needs to run reliably for many versions into the future. However if you"re just making a scratchpad in jupyter or the terminal, using .values to save a few milliseconds of typing is a permissable exception. You can always add the fit n finish later.



Towards Better Consistency: to_numpy()

In the spirit of better consistency throughout the API, a new method to_numpy has been introduced to extract the underlying NumPy array from DataFrames.

# Setup
df = pd.DataFrame(data={"A": [1, 2, 3], "B": [4, 5, 6], "C": [7, 8, 9]}, 
                  index=["a", "b", "c"])

# Convert the entire DataFrame
df.to_numpy()
# array([[1, 4, 7],
#        [2, 5, 8],
#        [3, 6, 9]])

# Convert specific columns
df[["A", "C"]].to_numpy()
# array([[1, 7],
#        [2, 8],
#        [3, 9]])

As mentioned above, this method is also defined on Index and Series objects (see here).

df.index.to_numpy()
# array(["a", "b", "c"], dtype=object)

df["A"].to_numpy()
#  array([1, 2, 3])

By default, a view is returned, so any modifications made will affect the original.

v = df.to_numpy()
v[0, 0] = -1
 
df
   A  B  C
a -1  4  7
b  2  5  8
c  3  6  9

If you need a copy instead, use to_numpy(copy=True).


pandas >= 1.0 update for ExtensionTypes

If you"re using pandas 1.x, chances are you"ll be dealing with extension types a lot more. You"ll have to be a little more careful that these extension types are correctly converted.

a = pd.array([1, 2, None], dtype="Int64")                                  
a                                                                          

<IntegerArray>
[1, 2, <NA>]
Length: 3, dtype: Int64 

# Wrong
a.to_numpy()                                                               
# array([1, 2, <NA>], dtype=object)  # yuck, objects

# Correct
a.to_numpy(dtype="float", na_value=np.nan)                                 
# array([ 1.,  2., nan])

# Also correct
a.to_numpy(dtype="int", na_value=-1)
# array([ 1,  2, -1])

This is called out in the docs.


If you need the dtypes in the result...

As shown in another answer, DataFrame.to_records is a good way to do this.

df.to_records()
# rec.array([("a", 1, 4, 7), ("b", 2, 5, 8), ("c", 3, 6, 9)],
#           dtype=[("index", "O"), ("A", "<i8"), ("B", "<i8"), ("C", "<i8")])

This cannot be done with to_numpy, unfortunately. However, as an alternative, you can use np.rec.fromrecords:

v = df.reset_index()
np.rec.fromrecords(v, names=v.columns.tolist())
# rec.array([("a", 1, 4, 7), ("b", 2, 5, 8), ("c", 3, 6, 9)],
#           dtype=[("index", "<U1"), ("A", "<i8"), ("B", "<i8"), ("C", "<i8")])

Performance wise, it"s nearly the same (actually, using rec.fromrecords is a bit faster).

df2 = pd.concat([df] * 10000)

%timeit df2.to_records()
%%timeit
v = df2.reset_index()
np.rec.fromrecords(v, names=v.columns.tolist())

12.9 ms ± 511 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
9.56 ms ± 291 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)


Rationale for Adding a New Method

to_numpy() (in addition to array) was added as a result of discussions under two GitHub issues GH19954 and GH23623.

Specifically, the docs mention the rationale:

[...] with .values it was unclear whether the returned value would be the actual array, some transformation of it, or one of pandas custom arrays (like Categorical). For example, with PeriodIndex, .values generates a new ndarray of period objects each time. [...]

to_numpy aims to improve the consistency of the API, which is a major step in the right direction. .values will not be deprecated in the current version, but I expect this may happen at some point in the future, so I would urge users to migrate towards the newer API, as soon as you can.



Critique of Other Solutions

DataFrame.values has inconsistent behaviour, as already noted.

DataFrame.get_values() is simply a wrapper around DataFrame.values, so everything said above applies.

DataFrame.as_matrix() is deprecated now, do NOT use!

Answer #5

You can use the DataFrame constructor with lists created by to_list:

import pandas as pd

d1 = {"teams": [["SF", "NYG"],["SF", "NYG"],["SF", "NYG"],
                ["SF", "NYG"],["SF", "NYG"],["SF", "NYG"],["SF", "NYG"]]}
df2 = pd.DataFrame(d1)
print (df2)
       teams
0  [SF, NYG]
1  [SF, NYG]
2  [SF, NYG]
3  [SF, NYG]
4  [SF, NYG]
5  [SF, NYG]
6  [SF, NYG]

df2[["team1","team2"]] = pd.DataFrame(df2.teams.tolist(), index= df2.index)
print (df2)
       teams team1 team2
0  [SF, NYG]    SF   NYG
1  [SF, NYG]    SF   NYG
2  [SF, NYG]    SF   NYG
3  [SF, NYG]    SF   NYG
4  [SF, NYG]    SF   NYG
5  [SF, NYG]    SF   NYG
6  [SF, NYG]    SF   NYG

And for a new DataFrame:

df3 = pd.DataFrame(df2["teams"].to_list(), columns=["team1","team2"])
print (df3)
  team1 team2
0    SF   NYG
1    SF   NYG
2    SF   NYG
3    SF   NYG
4    SF   NYG
5    SF   NYG
6    SF   NYG

A solution with apply(pd.Series) is very slow:

#7k rows
df2 = pd.concat([df2]*1000).reset_index(drop=True)

In [121]: %timeit df2["teams"].apply(pd.Series)
1.79 s ± 52.5 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [122]: %timeit pd.DataFrame(df2["teams"].to_list(), columns=["team1","team2"])
1.63 ms ± 54.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Answer #6

UPDATE
From v0.20, melt is a first order function, you can now use

df.melt(id_vars=["location", "name"], 
        var_name="Date", 
        value_name="Value")

  location    name        Date  Value
0        A  "test"    Jan-2010     12
1        B   "foo"    Jan-2010     18
2        A  "test"    Feb-2010     20
3        B   "foo"    Feb-2010     20
4        A  "test"  March-2010     30
5        B   "foo"  March-2010     25

OLD(ER) VERSIONS: <0.20

You can use pd.melt to get most of the way there, and then sort:

>>> df
  location  name  Jan-2010  Feb-2010  March-2010
0        A  test        12        20          30
1        B   foo        18        20          25
>>> df2 = pd.melt(df, id_vars=["location", "name"], 
                  var_name="Date", value_name="Value")
>>> df2
  location  name        Date  Value
0        A  test    Jan-2010     12
1        B   foo    Jan-2010     18
2        A  test    Feb-2010     20
3        B   foo    Feb-2010     20
4        A  test  March-2010     30
5        B   foo  March-2010     25
>>> df2 = df2.sort(["location", "name"])
>>> df2
  location  name        Date  Value
0        A  test    Jan-2010     12
2        A  test    Feb-2010     20
4        A  test  March-2010     30
1        B   foo    Jan-2010     18
3        B   foo    Feb-2010     20
5        B   foo  March-2010     25

(Might want to throw in a .reset_index(drop=True), just to keep the output clean.)

Note: pd.DataFrame.sort has been deprecated in favour of pd.DataFrame.sort_values.

Answer #7

I know object columns type makes the data hard to convert with a pandas function. When I received the data like this, the first thing that came to mind was to "flatten" or unnest the columns .

I am using pandas and python functions for this type of question. If you are worried about the speed of the above solutions, check user3483203"s answer, since it"s using numpy and most of the time numpy is faster . I recommend Cpython and numba if speed matters.


Method 0 [pandas >= 0.25]
Starting from pandas 0.25, if you only need to explode one column, you can use the pandas.DataFrame.explode function:

df.explode("B")

       A  B
    0  1  1
    1  1  2
    0  2  1
    1  2  2

Given a dataframe with an empty list or a NaN in the column. An empty list will not cause an issue, but a NaN will need to be filled with a list

df = pd.DataFrame({"A": [1, 2, 3, 4],"B": [[1, 2], [1, 2], [], np.nan]})
df.B = df.B.fillna({i: [] for i in df.index})  # replace NaN with []
df.explode("B")

   A    B
0  1    1
0  1    2
1  2    1
1  2    2
2  3  NaN
3  4  NaN

Method 1
apply + pd.Series (easy to understand but in terms of performance not recommended . )

df.set_index("A").B.apply(pd.Series).stack().reset_index(level=0).rename(columns={0:"B"})
Out[463]: 
   A  B
0  1  1
1  1  2
0  2  1
1  2  2

Method 2
Using repeat with DataFrame constructor , re-create your dataframe (good at performance, not good at multiple columns )

df=pd.DataFrame({"A":df.A.repeat(df.B.str.len()),"B":np.concatenate(df.B.values)})
df
Out[465]: 
   A  B
0  1  1
0  1  2
1  2  1
1  2  2

Method 2.1
for example besides A we have A.1 .....A.n. If we still use the method(Method 2) above it is hard for us to re-create the columns one by one .

Solution : join or merge with the index after "unnest" the single columns

s=pd.DataFrame({"B":np.concatenate(df.B.values)},index=df.index.repeat(df.B.str.len()))
s.join(df.drop("B",1),how="left")
Out[477]: 
   B  A
0  1  1
0  2  1
1  1  2
1  2  2

If you need the column order exactly the same as before, add reindex at the end.

s.join(df.drop("B",1),how="left").reindex(columns=df.columns)

Method 3
recreate the list

pd.DataFrame([[x] + [z] for x, y in df.values for z in y],columns=df.columns)
Out[488]: 
   A  B
0  1  1
1  1  2
2  2  1
3  2  2

If more than two columns, use

s=pd.DataFrame([[x] + [z] for x, y in zip(df.index,df.B) for z in y])
s.merge(df,left_on=0,right_index=True)
Out[491]: 
   0  1  A       B
0  0  1  1  [1, 2]
1  0  2  1  [1, 2]
2  1  1  2  [1, 2]
3  1  2  2  [1, 2]

Method 4
using reindex or loc

df.reindex(df.index.repeat(df.B.str.len())).assign(B=np.concatenate(df.B.values))
Out[554]: 
   A  B
0  1  1
0  1  2
1  2  1
1  2  2

#df.loc[df.index.repeat(df.B.str.len())].assign(B=np.concatenate(df.B.values))

Method 5
when the list only contains unique values:

df=pd.DataFrame({"A":[1,2],"B":[[1,2],[3,4]]})
from collections import ChainMap
d = dict(ChainMap(*map(dict.fromkeys, df["B"], df["A"])))
pd.DataFrame(list(d.items()),columns=df.columns[::-1])
Out[574]: 
   B  A
0  1  1
1  2  1
2  3  2
3  4  2

Method 6
using numpy for high performance:

newvalues=np.dstack((np.repeat(df.A.values,list(map(len,df.B.values))),np.concatenate(df.B.values)))
pd.DataFrame(data=newvalues[0],columns=df.columns)
   A  B
0  1  1
1  1  2
2  2  1
3  2  2

Method 7
using base function itertools cycle and chain: Pure python solution just for fun

from itertools import cycle,chain
l=df.values.tolist()
l1=[list(zip([x[0]], cycle(x[1])) if len([x[0]]) > len(x[1]) else list(zip(cycle([x[0]]), x[1]))) for x in l]
pd.DataFrame(list(chain.from_iterable(l1)),columns=df.columns)
   A  B
0  1  1
1  1  2
2  2  1
3  2  2

Generalizing to multiple columns

df=pd.DataFrame({"A":[1,2],"B":[[1,2],[3,4]],"C":[[1,2],[3,4]]})
df
Out[592]: 
   A       B       C
0  1  [1, 2]  [1, 2]
1  2  [3, 4]  [3, 4]

Self-def function:

def unnesting(df, explode):
    idx = df.index.repeat(df[explode[0]].str.len())
    df1 = pd.concat([
        pd.DataFrame({x: np.concatenate(df[x].values)}) for x in explode], axis=1)
    df1.index = idx

    return df1.join(df.drop(explode, 1), how="left")

        
unnesting(df,["B","C"])
Out[609]: 
   B  C  A
0  1  1  1
0  2  2  1
1  3  3  2
1  4  4  2

Column-wise Unnesting

All above method is talking about the vertical unnesting and explode , If you do need expend the list horizontal, Check with pd.DataFrame constructor

df.join(pd.DataFrame(df.B.tolist(),index=df.index).add_prefix("B_"))
Out[33]: 
   A       B       C  B_0  B_1
0  1  [1, 2]  [1, 2]    1    2
1  2  [3, 4]  [3, 4]    3    4

Updated function

def unnesting(df, explode, axis):
    if axis==1:
        idx = df.index.repeat(df[explode[0]].str.len())
        df1 = pd.concat([
            pd.DataFrame({x: np.concatenate(df[x].values)}) for x in explode], axis=1)
        df1.index = idx

        return df1.join(df.drop(explode, 1), how="left")
    else :
        df1 = pd.concat([
                         pd.DataFrame(df[x].tolist(), index=df.index).add_prefix(x) for x in explode], axis=1)
        return df1.join(df.drop(explode, 1), how="left")

Test Output

unnesting(df, ["B","C"], axis=0)
Out[36]: 
   B0  B1  C0  C1  A
0   1   2   1   2  1
1   3   4   3   4  2

Update 2021-02-17 with original explode function

def unnesting(df, explode, axis):
    if axis==1:
        df1 = pd.concat([df[x].explode() for x in explode], axis=1)
        return df1.join(df.drop(explode, 1), how="left")
    else :
        df1 = pd.concat([
                         pd.DataFrame(df[x].tolist(), index=df.index).add_prefix(x) for x in explode], axis=1)
        return df1.join(df.drop(explode, 1), how="left")

Answer #8

You can groupby on cols "A" and "B" and call size and then reset_index and rename the generated column:

In [26]:

df1.groupby(["A","B"]).size().reset_index().rename(columns={0:"count"})
Out[26]:
     A    B  count
0   no   no      1
1   no  yes      2
2  yes   no      4
3  yes  yes      3

update

A little explanation, by grouping on the 2 columns, this groups rows where A and B values are the same, we call size which returns the number of unique groups:

In[202]:
df1.groupby(["A","B"]).size()

Out[202]: 
A    B  
no   no     1
     yes    2
yes  no     4
     yes    3
dtype: int64

So now to restore the grouped columns, we call reset_index:

In[203]:
df1.groupby(["A","B"]).size().reset_index()

Out[203]: 
     A    B  0
0   no   no  1
1   no  yes  2
2  yes   no  4
3  yes  yes  3

This restores the indices but the size aggregation is turned into a generated column 0, so we have to rename this:

In[204]:
df1.groupby(["A","B"]).size().reset_index().rename(columns={0:"count"})

Out[204]: 
     A    B  count
0   no   no      1
1   no  yes      2
2  yes   no      4
3  yes  yes      3

groupby does accept the arg as_index which we could have set to False so it doesn"t make the grouped columns the index, but this generates a series and you"d still have to restore the indices and so on....:

In[205]:
df1.groupby(["A","B"], as_index=False).size()

Out[205]: 
A    B  
no   no     1
     yes    2
yes  no     4
     yes    3
dtype: int64

Answer #9

You can reset the index using reset_index to get back a default index of 0, 1, 2, ..., n-1 (and use drop=True to indicate you want to drop the existing index instead of adding it as an additional column to your dataframe):

In [19]: df2 = df2.reset_index(drop=True)

In [20]: df2
Out[20]:
   x  y
0  0  0
1  0  1
2  0  2
3  1  0
4  1  1
5  1  2
6  2  0
7  2  1
8  2  2

Answer #10

TLDR; No, for loops are not blanket "bad", at least, not always. It is probably more accurate to say that some vectorized operations are slower than iterating, versus saying that iteration is faster than some vectorized operations. Knowing when and why is key to getting the most performance out of your code. In a nutshell, these are the situations where it is worth considering an alternative to vectorized pandas functions:

  1. When your data is small (...depending on what you"re doing),
  2. When dealing with object/mixed dtypes
  3. When using the str/regex accessor functions

Let"s examine these situations individually.


Iteration v/s Vectorization on Small Data

Pandas follows a "Convention Over Configuration" approach in its API design. This means that the same API has been fitted to cater to a broad range of data and use cases.

When a pandas function is called, the following things (among others) must internally be handled by the function, to ensure working

  1. Index/axis alignment
  2. Handling mixed datatypes
  3. Handling missing data

Almost every function will have to deal with these to varying extents, and this presents an overhead. The overhead is less for numeric functions (for example, Series.add), while it is more pronounced for string functions (for example, Series.str.replace).

for loops, on the other hand, are faster then you think. What"s even better is list comprehensions (which create lists through for loops) are even faster as they are optimized iterative mechanisms for list creation.

List comprehensions follow the pattern

[f(x) for x in seq]

Where seq is a pandas series or DataFrame column. Or, when operating over multiple columns,

[f(x, y) for x, y in zip(seq1, seq2)]

Where seq1 and seq2 are columns.

Numeric Comparison
Consider a simple boolean indexing operation. The list comprehension method has been timed against Series.ne (!=) and query. Here are the functions:

# Boolean indexing with Numeric value comparison.
df[df.A != df.B]                            # vectorized !=
df.query("A != B")                          # query (numexpr)
df[[x != y for x, y in zip(df.A, df.B)]]    # list comp

For simplicity, I have used the perfplot package to run all the timeit tests in this post. The timings for the operations above are below:

enter image description here

The list comprehension outperforms query for moderately sized N, and even outperforms the vectorized not equals comparison for tiny N. Unfortunately, the list comprehension scales linearly, so it does not offer much performance gain for larger N.

Note
It is worth mentioning that much of the benefit of list comprehension come from not having to worry about the index alignment, but this means that if your code is dependent on indexing alignment, this will break. In some cases, vectorised operations over the underlying NumPy arrays can be considered as bringing in the "best of both worlds", allowing for vectorisation without all the unneeded overhead of the pandas functions. This means that you can rewrite the operation above as

df[df.A.values != df.B.values]

Which outperforms both the pandas and list comprehension equivalents:

NumPy vectorization is out of the scope of this post, but it is definitely worth considering, if performance matters.

Value Counts
Taking another example - this time, with another vanilla python construct that is faster than a for loop - collections.Counter. A common requirement is to compute the value counts and return the result as a dictionary. This is done with value_counts, np.unique, and Counter:

# Value Counts comparison.
ser.value_counts(sort=False).to_dict()           # value_counts
dict(zip(*np.unique(ser, return_counts=True)))   # np.unique
Counter(ser)                                     # Counter

enter image description here

The results are more pronounced, Counter wins out over both vectorized methods for a larger range of small N (~3500).

Note
More trivia (courtesy @user2357112). The Counter is implemented with a C accelerator, so while it still has to work with python objects instead of the underlying C datatypes, it is still faster than a for loop. Python power!

Of course, the take away from here is that the performance depends on your data and use case. The point of these examples is to convince you not to rule out these solutions as legitimate options. If these still don"t give you the performance you need, there is always cython and numba. Let"s add this test into the mix.

from numba import njit, prange

@njit(parallel=True)
def get_mask(x, y):
    result = [False] * len(x)
    for i in prange(len(x)):
        result[i] = x[i] != y[i]

    return np.array(result)

df[get_mask(df.A.values, df.B.values)] # numba

enter image description here

Numba offers JIT compilation of loopy python code to very powerful vectorized code. Understanding how to make numba work involves a learning curve.


Operations with Mixed/object dtypes

String-based Comparison
Revisiting the filtering example from the first section, what if the columns being compared are strings? Consider the same 3 functions above, but with the input DataFrame cast to string.

# Boolean indexing with string value comparison.
df[df.A != df.B]                            # vectorized !=
df.query("A != B")                          # query (numexpr)
df[[x != y for x, y in zip(df.A, df.B)]]    # list comp

enter image description here

So, what changed? The thing to note here is that string operations are inherently difficult to vectorize. Pandas treats strings as objects, and all operations on objects fall back to a slow, loopy implementation.

Now, because this loopy implementation is surrounded by all the overhead mentioned above, there is a constant magnitude difference between these solutions, even though they scale the same.

When it comes to operations on mutable/complex objects, there is no comparison. List comprehension outperforms all operations involving dicts and lists.

Accessing Dictionary Value(s) by Key
Here are timings for two operations that extract a value from a column of dictionaries: map and the list comprehension. The setup is in the Appendix, under the heading "Code Snippets".

# Dictionary value extraction.
ser.map(operator.itemgetter("value"))     # map
pd.Series([x.get("value") for x in ser])  # list comprehension

enter image description here

Positional List Indexing
Timings for 3 operations that extract the 0th element from a list of columns (handling exceptions), map, str.get accessor method, and the list comprehension:

# List positional indexing. 
def get_0th(lst):
    try:
        return lst[0]
    # Handle empty lists and NaNs gracefully.
    except (IndexError, TypeError):
        return np.nan

ser.map(get_0th)                                          # map
ser.str[0]                                                # str accessor
pd.Series([x[0] if len(x) > 0 else np.nan for x in ser])  # list comp
pd.Series([get_0th(x) for x in ser])                      # list comp safe

Note
If the index matters, you would want to do:

pd.Series([...], index=ser.index)

When reconstructing the series.

enter image description here

List Flattening
A final example is flattening lists. This is another common problem, and demonstrates just how powerful pure python is here.

# Nested list flattening.
pd.DataFrame(ser.tolist()).stack().reset_index(drop=True)  # stack
pd.Series(list(chain.from_iterable(ser.tolist())))         # itertools.chain
pd.Series([y for x in ser for y in x])                     # nested list comp

enter image description here

Both itertools.chain.from_iterable and the nested list comprehension are pure python constructs, and scale much better than the stack solution.

These timings are a strong indication of the fact that pandas is not equipped to work with mixed dtypes, and that you should probably refrain from using it to do so. Wherever possible, data should be present as scalar values (ints/floats/strings) in separate columns.

Lastly, the applicability of these solutions depend widely on your data. So, the best thing to do would be to test these operations on your data before deciding what to go with. Notice how I have not timed apply on these solutions, because it would skew the graph (yes, it"s that slow).


Regex Operations, and .str Accessor Methods

Pandas can apply regex operations such as str.contains, str.extract, and str.extractall, as well as other "vectorized" string operations (such as str.split, str.find,str.translate`, and so on) on string columns. These functions are slower than list comprehensions, and are meant to be more convenience functions than anything else.

It is usually much faster to pre-compile a regex pattern and iterate over your data with re.compile (also see Is it worth using Python's re.compile?). The list comp equivalent to str.contains looks something like this:

p = re.compile(...)
ser2 = pd.Series([x for x in ser if p.search(x)])

Or,

ser2 = ser[[bool(p.search(x)) for x in ser]]

If you need to handle NaNs, you can do something like

ser[[bool(p.search(x)) if pd.notnull(x) else False for x in ser]]

The list comp equivalent to str.extract (without groups) will look something like:

df["col2"] = [p.search(x).group(0) for x in df["col"]]

If you need to handle no-matches and NaNs, you can use a custom function (still faster!):

def matcher(x):
    m = p.search(str(x))
    if m:
        return m.group(0)
    return np.nan

df["col2"] = [matcher(x) for x in df["col"]]

The matcher function is very extensible. It can be fitted to return a list for each capture group, as needed. Just extract query the group or groups attribute of the matcher object.

For str.extractall, change p.search to p.findall.

String Extraction
Consider a simple filtering operation. The idea is to extract 4 digits if it is preceded by an upper case letter.

# Extracting strings.
p = re.compile(r"(?<=[A-Z])(d{4})")
def matcher(x):
    m = p.search(x)
    if m:
        return m.group(0)
    return np.nan

ser.str.extract(r"(?<=[A-Z])(d{4})", expand=False)   #  str.extract
pd.Series([matcher(x) for x in ser])                  #  list comprehension

enter image description here

More Examples
Full disclosure - I am the author (in part or whole) of these posts listed below.


Conclusion

As shown from the examples above, iteration shines when working with small rows of DataFrames, mixed datatypes, and regular expressions.

The speedup you get depends on your data and your problem, so your mileage may vary. The best thing to do is to carefully run tests and see if the payout is worth the effort.

The "vectorized" functions shine in their simplicity and readability, so if performance is not critical, you should definitely prefer those.

Another side note, certain string operations deal with constraints that favour the use of NumPy. Here are two examples where careful NumPy vectorization outperforms python:

Additionally, sometimes just operating on the underlying arrays via .values as opposed to on the Series or DataFrames can offer a healthy enough speedup for most usual scenarios (see the Note in the Numeric Comparison section above). So, for example df[df.A.values != df.B.values] would show instant performance boosts over df[df.A != df.B]. Using .values may not be appropriate in every situation, but it is a useful hack to know.

As mentioned above, it"s up to you to decide whether these solutions are worth the trouble of implementing.


Appendix: Code Snippets

import perfplot  
import operator 
import pandas as pd
import numpy as np
import re

from collections import Counter
from itertools import chain

# Boolean indexing with Numeric value comparison.
perfplot.show(
    setup=lambda n: pd.DataFrame(np.random.choice(1000, (n, 2)), columns=["A","B"]),
    kernels=[
        lambda df: df[df.A != df.B],
        lambda df: df.query("A != B"),
        lambda df: df[[x != y for x, y in zip(df.A, df.B)]],
        lambda df: df[get_mask(df.A.values, df.B.values)]
    ],
    labels=["vectorized !=", "query (numexpr)", "list comp", "numba"],
    n_range=[2**k for k in range(0, 15)],
    xlabel="N"
)

# Value Counts comparison.
perfplot.show(
    setup=lambda n: pd.Series(np.random.choice(1000, n)),
    kernels=[
        lambda ser: ser.value_counts(sort=False).to_dict(),
        lambda ser: dict(zip(*np.unique(ser, return_counts=True))),
        lambda ser: Counter(ser),
    ],
    labels=["value_counts", "np.unique", "Counter"],
    n_range=[2**k for k in range(0, 15)],
    xlabel="N",
    equality_check=lambda x, y: dict(x) == dict(y)
)

# Boolean indexing with string value comparison.
perfplot.show(
    setup=lambda n: pd.DataFrame(np.random.choice(1000, (n, 2)), columns=["A","B"], dtype=str),
    kernels=[
        lambda df: df[df.A != df.B],
        lambda df: df.query("A != B"),
        lambda df: df[[x != y for x, y in zip(df.A, df.B)]],
    ],
    labels=["vectorized !=", "query (numexpr)", "list comp"],
    n_range=[2**k for k in range(0, 15)],
    xlabel="N",
    equality_check=None
)

# Dictionary value extraction.
ser1 = pd.Series([{"key": "abc", "value": 123}, {"key": "xyz", "value": 456}])
perfplot.show(
    setup=lambda n: pd.concat([ser1] * n, ignore_index=True),
    kernels=[
        lambda ser: ser.map(operator.itemgetter("value")),
        lambda ser: pd.Series([x.get("value") for x in ser]),
    ],
    labels=["map", "list comprehension"],
    n_range=[2**k for k in range(0, 15)],
    xlabel="N",
    equality_check=None
)

# List positional indexing. 
ser2 = pd.Series([["a", "b", "c"], [1, 2], []])        
perfplot.show(
    setup=lambda n: pd.concat([ser2] * n, ignore_index=True),
    kernels=[
        lambda ser: ser.map(get_0th),
        lambda ser: ser.str[0],
        lambda ser: pd.Series([x[0] if len(x) > 0 else np.nan for x in ser]),
        lambda ser: pd.Series([get_0th(x) for x in ser]),
    ],
    labels=["map", "str accessor", "list comprehension", "list comp safe"],
    n_range=[2**k for k in range(0, 15)],
    xlabel="N",
    equality_check=None
)

# Nested list flattening.
ser3 = pd.Series([["a", "b", "c"], ["d", "e"], ["f", "g"]])
perfplot.show(
    setup=lambda n: pd.concat([ser2] * n, ignore_index=True),
    kernels=[
        lambda ser: pd.DataFrame(ser.tolist()).stack().reset_index(drop=True),
        lambda ser: pd.Series(list(chain.from_iterable(ser.tolist()))),
        lambda ser: pd.Series([y for x in ser for y in x]),
    ],
    labels=["stack", "itertools.chain", "nested list comp"],
    n_range=[2**k for k in range(0, 15)],
    xlabel="N",    
    equality_check=None

)

# Extracting strings.
ser4 = pd.Series(["foo xyz", "test A1234", "D3345 xtz"])
perfplot.show(
    setup=lambda n: pd.concat([ser4] * n, ignore_index=True),
    kernels=[
        lambda ser: ser.str.extract(r"(?<=[A-Z])(d{4})", expand=False),
        lambda ser: pd.Series([matcher(x) for x in ser])
    ],
    labels=["str.extract", "list comprehension"],
    n_range=[2**k for k in range(0, 15)],
    xlabel="N",
    equality_check=None
)

Python | Pandas DataFrame.reset_index (): StackOverflow Questions

How to reset index in a pandas dataframe?

I have a dataframe from which I remove some rows. As a result, I get a dataframe in which index is something like that: [1,5,6,10,11] and I would like to reset it to [0,1,2,3,4]. How can I do it?


The following seems to work:

df = df.reset_index()
del df["index"]

The following does not work:

df = df.reindex()

Answer #1

Quick Answer:

The simplest way to get row counts per group is by calling .size(), which returns a Series:

df.groupby(["col1","col2"]).size()


Usually you want this result as a DataFrame (instead of a Series) so you can do:

df.groupby(["col1", "col2"]).size().reset_index(name="counts")


If you want to find out how to calculate the row counts and other statistics for each group continue reading below.


Detailed example:

Consider the following example dataframe:

In [2]: df
Out[2]: 
  col1 col2  col3  col4  col5  col6
0    A    B  0.20 -0.61 -0.49  1.49
1    A    B -1.53 -1.01 -0.39  1.82
2    A    B -0.44  0.27  0.72  0.11
3    A    B  0.28 -1.32  0.38  0.18
4    C    D  0.12  0.59  0.81  0.66
5    C    D -0.13 -1.65 -1.64  0.50
6    C    D -1.42 -0.11 -0.18 -0.44
7    E    F -0.00  1.42 -0.26  1.17
8    E    F  0.91 -0.47  1.35 -0.34
9    G    H  1.48 -0.63 -1.14  0.17

First let"s use .size() to get the row counts:

In [3]: df.groupby(["col1", "col2"]).size()
Out[3]: 
col1  col2
A     B       4
C     D       3
E     F       2
G     H       1
dtype: int64

Then let"s use .size().reset_index(name="counts") to get the row counts:

In [4]: df.groupby(["col1", "col2"]).size().reset_index(name="counts")
Out[4]: 
  col1 col2  counts
0    A    B       4
1    C    D       3
2    E    F       2
3    G    H       1


Including results for more statistics

When you want to calculate statistics on grouped data, it usually looks like this:

In [5]: (df
   ...: .groupby(["col1", "col2"])
   ...: .agg({
   ...:     "col3": ["mean", "count"], 
   ...:     "col4": ["median", "min", "count"]
   ...: }))
Out[5]: 
            col4                  col3      
          median   min count      mean count
col1 col2                                   
A    B    -0.810 -1.32     4 -0.372500     4
C    D    -0.110 -1.65     3 -0.476667     3
E    F     0.475 -0.47     2  0.455000     2
G    H    -0.630 -0.63     1  1.480000     1

The result above is a little annoying to deal with because of the nested column labels, and also because row counts are on a per column basis.

To gain more control over the output I usually split the statistics into individual aggregations that I then combine using join. It looks like this:

In [6]: gb = df.groupby(["col1", "col2"])
   ...: counts = gb.size().to_frame(name="counts")
   ...: (counts
   ...:  .join(gb.agg({"col3": "mean"}).rename(columns={"col3": "col3_mean"}))
   ...:  .join(gb.agg({"col4": "median"}).rename(columns={"col4": "col4_median"}))
   ...:  .join(gb.agg({"col4": "min"}).rename(columns={"col4": "col4_min"}))
   ...:  .reset_index()
   ...: )
   ...: 
Out[6]: 
  col1 col2  counts  col3_mean  col4_median  col4_min
0    A    B       4  -0.372500       -0.810     -1.32
1    C    D       3  -0.476667       -0.110     -1.65
2    E    F       2   0.455000        0.475     -0.47
3    G    H       1   1.480000       -0.630     -0.63



Footnotes

The code used to generate the test data is shown below:

In [1]: import numpy as np
   ...: import pandas as pd 
   ...: 
   ...: keys = np.array([
   ...:         ["A", "B"],
   ...:         ["A", "B"],
   ...:         ["A", "B"],
   ...:         ["A", "B"],
   ...:         ["C", "D"],
   ...:         ["C", "D"],
   ...:         ["C", "D"],
   ...:         ["E", "F"],
   ...:         ["E", "F"],
   ...:         ["G", "H"] 
   ...:         ])
   ...: 
   ...: df = pd.DataFrame(
   ...:     np.hstack([keys,np.random.randn(10,4).round(2)]), 
   ...:     columns = ["col1", "col2", "col3", "col4", "col5", "col6"]
   ...: )
   ...: 
   ...: df[["col3", "col4", "col5", "col6"]] = 
   ...:     df[["col3", "col4", "col5", "col6"]].astype(float)
   ...: 


Disclaimer:

If some of the columns that you are aggregating have null values, then you really want to be looking at the group row counts as an independent aggregation for each column. Otherwise you may be misled as to how many records are actually being used to calculate things like the mean because pandas will drop NaN entries in the mean calculation without telling you about it.

Answer #2

The idiomatic way to do this with Pandas is to use the .sample method of your dataframe to sample all rows without replacement:

df.sample(frac=1)

The frac keyword argument specifies the fraction of rows to return in the random sample, so frac=1 means return all rows (in random order).


Note: If you wish to shuffle your dataframe in-place and reset the index, you could do e.g.

df = df.sample(frac=1).reset_index(drop=True)

Here, specifying drop=True prevents .reset_index from creating a column containing the old index entries.

Follow-up note: Although it may not look like the above operation is in-place, python/pandas is smart enough not to do another malloc for the shuffled object. That is, even though the reference object has changed (by which I mean id(df_old) is not the same as id(df_new)), the underlying C object is still the same. To show that this is indeed the case, you could run a simple memory profiler:

$ python3 -m memory_profiler .	est.py
Filename: .	est.py

Line #    Mem usage    Increment   Line Contents
================================================
     5     68.5 MiB     68.5 MiB   @profile
     6                             def shuffle():
     7    847.8 MiB    779.3 MiB       df = pd.DataFrame(np.random.randn(100, 1000000))
     8    847.9 MiB      0.1 MiB       df = df.sample(frac=1).reset_index(drop=True)

Answer #3

I would suggest using the duplicated method on the Pandas Index itself:

df3 = df3[~df3.index.duplicated(keep="first")]

While all the other methods work, .drop_duplicates is by far the least performant for the provided example. Furthermore, while the groupby method is only slightly less performant, I find the duplicated method to be more readable.

Using the sample data provided:

>>> %timeit df3.reset_index().drop_duplicates(subset="index", keep="first").set_index("index")
1000 loops, best of 3: 1.54 ms per loop

>>> %timeit df3.groupby(df3.index).first()
1000 loops, best of 3: 580 µs per loop

>>> %timeit df3[~df3.index.duplicated(keep="first")]
1000 loops, best of 3: 307 µs per loop

Note that you can keep the last element by changing the keep argument to "last".

It should also be noted that this method works with MultiIndex as well (using df1 as specified in Paul"s example):

>>> %timeit df1.groupby(level=df1.index.names).last()
1000 loops, best of 3: 771 µs per loop

>>> %timeit df1[~df1.index.duplicated(keep="last")]
1000 loops, best of 3: 365 µs per loop

Answer #4

df.to_numpy() is better than df.values, here"s why.*

It"s time to deprecate your usage of values and as_matrix().

pandas v0.24.0 introduced two new methods for obtaining NumPy arrays from pandas objects:

  1. to_numpy(), which is defined on Index, Series, and DataFrame objects, and
  2. array, which is defined on Index and Series objects only.

If you visit the v0.24 docs for .values, you will see a big red warning that says:

Warning: We recommend using DataFrame.to_numpy() instead.

See this section of the v0.24.0 release notes, and this answer for more information.

* - to_numpy() is my recommended method for any production code that needs to run reliably for many versions into the future. However if you"re just making a scratchpad in jupyter or the terminal, using .values to save a few milliseconds of typing is a permissable exception. You can always add the fit n finish later.



Towards Better Consistency: to_numpy()

In the spirit of better consistency throughout the API, a new method to_numpy has been introduced to extract the underlying NumPy array from DataFrames.

# Setup
df = pd.DataFrame(data={"A": [1, 2, 3], "B": [4, 5, 6], "C": [7, 8, 9]}, 
                  index=["a", "b", "c"])

# Convert the entire DataFrame
df.to_numpy()
# array([[1, 4, 7],
#        [2, 5, 8],
#        [3, 6, 9]])

# Convert specific columns
df[["A", "C"]].to_numpy()
# array([[1, 7],
#        [2, 8],
#        [3, 9]])

As mentioned above, this method is also defined on Index and Series objects (see here).

df.index.to_numpy()
# array(["a", "b", "c"], dtype=object)

df["A"].to_numpy()
#  array([1, 2, 3])

By default, a view is returned, so any modifications made will affect the original.

v = df.to_numpy()
v[0, 0] = -1
 
df
   A  B  C
a -1  4  7
b  2  5  8
c  3  6  9

If you need a copy instead, use to_numpy(copy=True).


pandas >= 1.0 update for ExtensionTypes

If you"re using pandas 1.x, chances are you"ll be dealing with extension types a lot more. You"ll have to be a little more careful that these extension types are correctly converted.

a = pd.array([1, 2, None], dtype="Int64")                                  
a                                                                          

<IntegerArray>
[1, 2, <NA>]
Length: 3, dtype: Int64 

# Wrong
a.to_numpy()                                                               
# array([1, 2, <NA>], dtype=object)  # yuck, objects

# Correct
a.to_numpy(dtype="float", na_value=np.nan)                                 
# array([ 1.,  2., nan])

# Also correct
a.to_numpy(dtype="int", na_value=-1)
# array([ 1,  2, -1])

This is called out in the docs.


If you need the dtypes in the result...

As shown in another answer, DataFrame.to_records is a good way to do this.

df.to_records()
# rec.array([("a", 1, 4, 7), ("b", 2, 5, 8), ("c", 3, 6, 9)],
#           dtype=[("index", "O"), ("A", "<i8"), ("B", "<i8"), ("C", "<i8")])

This cannot be done with to_numpy, unfortunately. However, as an alternative, you can use np.rec.fromrecords:

v = df.reset_index()
np.rec.fromrecords(v, names=v.columns.tolist())
# rec.array([("a", 1, 4, 7), ("b", 2, 5, 8), ("c", 3, 6, 9)],
#           dtype=[("index", "<U1"), ("A", "<i8"), ("B", "<i8"), ("C", "<i8")])

Performance wise, it"s nearly the same (actually, using rec.fromrecords is a bit faster).

df2 = pd.concat([df] * 10000)

%timeit df2.to_records()
%%timeit
v = df2.reset_index()
np.rec.fromrecords(v, names=v.columns.tolist())

12.9 ms ± 511 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
9.56 ms ± 291 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)


Rationale for Adding a New Method

to_numpy() (in addition to array) was added as a result of discussions under two GitHub issues GH19954 and GH23623.

Specifically, the docs mention the rationale:

[...] with .values it was unclear whether the returned value would be the actual array, some transformation of it, or one of pandas custom arrays (like Categorical). For example, with PeriodIndex, .values generates a new ndarray of period objects each time. [...]

to_numpy aims to improve the consistency of the API, which is a major step in the right direction. .values will not be deprecated in the current version, but I expect this may happen at some point in the future, so I would urge users to migrate towards the newer API, as soon as you can.



Critique of Other Solutions

DataFrame.values has inconsistent behaviour, as already noted.

DataFrame.get_values() is simply a wrapper around DataFrame.values, so everything said above applies.

DataFrame.as_matrix() is deprecated now, do NOT use!

Answer #5

You can use the DataFrame constructor with lists created by to_list:

import pandas as pd

d1 = {"teams": [["SF", "NYG"],["SF", "NYG"],["SF", "NYG"],
                ["SF", "NYG"],["SF", "NYG"],["SF", "NYG"],["SF", "NYG"]]}
df2 = pd.DataFrame(d1)
print (df2)
       teams
0  [SF, NYG]
1  [SF, NYG]
2  [SF, NYG]
3  [SF, NYG]
4  [SF, NYG]
5  [SF, NYG]
6  [SF, NYG]

df2[["team1","team2"]] = pd.DataFrame(df2.teams.tolist(), index= df2.index)
print (df2)
       teams team1 team2
0  [SF, NYG]    SF   NYG
1  [SF, NYG]    SF   NYG
2  [SF, NYG]    SF   NYG
3  [SF, NYG]    SF   NYG
4  [SF, NYG]    SF   NYG
5  [SF, NYG]    SF   NYG
6  [SF, NYG]    SF   NYG

And for a new DataFrame:

df3 = pd.DataFrame(df2["teams"].to_list(), columns=["team1","team2"])
print (df3)
  team1 team2
0    SF   NYG
1    SF   NYG
2    SF   NYG
3    SF   NYG
4    SF   NYG
5    SF   NYG
6    SF   NYG

A solution with apply(pd.Series) is very slow:

#7k rows
df2 = pd.concat([df2]*1000).reset_index(drop=True)

In [121]: %timeit df2["teams"].apply(pd.Series)
1.79 s ± 52.5 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [122]: %timeit pd.DataFrame(df2["teams"].to_list(), columns=["team1","team2"])
1.63 ms ± 54.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Answer #6

UPDATE
From v0.20, melt is a first order function, you can now use

df.melt(id_vars=["location", "name"], 
        var_name="Date", 
        value_name="Value")

  location    name        Date  Value
0        A  "test"    Jan-2010     12
1        B   "foo"    Jan-2010     18
2        A  "test"    Feb-2010     20
3        B   "foo"    Feb-2010     20
4        A  "test"  March-2010     30
5        B   "foo"  March-2010     25

OLD(ER) VERSIONS: <0.20

You can use pd.melt to get most of the way there, and then sort:

>>> df
  location  name  Jan-2010  Feb-2010  March-2010
0        A  test        12        20          30
1        B   foo        18        20          25
>>> df2 = pd.melt(df, id_vars=["location", "name"], 
                  var_name="Date", value_name="Value")
>>> df2
  location  name        Date  Value
0        A  test    Jan-2010     12
1        B   foo    Jan-2010     18
2        A  test    Feb-2010     20
3        B   foo    Feb-2010     20
4        A  test  March-2010     30
5        B   foo  March-2010     25
>>> df2 = df2.sort(["location", "name"])
>>> df2
  location  name        Date  Value
0        A  test    Jan-2010     12
2        A  test    Feb-2010     20
4        A  test  March-2010     30
1        B   foo    Jan-2010     18
3        B   foo    Feb-2010     20
5        B   foo  March-2010     25

(Might want to throw in a .reset_index(drop=True), just to keep the output clean.)

Note: pd.DataFrame.sort has been deprecated in favour of pd.DataFrame.sort_values.

Answer #7

I know object columns type makes the data hard to convert with a pandas function. When I received the data like this, the first thing that came to mind was to "flatten" or unnest the columns .

I am using pandas and python functions for this type of question. If you are worried about the speed of the above solutions, check user3483203"s answer, since it"s using numpy and most of the time numpy is faster . I recommend Cpython and numba if speed matters.


Method 0 [pandas >= 0.25]
Starting from pandas 0.25, if you only need to explode one column, you can use the pandas.DataFrame.explode function:

df.explode("B")

       A  B
    0  1  1
    1  1  2
    0  2  1
    1  2  2

Given a dataframe with an empty list or a NaN in the column. An empty list will not cause an issue, but a NaN will need to be filled with a list

df = pd.DataFrame({"A": [1, 2, 3, 4],"B": [[1, 2], [1, 2], [], np.nan]})
df.B = df.B.fillna({i: [] for i in df.index})  # replace NaN with []
df.explode("B")

   A    B
0  1    1
0  1    2
1  2    1
1  2    2
2  3  NaN
3  4  NaN

Method 1
apply + pd.Series (easy to understand but in terms of performance not recommended . )

df.set_index("A").B.apply(pd.Series).stack().reset_index(level=0).rename(columns={0:"B"})
Out[463]: 
   A  B
0  1  1
1  1  2
0  2  1
1  2  2

Method 2
Using repeat with DataFrame constructor , re-create your dataframe (good at performance, not good at multiple columns )

df=pd.DataFrame({"A":df.A.repeat(df.B.str.len()),"B":np.concatenate(df.B.values)})
df
Out[465]: 
   A  B
0  1  1
0  1  2
1  2  1
1  2  2

Method 2.1
for example besides A we have A.1 .....A.n. If we still use the method(Method 2) above it is hard for us to re-create the columns one by one .

Solution : join or merge with the index after "unnest" the single columns

s=pd.DataFrame({"B":np.concatenate(df.B.values)},index=df.index.repeat(df.B.str.len()))
s.join(df.drop("B",1),how="left")
Out[477]: 
   B  A
0  1  1
0  2  1
1  1  2
1  2  2

If you need the column order exactly the same as before, add reindex at the end.

s.join(df.drop("B",1),how="left").reindex(columns=df.columns)

Method 3
recreate the list

pd.DataFrame([[x] + [z] for x, y in df.values for z in y],columns=df.columns)
Out[488]: 
   A  B
0  1  1
1  1  2
2  2  1
3  2  2

If more than two columns, use

s=pd.DataFrame([[x] + [z] for x, y in zip(df.index,df.B) for z in y])
s.merge(df,left_on=0,right_index=True)
Out[491]: 
   0  1  A       B
0  0  1  1  [1, 2]
1  0  2  1  [1, 2]
2  1  1  2  [1, 2]
3  1  2  2  [1, 2]

Method 4
using reindex or loc

df.reindex(df.index.repeat(df.B.str.len())).assign(B=np.concatenate(df.B.values))
Out[554]: 
   A  B
0  1  1
0  1  2
1  2  1
1  2  2

#df.loc[df.index.repeat(df.B.str.len())].assign(B=np.concatenate(df.B.values))

Method 5
when the list only contains unique values:

df=pd.DataFrame({"A":[1,2],"B":[[1,2],[3,4]]})
from collections import ChainMap
d = dict(ChainMap(*map(dict.fromkeys, df["B"], df["A"])))
pd.DataFrame(list(d.items()),columns=df.columns[::-1])
Out[574]: 
   B  A
0  1  1
1  2  1
2  3  2
3  4  2

Method 6
using numpy for high performance:

newvalues=np.dstack((np.repeat(df.A.values,list(map(len,df.B.values))),np.concatenate(df.B.values)))
pd.DataFrame(data=newvalues[0],columns=df.columns)
   A  B
0  1  1
1  1  2
2  2  1
3  2  2

Method 7
using base function itertools cycle and chain: Pure python solution just for fun

from itertools import cycle,chain
l=df.values.tolist()
l1=[list(zip([x[0]], cycle(x[1])) if len([x[0]]) > len(x[1]) else list(zip(cycle([x[0]]), x[1]))) for x in l]
pd.DataFrame(list(chain.from_iterable(l1)),columns=df.columns)
   A  B
0  1  1
1  1  2
2  2  1
3  2  2

Generalizing to multiple columns

df=pd.DataFrame({"A":[1,2],"B":[[1,2],[3,4]],"C":[[1,2],[3,4]]})
df
Out[592]: 
   A       B       C
0  1  [1, 2]  [1, 2]
1  2  [3, 4]  [3, 4]

Self-def function:

def unnesting(df, explode):
    idx = df.index.repeat(df[explode[0]].str.len())
    df1 = pd.concat([
        pd.DataFrame({x: np.concatenate(df[x].values)}) for x in explode], axis=1)
    df1.index = idx

    return df1.join(df.drop(explode, 1), how="left")

        
unnesting(df,["B","C"])
Out[609]: 
   B  C  A
0  1  1  1
0  2  2  1
1  3  3  2
1  4  4  2

Column-wise Unnesting

All above method is talking about the vertical unnesting and explode , If you do need expend the list horizontal, Check with pd.DataFrame constructor

df.join(pd.DataFrame(df.B.tolist(),index=df.index).add_prefix("B_"))
Out[33]: 
   A       B       C  B_0  B_1
0  1  [1, 2]  [1, 2]    1    2
1  2  [3, 4]  [3, 4]    3    4

Updated function

def unnesting(df, explode, axis):
    if axis==1:
        idx = df.index.repeat(df[explode[0]].str.len())
        df1 = pd.concat([
            pd.DataFrame({x: np.concatenate(df[x].values)}) for x in explode], axis=1)
        df1.index = idx

        return df1.join(df.drop(explode, 1), how="left")
    else :
        df1 = pd.concat([
                         pd.DataFrame(df[x].tolist(), index=df.index).add_prefix(x) for x in explode], axis=1)
        return df1.join(df.drop(explode, 1), how="left")

Test Output

unnesting(df, ["B","C"], axis=0)
Out[36]: 
   B0  B1  C0  C1  A
0   1   2   1   2  1
1   3   4   3   4  2

Update 2021-02-17 with original explode function

def unnesting(df, explode, axis):
    if axis==1:
        df1 = pd.concat([df[x].explode() for x in explode], axis=1)
        return df1.join(df.drop(explode, 1), how="left")
    else :
        df1 = pd.concat([
                         pd.DataFrame(df[x].tolist(), index=df.index).add_prefix(x) for x in explode], axis=1)
        return df1.join(df.drop(explode, 1), how="left")

Answer #8

You can groupby on cols "A" and "B" and call size and then reset_index and rename the generated column:

In [26]:

df1.groupby(["A","B"]).size().reset_index().rename(columns={0:"count"})
Out[26]:
     A    B  count
0   no   no      1
1   no  yes      2
2  yes   no      4
3  yes  yes      3

update

A little explanation, by grouping on the 2 columns, this groups rows where A and B values are the same, we call size which returns the number of unique groups:

In[202]:
df1.groupby(["A","B"]).size()

Out[202]: 
A    B  
no   no     1
     yes    2
yes  no     4
     yes    3
dtype: int64

So now to restore the grouped columns, we call reset_index:

In[203]:
df1.groupby(["A","B"]).size().reset_index()

Out[203]: 
     A    B  0
0   no   no  1
1   no  yes  2
2  yes   no  4
3  yes  yes  3

This restores the indices but the size aggregation is turned into a generated column 0, so we have to rename this:

In[204]:
df1.groupby(["A","B"]).size().reset_index().rename(columns={0:"count"})

Out[204]: 
     A    B  count
0   no   no      1
1   no  yes      2
2  yes   no      4
3  yes  yes      3

groupby does accept the arg as_index which we could have set to False so it doesn"t make the grouped columns the index, but this generates a series and you"d still have to restore the indices and so on....:

In[205]:
df1.groupby(["A","B"], as_index=False).size()

Out[205]: 
A    B  
no   no     1
     yes    2
yes  no     4
     yes    3
dtype: int64

Answer #9

You can reset the index using reset_index to get back a default index of 0, 1, 2, ..., n-1 (and use drop=True to indicate you want to drop the existing index instead of adding it as an additional column to your dataframe):

In [19]: df2 = df2.reset_index(drop=True)

In [20]: df2
Out[20]:
   x  y
0  0  0
1  0  1
2  0  2
3  1  0
4  1  1
5  1  2
6  2  0
7  2  1
8  2  2

Answer #10

TLDR; No, for loops are not blanket "bad", at least, not always. It is probably more accurate to say that some vectorized operations are slower than iterating, versus saying that iteration is faster than some vectorized operations. Knowing when and why is key to getting the most performance out of your code. In a nutshell, these are the situations where it is worth considering an alternative to vectorized pandas functions:

  1. When your data is small (...depending on what you"re doing),
  2. When dealing with object/mixed dtypes
  3. When using the str/regex accessor functions

Let"s examine these situations individually.


Iteration v/s Vectorization on Small Data

Pandas follows a "Convention Over Configuration" approach in its API design. This means that the same API has been fitted to cater to a broad range of data and use cases.

When a pandas function is called, the following things (among others) must internally be handled by the function, to ensure working

  1. Index/axis alignment
  2. Handling mixed datatypes
  3. Handling missing data

Almost every function will have to deal with these to varying extents, and this presents an overhead. The overhead is less for numeric functions (for example, Series.add), while it is more pronounced for string functions (for example, Series.str.replace).

for loops, on the other hand, are faster then you think. What"s even better is list comprehensions (which create lists through for loops) are even faster as they are optimized iterative mechanisms for list creation.

List comprehensions follow the pattern

[f(x) for x in seq]

Where seq is a pandas series or DataFrame column. Or, when operating over multiple columns,

[f(x, y) for x, y in zip(seq1, seq2)]

Where seq1 and seq2 are columns.

Numeric Comparison
Consider a simple boolean indexing operation. The list comprehension method has been timed against Series.ne (!=) and query. Here are the functions:

# Boolean indexing with Numeric value comparison.
df[df.A != df.B]                            # vectorized !=
df.query("A != B")                          # query (numexpr)
df[[x != y for x, y in zip(df.A, df.B)]]    # list comp

For simplicity, I have used the perfplot package to run all the timeit tests in this post. The timings for the operations above are below:

enter image description here

The list comprehension outperforms query for moderately sized N, and even outperforms the vectorized not equals comparison for tiny N. Unfortunately, the list comprehension scales linearly, so it does not offer much performance gain for larger N.

Note
It is worth mentioning that much of the benefit of list comprehension come from not having to worry about the index alignment, but this means that if your code is dependent on indexing alignment, this will break. In some cases, vectorised operations over the underlying NumPy arrays can be considered as bringing in the "best of both worlds", allowing for vectorisation without all the unneeded overhead of the pandas functions. This means that you can rewrite the operation above as

df[df.A.values != df.B.values]

Which outperforms both the pandas and list comprehension equivalents:

NumPy vectorization is out of the scope of this post, but it is definitely worth considering, if performance matters.

Value Counts
Taking another example - this time, with another vanilla python construct that is faster than a for loop - collections.Counter. A common requirement is to compute the value counts and return the result as a dictionary. This is done with value_counts, np.unique, and Counter:

# Value Counts comparison.
ser.value_counts(sort=False).to_dict()           # value_counts
dict(zip(*np.unique(ser, return_counts=True)))   # np.unique
Counter(ser)                                     # Counter

enter image description here

The results are more pronounced, Counter wins out over both vectorized methods for a larger range of small N (~3500).

Note
More trivia (courtesy @user2357112). The Counter is implemented with a C accelerator, so while it still has to work with python objects instead of the underlying C datatypes, it is still faster than a for loop. Python power!

Of course, the take away from here is that the performance depends on your data and use case. The point of these examples is to convince you not to rule out these solutions as legitimate options. If these still don"t give you the performance you need, there is always cython and numba. Let"s add this test into the mix.

from numba import njit, prange

@njit(parallel=True)
def get_mask(x, y):
    result = [False] * len(x)
    for i in prange(len(x)):
        result[i] = x[i] != y[i]

    return np.array(result)

df[get_mask(df.A.values, df.B.values)] # numba

enter image description here

Numba offers JIT compilation of loopy python code to very powerful vectorized code. Understanding how to make numba work involves a learning curve.


Operations with Mixed/object dtypes

String-based Comparison
Revisiting the filtering example from the first section, what if the columns being compared are strings? Consider the same 3 functions above, but with the input DataFrame cast to string.

# Boolean indexing with string value comparison.
df[df.A != df.B]                            # vectorized !=
df.query("A != B")                          # query (numexpr)
df[[x != y for x, y in zip(df.A, df.B)]]    # list comp

enter image description here

So, what changed? The thing to note here is that string operations are inherently difficult to vectorize. Pandas treats strings as objects, and all operations on objects fall back to a slow, loopy implementation.

Now, because this loopy implementation is surrounded by all the overhead mentioned above, there is a constant magnitude difference between these solutions, even though they scale the same.

When it comes to operations on mutable/complex objects, there is no comparison. List comprehension outperforms all operations involving dicts and lists.

Accessing Dictionary Value(s) by Key
Here are timings for two operations that extract a value from a column of dictionaries: map and the list comprehension. The setup is in the Appendix, under the heading "Code Snippets".

# Dictionary value extraction.
ser.map(operator.itemgetter("value"))     # map
pd.Series([x.get("value") for x in ser])  # list comprehension

enter image description here

Positional List Indexing
Timings for 3 operations that extract the 0th element from a list of columns (handling exceptions), map, str.get accessor method, and the list comprehension:

# List positional indexing. 
def get_0th(lst):
    try:
        return lst[0]
    # Handle empty lists and NaNs gracefully.
    except (IndexError, TypeError):
        return np.nan

ser.map(get_0th)                                          # map
ser.str[0]                                                # str accessor
pd.Series([x[0] if len(x) > 0 else np.nan for x in ser])  # list comp
pd.Series([get_0th(x) for x in ser])                      # list comp safe

Note
If the index matters, you would want to do:

pd.Series([...], index=ser.index)

When reconstructing the series.

enter image description here

List Flattening
A final example is flattening lists. This is another common problem, and demonstrates just how powerful pure python is here.

# Nested list flattening.
pd.DataFrame(ser.tolist()).stack().reset_index(drop=True)  # stack
pd.Series(list(chain.from_iterable(ser.tolist())))         # itertools.chain
pd.Series([y for x in ser for y in x])                     # nested list comp

enter image description here

Both itertools.chain.from_iterable and the nested list comprehension are pure python constructs, and scale much better than the stack solution.

These timings are a strong indication of the fact that pandas is not equipped to work with mixed dtypes, and that you should probably refrain from using it to do so. Wherever possible, data should be present as scalar values (ints/floats/strings) in separate columns.

Lastly, the applicability of these solutions depend widely on your data. So, the best thing to do would be to test these operations on your data before deciding what to go with. Notice how I have not timed apply on these solutions, because it would skew the graph (yes, it"s that slow).


Regex Operations, and .str Accessor Methods

Pandas can apply regex operations such as str.contains, str.extract, and str.extractall, as well as other "vectorized" string operations (such as str.split, str.find,str.translate`, and so on) on string columns. These functions are slower than list comprehensions, and are meant to be more convenience functions than anything else.

It is usually much faster to pre-compile a regex pattern and iterate over your data with re.compile (also see Is it worth using Python's re.compile?). The list comp equivalent to str.contains looks something like this:

p = re.compile(...)
ser2 = pd.Series([x for x in ser if p.search(x)])

Or,

ser2 = ser[[bool(p.search(x)) for x in ser]]

If you need to handle NaNs, you can do something like

ser[[bool(p.search(x)) if pd.notnull(x) else False for x in ser]]

The list comp equivalent to str.extract (without groups) will look something like:

df["col2"] = [p.search(x).group(0) for x in df["col"]]

If you need to handle no-matches and NaNs, you can use a custom function (still faster!):

def matcher(x):
    m = p.search(str(x))
    if m:
        return m.group(0)
    return np.nan

df["col2"] = [matcher(x) for x in df["col"]]

The matcher function is very extensible. It can be fitted to return a list for each capture group, as needed. Just extract query the group or groups attribute of the matcher object.

For str.extractall, change p.search to p.findall.

String Extraction
Consider a simple filtering operation. The idea is to extract 4 digits if it is preceded by an upper case letter.

# Extracting strings.
p = re.compile(r"(?<=[A-Z])(d{4})")
def matcher(x):
    m = p.search(x)
    if m:
        return m.group(0)
    return np.nan

ser.str.extract(r"(?<=[A-Z])(d{4})", expand=False)   #  str.extract
pd.Series([matcher(x) for x in ser])                  #  list comprehension

enter image description here

More Examples
Full disclosure - I am the author (in part or whole) of these posts listed below.


Conclusion

As shown from the examples above, iteration shines when working with small rows of DataFrames, mixed datatypes, and regular expressions.

The speedup you get depends on your data and your problem, so your mileage may vary. The best thing to do is to carefully run tests and see if the payout is worth the effort.

The "vectorized" functions shine in their simplicity and readability, so if performance is not critical, you should definitely prefer those.

Another side note, certain string operations deal with constraints that favour the use of NumPy. Here are two examples where careful NumPy vectorization outperforms python:

Additionally, sometimes just operating on the underlying arrays via .values as opposed to on the Series or DataFrames can offer a healthy enough speedup for most usual scenarios (see the Note in the Numeric Comparison section above). So, for example df[df.A.values != df.B.values] would show instant performance boosts over df[df.A != df.B]. Using .values may not be appropriate in every situation, but it is a useful hack to know.

As mentioned above, it"s up to you to decide whether these solutions are worth the trouble of implementing.


Appendix: Code Snippets

import perfplot  
import operator 
import pandas as pd
import numpy as np
import re

from collections import Counter
from itertools import chain

# Boolean indexing with Numeric value comparison.
perfplot.show(
    setup=lambda n: pd.DataFrame(np.random.choice(1000, (n, 2)), columns=["A","B"]),
    kernels=[
        lambda df: df[df.A != df.B],
        lambda df: df.query("A != B"),
        lambda df: df[[x != y for x, y in zip(df.A, df.B)]],
        lambda df: df[get_mask(df.A.values, df.B.values)]
    ],
    labels=["vectorized !=", "query (numexpr)", "list comp", "numba"],
    n_range=[2**k for k in range(0, 15)],
    xlabel="N"
)

# Value Counts comparison.
perfplot.show(
    setup=lambda n: pd.Series(np.random.choice(1000, n)),
    kernels=[
        lambda ser: ser.value_counts(sort=False).to_dict(),
        lambda ser: dict(zip(*np.unique(ser, return_counts=True))),
        lambda ser: Counter(ser),
    ],
    labels=["value_counts", "np.unique", "Counter"],
    n_range=[2**k for k in range(0, 15)],
    xlabel="N",
    equality_check=lambda x, y: dict(x) == dict(y)
)

# Boolean indexing with string value comparison.
perfplot.show(
    setup=lambda n: pd.DataFrame(np.random.choice(1000, (n, 2)), columns=["A","B"], dtype=str),
    kernels=[
        lambda df: df[df.A != df.B],
        lambda df: df.query("A != B"),
        lambda df: df[[x != y for x, y in zip(df.A, df.B)]],
    ],
    labels=["vectorized !=", "query (numexpr)", "list comp"],
    n_range=[2**k for k in range(0, 15)],
    xlabel="N",
    equality_check=None
)

# Dictionary value extraction.
ser1 = pd.Series([{"key": "abc", "value": 123}, {"key": "xyz", "value": 456}])
perfplot.show(
    setup=lambda n: pd.concat([ser1] * n, ignore_index=True),
    kernels=[
        lambda ser: ser.map(operator.itemgetter("value")),
        lambda ser: pd.Series([x.get("value") for x in ser]),
    ],
    labels=["map", "list comprehension"],
    n_range=[2**k for k in range(0, 15)],
    xlabel="N",
    equality_check=None
)

# List positional indexing. 
ser2 = pd.Series([["a", "b", "c"], [1, 2], []])        
perfplot.show(
    setup=lambda n: pd.concat([ser2] * n, ignore_index=True),
    kernels=[
        lambda ser: ser.map(get_0th),
        lambda ser: ser.str[0],
        lambda ser: pd.Series([x[0] if len(x) > 0 else np.nan for x in ser]),
        lambda ser: pd.Series([get_0th(x) for x in ser]),
    ],
    labels=["map", "str accessor", "list comprehension", "list comp safe"],
    n_range=[2**k for k in range(0, 15)],
    xlabel="N",
    equality_check=None
)

# Nested list flattening.
ser3 = pd.Series([["a", "b", "c"], ["d", "e"], ["f", "g"]])
perfplot.show(
    setup=lambda n: pd.concat([ser2] * n, ignore_index=True),
    kernels=[
        lambda ser: pd.DataFrame(ser.tolist()).stack().reset_index(drop=True),
        lambda ser: pd.Series(list(chain.from_iterable(ser.tolist()))),
        lambda ser: pd.Series([y for x in ser for y in x]),
    ],
    labels=["stack", "itertools.chain", "nested list comp"],
    n_range=[2**k for k in range(0, 15)],
    xlabel="N",    
    equality_check=None

)

# Extracting strings.
ser4 = pd.Series(["foo xyz", "test A1234", "D3345 xtz"])
perfplot.show(
    setup=lambda n: pd.concat([ser4] * n, ignore_index=True),
    kernels=[
        lambda ser: ser.str.extract(r"(?<=[A-Z])(d{4})", expand=False),
        lambda ser: pd.Series([matcher(x) for x in ser])
    ],
    labels=["str.extract", "list comprehension"],
    n_range=[2**k for k in range(0, 15)],
    xlabel="N",
    equality_check=None
)