Python | Pandas Index.value_counts ()

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Index.value_counts() Pandas Index.value_counts() returns an object containing the number of unique values. The resulting object will be in descending order, so the first item is the most common item. Excludes NA values ‚Äã‚Äãby default.

Syntax: Index.value_counts (normalize = False, sort = True, ascending = False, bins = None, dropna = True)

Parameters:
normalize: If True then the object returned will contain the relative frequencies of the unique values.
sort: Sort by values ‚Äã‚Äã
ascending: Sort in ascending order
bins: Rather than count values, group them into half- open bins, a convenience for pd.cut, only works with numeric data
dropna: Don’t include counts of NaN.

Returns: counts: Series

Example # 1: Use Index.value_counts () to count the number of unique values ‚Äã‚Äãin a given index.

# import pandas as pd

import pandas as pd


# Create index

idx = pd.Index ([ ’Harry’ , ’Mike’ , ’ Arther’ , ’Nick’ ,

’ Harry’ , ’Arther’ ], name = ’Student’ )


# Print index

print (idx)

Output:

Let’s find the count of all unique values ‚Äã‚Äãin the index.

# find the number of unique values ‚Äã‚Äãin the index
idx.value_counts ()

Output:

The function returned a counter of all unique values ‚Äã‚Äãin the given index. Note that the object returned by the function contains occurrences of values ‚Äã‚Äãin descending order.

Example # 2: Use Index.value_counts () to find the count all unique values ‚Äã‚Äãin this index.

# import pandas as pd

import pandas as pd


# Create Index

idx = pd .Index ([ 21 , 10 , 30 , 40 , 50 , 10 , 50 ])


# Print index

print (idx)

Output:

Let’s count the occurrence of all unique values ‚Äã‚Äãin the index.

# to count all
# unique values in the index.
idx.value_counts ()

Output:

The function returned a counter of all unique values in the index.

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Finding the index of an item in a list

5 answers

Given a list ["foo", "bar", "baz"] and an item in the list "bar", how do I get its index (1) in Python?

3740

Answer #1

>>> ["foo", "bar", "baz"].index("bar")
1

Reference: Data Structures > More on Lists

Caveats follow

Note that while this is perhaps the cleanest way to answer the question as asked, index is a rather weak component of the list API, and I can"t remember the last time I used it in anger. It"s been pointed out to me in the comments that because this answer is heavily referenced, it should be made more complete. Some caveats about list.index follow. It is probably worth initially taking a look at the documentation for it:

list.index(x[, start[, end]])

Return zero-based index in the list of the first item whose value is equal to x. Raises a ValueError if there is no such item.

The optional arguments start and end are interpreted as in the slice notation and are used to limit the search to a particular subsequence of the list. The returned index is computed relative to the beginning of the full sequence rather than the start argument.

Linear time-complexity in list length

An index call checks every element of the list in order, until it finds a match. If your list is long, and you don"t know roughly where in the list it occurs, this search could become a bottleneck. In that case, you should consider a different data structure. Note that if you know roughly where to find the match, you can give index a hint. For instance, in this snippet, l.index(999_999, 999_990, 1_000_000) is roughly five orders of magnitude faster than straight l.index(999_999), because the former only has to search 10 entries, while the latter searches a million:

>>> import timeit
>>> timeit.timeit("l.index(999_999)", setup="l = list(range(0, 1_000_000))", number=1000)
9.356267921015387
>>> timeit.timeit("l.index(999_999, 999_990, 1_000_000)", setup="l = list(range(0, 1_000_000))", number=1000)
0.0004404920036904514
 

Only returns the index of the first match to its argument

A call to index searches through the list in order until it finds a match, and stops there. If you expect to need indices of more matches, you should use a list comprehension, or generator expression.

>>> [1, 1].index(1)
0
>>> [i for i, e in enumerate([1, 2, 1]) if e == 1]
[0, 2]
>>> g = (i for i, e in enumerate([1, 2, 1]) if e == 1)
>>> next(g)
0
>>> next(g)
2

Most places where I once would have used index, I now use a list comprehension or generator expression because they"re more generalizable. So if you"re considering reaching for index, take a look at these excellent Python features.

Throws if element not present in list

A call to index results in a ValueError if the item"s not present.

>>> [1, 1].index(2)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: 2 is not in list

If the item might not be present in the list, you should either

  1. Check for it first with item in my_list (clean, readable approach), or
  2. Wrap the index call in a try/except block which catches ValueError (probably faster, at least when the list to search is long, and the item is usually present.)

3740

Answer #2

One thing that is really helpful in learning Python is to use the interactive help function:

>>> help(["foo", "bar", "baz"])
Help on list object:

class list(object)
 ...

 |
 |  index(...)
 |      L.index(value, [start, [stop]]) -> integer -- return first index of value
 |

which will often lead you to the method you are looking for.

3740

Answer #3

The majority of answers explain how to find a single index, but their methods do not return multiple indexes if the item is in the list multiple times. Use enumerate():

for i, j in enumerate(["foo", "bar", "baz"]):
    if j == "bar":
        print(i)

The index() function only returns the first occurrence, while enumerate() returns all occurrences.

As a list comprehension:

[i for i, j in enumerate(["foo", "bar", "baz"]) if j == "bar"]

Here"s also another small solution with itertools.count() (which is pretty much the same approach as enumerate):

from itertools import izip as zip, count # izip for maximum efficiency
[i for i, j in zip(count(), ["foo", "bar", "baz"]) if j == "bar"]

This is more efficient for larger lists than using enumerate():

$ python -m timeit -s "from itertools import izip as zip, count" "[i for i, j in zip(count(), ["foo", "bar", "baz"]*500) if j == "bar"]"
10000 loops, best of 3: 174 usec per loop
$ python -m timeit "[i for i, j in enumerate(["foo", "bar", "baz"]*500) if j == "bar"]"
10000 loops, best of 3: 196 usec per loop

We hope this article has helped you to resolve the problem. Apart from Python | Pandas Index.value_counts (), check other Counters-related topics.

Want to excel in Python? See our review of the best Python online courses 2022. If you are interested in Data Science, check also how to learn programming in R.

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Ken Chamberlet

Tallinn | 2022-12-10

I was preparing for my coding interview, thanks for clarifying this - Python | Pandas Index.value_counts () in Python is not the simplest one. I just hope that will not emerge anymore

Marie Jackson

Texas | 2022-12-10

Simply put and clear. Thank you for sharing. Python | Pandas Index.value_counts () and other issues with find was always my weak point 😁. Will use it in my bachelor thesis

Chen Jackson

Paris | 2022-12-10

open is always a bit confusing 😭 Python | Pandas Index.value_counts () is not the only problem I encountered. I just hope that will not emerge anymore

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