Python | Pandas Series.clip_lower ()

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Pandas Series.clip_lower() is used for Series.clip_lower() values ‚Äã‚Äãbelow the passed smallest value. A threshold value is passed as a parameter, and all values ‚Äã‚Äãin the sequence that are less than the threshold values ‚Äã‚Äãbecome equal to it.

Syntax: Series.clip_lower (threshold, axis = None , inplace = False)

threshold: numeric or list like, Sets minimum threshold value and in case of list, sets separate threshold values ‚Äã‚Äãfor each value in caller series (Given list size is same)
axis: 0 or ’index’ to apply method by rows and 1 or ’columns’ to apply by columns
inplace: Make changes in the caller series itself. (Overwrite with new values)

Return type: Series with updated values ‚Äã‚Äã

To load the dataset used in the following example, click here.

In the following examples, the data frame used contains data for some NBA players. An image of the data frame before any operations is attached below.

Example # 1 : Applied to single value series

In this example, a minimum threshold of 26 is passed as a parameter to the .clip_lower () method. This method is called on the Age column of the data frame, and the new values ‚Äã‚Äãare stored in the Age_new column. Before performing any operations, null lines are removed using .dropna ()

# pandas module import

import pandas as pd

# create data frame

data = pd.read_csv ( " " )

# remove null values ‚Äã‚Äãto avoid errors

data.dropna (inplace = True )

# setting the threshold

threshold = 26.0

# apply the method and go to the new column

data [ "Age_new" ] = data [ "Age" ]. clip_lower (threshold)

# display

As shown in the output image, the minimum value of the Age_new column is 26. All values ‚Äã‚Äãless than 26 have been increased to 26 and saved as a new th column.

Example # 2: Applied to series with a list type value

In this example, the first 10 rows of the Age column are retrieved and stored using the .head () method. After that, a list of the same length is created, which is passed to the threshold parameter of the .clip_lower () method to set a separate threshold value for each value in the series. The returned values ‚Äã‚Äãare stored in a new column "clipped_values".

# pandas module import

import pandas as pd

# regex module import

import re

# create data frame

data = pd.read_csv ( " " )

# remove null values ‚Äã‚Äãto avoid errors

data.dropna (inplace = True )

# returns top lines

new_data = data.head ( 10 ). copy ()

# list for individual thresholds

threshold = [ 27 , 23 , 19 , 30 , 26 , 22 , 22 , 41 , 11 , 33 ]

# applying the method and returning to a new column

new_data [ "Clipped values" ] = new_data [ "Age" ]. clip_lower (threshold = threshold)

# display

As display But on the output image, each value in the sequence had a different threshold value according to the passed list, and therefore results were returned according to the separate threshold value of each item.

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Python | Pandas Series.clip_lower () clip: Questions

How do I copy a string to the clipboard?

2 answers

Dancrew32 By Dancrew32

I"m trying to make a basic Windows application that builds a string out of user input and then adds it to the clipboard. How do I copy a string to the clipboard using Python?


Answer #1

Actually, pywin32 and ctypes seem to be an overkill for this simple task. Tkinter is a cross-platform GUI framework, which ships with Python by default and has clipboard accessing methods along with other cool stuff.

If all you need is to put some text to system clipboard, this will do it:

from Tkinter import Tk
r = Tk()
r.clipboard_append("i can has clipboardz?")
r.update() # now it stays on the clipboard after the window is closed

And that"s all, no need to mess around with platform-specific third-party libraries.

If you are using Python 3, replace TKinter with tkinter.

Python | Pandas Series.clip_lower () clip: Questions

Python script to copy text to clipboard

2 answers

I just need a python script that copies text to the clipboard.

After the script gets executed i need the output of the text to be pasted to another source. Is it possible to write a python script that does this job?


Answer #1

See Pyperclip. Example (taken from Pyperclip site):

import pyperclip
pyperclip.copy("The text to be copied to the clipboard.")
spam = pyperclip.paste()

Also, see Xerox. But it appears to have more dependencies.

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?


Answer #1

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

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

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)
>>> [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)
>>> next(g)

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.)


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.


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":

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


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