Extract string in list based on character in Python

Question

I have a list of filenames in Python that looks like this, except that the list is much longer:

filenames = ['BETON\map (120).png',
 'BETON\map (125).png',
 'BETON\map (134).png',
 'BETON\map (137).png',
 'TUILES\map (885).png',
 'TUILES\map (892).png',
 'TUILES\map (924).png',
 'TUILES\map (936).png',
 'TUILES\map (954).png',
 'TUILES\map (957).png',
 'TUILES\map (97).png',
 'TUILES\map (974).png',
 'TUILES\map (987).png']

I would like to only keep the first part of the filename strings in my list in order to only keep its type, like so:

filenames = ['BETON',
     'BETON',
     'BETON',
     'BETON',
     'TUILES',
     'TUILES',
     'TUILES',
     'TUILES',
     'TUILES',
     'TUILES',
     'TUILES',
     'TUILES',
     'TUILES']

I have been using a workaround grabbing the first 5 elements

def Extract(files):
    return [item[:5] for item in files]
     
# Driver code
files2 = Extract(files)

However, it’s becoming an issue as I have many more types (indicated in the list of filenames) coming with varying lengths in and I cannot just take the first elements.
How can I extract as soon as it spots the backslash \ ?

Many thanks!

Asked By: Julie-Anne

||

Source

Answer 1

Split the filenames on a backslash, and take only the first item from the split.

filenames = [n.split('\')[0] for n in filenames]

Answered By: John Gordon

Answer 2

string.split()

Yeah, you indeed can split every string and take only the part you need.

Try this:

for index in range(len(filenames)):
    # Only take the name
    filenames[index] = filenames[index].split('\')[0]

This code above doesn’t distingues by file name lenght but it just take the string before the character you pass to the split function. ” in your case.

Answered By: mjesbar

Answer 3

Alternate solution producing output as desired by OP.
Use python’s operator.methodcaller() function

The methodcaller() assists in maintaining your code when you start to define functions, as your use-case expands. With methodcaller(), you can use list comprehension or map, or even lambda (if required).
Please note that you’ll need to import the methodcaller from the operator library. There is nothing to install!

methodcaller
Return a callable object that calls the method name on its operand. If additional arguments and/or keyword arguments are given, they will be given to the method as well.

## import methodcaller
from operator import methodcaller

method_to_use = 'split'
arg1 = '\'

## use methodcaller with list comprehension to split filenames and return the first split
## similar to @John Gordon above
#[methodcaller('split', '\')(n)[0] for n in filenames]

[methodcaller(method_to_use, arg1)(n)[0] for n in filenames]

##alternatively
#'''
## call methodcaller
#f = methodcaller('split', '\')
f = methodcaller(method_to_use, arg1)

filenames_split_caller = [f(n)[0] for n in filenames]
filenames_split_caller
#'''

methodcaller() with map()

[e[0] for e in list(map(methodcaller(method_to_use, arg1), filenames))]
## get the first element of the first list (list of list after split)

[Edit]
The ‘three’ codes above each provide desired output.

I would like to only keep the first part of the filename strings in my list in order to only keep its type.

PS: The author’s context is not properly addressed by the ‘duplicate’.

Answered By: semmyk-research

Answer 4

import os
list(map(os.path.dirname, filenames))

[‘BETON’, ‘BETON’, ‘BETON’, ‘BETON’, ‘TUILES’, ‘TUILES’, ‘TUILES’, ‘TUILES’, ‘TUILES’, ‘TUILES’, ‘TUILES’, ‘TUILES’, ‘TUILES’]

Answered By: pinky

Extract string in list based on character in Python

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