Finding files in a list by partial name in Python

Question

I have a list of full file paths:

filelist = [
    "C:\Folder1\Files\fileabc.txt",
    "C:\Folder1\Files\filedef.txt",
    "C:\Folder2\Data\file123.txt"
]

I want to find a file in the list by its basename, with the extension, but without specifying full path.

I've tried something like this:

name = "filedef.txt"

if name in filelist:
   print "Found"

But it doesn't work.

Any hints?

Danil Speransky · Accepted Answer · 2017-08-27 22:07:28Z

1

You need to do two things. First, iterate through the array. Second, escape \ special character.

paths = [r'C:\Folder1\Files\fileabc.txt', r'C:\Folder1\Files\filedef.txt', r'C:\Folder2\Data\file123.txt']

name = 'filedef.txt'

for path in paths:
   if name in path:
       print('Found', path)

answered Aug 27, 2017 at 22:07

Danil Speransky

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2 Comments

maurobio Over a year ago

Couldn't figure out how to escape the '\' for the path in the path list. A similar question has been presented here: stackoverflow.com/questions/21605526/… but the answers were not quite helpful for me.

maurobio Over a year ago

I did understand, but my problem is that I get the names in the file list from a GUI interface and populate the list as follows: filelist.append(str(filename)), where filename is returned from a standard file dialog. Cannot figure out how to escape the backlashes in the name in that case.

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