CODEWITHSUNDEEP
pythonfileoperating-system
JoeFromAccounting
JoeFromAccounting

Reputation: 145

Python os.path.exists() ignore part of file name

I have a directory that looks like the following:

> myDirectory
    > L1.zip
    > L2_abc.zip

I want to search through the directory to return if a file exists, but I will only have the first part of the zip file name (L1 or L2). How would I go about checking if the file exists?

The results should look a little like the following:

>>> file_exists("L1")
true 
>>> file_exists("L2")
true

I am currently just using os.path.exists(), but I don't know how to ignore the _abc part of the file name.

Upvotes: 1

Views: 2587

Answers (3)

Eugene Yarmash
Eugene Yarmash

Reputation: 150031

You could use pathlib:

from pathlib import Path

def file_exists(prefix):
    return any(Path("myDirectory").glob(prefix + "*"))

Upvotes: 0

Mureinik
Mureinik

Reputation: 311998

Using glob, and checking if the result output is empty or not should do the trick:

from glob import glob

def file_exists(filename):
    return bool(glob(filename + '.*'))

Upvotes: 2

Alter
Alter

Reputation: 3464

you can use listdir and do a custom check. Here's one way that only matches if the file/dir starts with L2

matches = [f for f in os.listdir() if f.startswith("L2")]
print(matches)

Upvotes: 3

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