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

Question

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.

Answer 1

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)

Answer 2

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 + '.*'))

Answer 3

You could use pathlib :

from pathlib import Path

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