How to select the first file in a directory?

Question

I'm trying to process some files in a directory without knowing their name, and one by one. So I've used os.listdir(path) to list files.

So I have to list files at each call of my function. The problem is when there is a lot of files (like 2000), it takes a loooooong time to list each file and I just want the first one.

Is there any solution to get the first name without listing each files ?

Answer 1

os.listdir(path)[0]

It would be faster than 'listing' (printing?) each filename, but it still has to load all of the filenames into memory. Also, which file is the first file, do you only want whichever one comes first or is there a specific one you want, because that is different.

Answer 2

If your goal is to process each file name, use os.walk() generator:

Help on function walk in module os:

walk(top, topdown=True, onerror=None, followlinks=False)
    Directory tree generator.

Answer 3

It seems like you're trying to process the files en masse, and you'll be iterating through all the files at some point. Instead of having calling the method every time that you enter your function, why not have a global parameter so that you only load the list once? So, for example, instead of:

import os
def foo(path):
    os.listdir(path)[0]

you have:

import os
fnames = os.listdir(path)
def foo(path):
    fnames[0]

Answer 4

To get the first filename without having to scan the whole directory, you have to use the walk function to get the generator and then you can use next() to get the first value of the generator.

folder_walk = os.walk(path)
first_file_in_folder = next(folder_walk)[2][0]

print(first_file_in_folder)
# "firstFile.jpg"