Read first file in python

Question

I'm trying to make a script that iterates through folders in a directory and saves the very first one in the folder to a variable. I'm trying to do this because I have folders in a directory with hundreads of .exr files but I only need the first file in the folder to be saved to the directory. I can get it to print out all the files in each folder but thats a bit too much info than I need. Is there an easy way to do this using something like os.walk? This is what I'm working with so far

import os
    
def main():

    dirName = r"F:\FOLDERNAME"

    #Get the List of all files in the directory tree at given path
    listOfFiles = getListOfFiles(dirName)

    #Print the files
    for elem in listOfFiles:
    
        print(elem)

if __name__ =='__main__':
    main()

Thanks yall!

First by what criteria? Alphabetical name, age, file length? — Barmar, Jun 03 '22 at 23:00

score 0 · Answer 1 · answered Jun 03 '22 at 23:02

If you already:

can get it to print out all the files in each folder

And you only want :

the very first one

Then all you need is something like:

first_file = list_of_files[0]

Looks like you're still learning the very basics of coding. I suggest looking up lessons/tutorials on lists, list indexing, and iterables.

score 0 · Answer 2 · answered Jun 03 '22 at 23:03

0

If you use os.walk(), it returns a list of files in each directory. You can print the first one.

for root, dirs, files in os.walk(dirName):
    if files:
        print(os.path.join(root, files[0]))

answered Jun 03 '22 at 23:03

Barmar

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Read first file in python

2 Answers2