使用Python计算所有文件夹/子文件夹中的所有文件

Question

Which is the most efficient way to count all files in all folders and subfolders in Python? I want to use this on Linux systems.

Example output:

(Path files)

/ 2

/bin 100

/boot 20

/boot/efi/EFI/redhat 1

....

/root 34

....

Paths without a file should be ignored.

Thanks.

Answer 1

import os

print [(item[0], len(item[2])) for item in os.walk('/path') if item[2]]

It returns a list of tuples of folders/subfolders and files count in /path .

OR

import os

for item in os.walk('/path'):
    if item[2]:
        print item[0], len(item[2])

It prints folders/subfolders and files count in /path .

If you want try faster solution, then you had to try to combine:

os.scandir() # from python 3.5.2

iterate recursively and use:

from itertools import count

counter = count()
counter.next() # returns at first 0, next 1, 2, 3 ...

if counter.next() > 1000:
    print 'dir with file count over 1000' # and use continue in for loop

Maybe that will be faster, because I think in os.walk function are unnecessary things for you.

Answer 2

You can do it with os.walk() ;

import os

for root, dirs, files in os.walk('/some/path'):
    if files:
        print('{0} {1}'.format(root, len(files)))

Note that this will also include hidden files, ie those that begin with a dot ( . ).