一行和递归:

def count_files(path):
    return sum([len(files) for _, _, files in os.walk(path)])

count_files('path/to/dir')

一行和递归:

def count_files(path):
    return sum([len(files) for _, _, files in os.walk(path)])

count_files('path/to/dir')

一个答案与pathlib和没有加载到内存的整个列表:

from pathlib import Path

path = Path('.')

print(sum(1 for _ in path.glob('*')))  # Files and folders, not recursive
print(sum(1 for _ in path.glob('**/*')))  # Files and folders, recursive

print(sum(1 for x in path.glob('*') if x.is_file()))  # Only files, not recursive
print(sum(1 for x in path.glob('**/*') if x.is_file()))  # Only files, recursive

对于所有类型的文件，包括子目录(Python 2):

import os

lst = os.listdir(directory) # your directory path
number_files = len(lst)
print number_files

只有文件(避免子目录):

import os

onlyfiles = next(os.walk(directory))[2] #directory is your directory path as string
print len(onlyfiles)

如果您将使用操作系统的标准shell，则可以比使用纯python方式更快地获得结果。

Windows示例:

import os
import subprocess

def get_num_files(path):
    cmd = 'DIR \"%s\" /A-D /B /S | FIND /C /V ""' % path
    return int(subprocess.check_output(cmd, shell=True))

def count_em(valid_path):
   x = 0
   for root, dirs, files in os.walk(valid_path):
       for f in files:
            x = x+1
print "There are", x, "files in this directory."
return x

摘自本文