卢克的代码重新格式化。

import os

print len(os.walk('/usr/lib').next()[2])

import os

def count_files(in_directory):
    joiner= (in_directory + os.path.sep).__add__
    return sum(
        os.path.isfile(filename)
        for filename
        in map(joiner, os.listdir(in_directory))
    )

>>> count_files("/usr/lib")
1797
>>> len(os.listdir("/usr/lib"))
2049

我这样做了，这返回了文件夹(Attack_Data)中的文件数量…这很好。

import os
def fcount(path):
    #Counts the number of files in a directory
    count = 0
    for f in os.listdir(path):
        if os.path.isfile(os.path.join(path, f)):
            count += 1

    return count
path = r"C:\Users\EE EKORO\Desktop\Attack_Data" #Read files in folder
print (fcount(path))

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

这就是fnmatch非常方便的地方:

import fnmatch

print len(fnmatch.filter(os.listdir(dirpath), '*.txt'))

详情:http://docs.python.org/2/library/fnmatch.html

我找到了另一个可能是正确的公认答案。

for root, dirs, files in os.walk(input_path):    
for name in files:
    if os.path.splitext(name)[1] == '.TXT' or os.path.splitext(name)[1] == '.txt':
        datafiles.append(os.path.join(root,name)) 


print len(files)