def directory(path,extension):
  list_dir = []
  list_dir = os.listdir(path)
  count = 0
  for file in list_dir:
    if file.endswith(extension): # eg: '.txt'
      count += 1
  return count

这是一个简单的解决方案，可以计算包含子文件夹的目录中的文件数量。它可能会派上用场:

import os
from pathlib import Path

def count_files(rootdir):
    '''counts the number of files in each subfolder in a directory'''
    for path in pathlib.Path(rootdir).iterdir():
        if path.is_dir():
            print("There are " + str(len([name for name in os.listdir(path) \
            if os.path.isfile(os.path.join(path, name))])) + " files in " + \
            str(path.name))
            
 
count_files(data_dir) # data_dir is the directory you want files counted.

你应该得到一个类似这样的输出(当然，占位符改变了):

There are {number of files} files in {name of sub-folder1}
There are {number of files} files in {name of sub-folder2}

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))

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

Os.listdir()将比使用glob.glob更有效。要测试文件名是否为普通文件(而不是目录或其他实体)，请使用os.path.isfile():

import os, os.path

# simple version for working with CWD
print len([name for name in os.listdir('.') if os.path.isfile(name)])

# path joining version for other paths
DIR = '/tmp'
print len([name for name in os.listdir(DIR) if os.path.isfile(os.path.join(DIR, name))])

很简单:

print(len([iq for iq in os.scandir('PATH')]))

它只是简单地计算目录中的文件数量，我使用了列表理解技术来遍历特定目录，返回所有文件。"len(返回列表)"返回文件数。