如何只计算目录中的文件?这将目录本身计算为一个文件:

len(glob.glob('*'))

当前回答

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}

其他回答

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

我解决了这个问题,同时通过谷歌Colab计算谷歌驱动器目录中的文件数量,通过将自己定向到目录文件夹by

import os                                                                                                
%cd /content/drive/My Drive/  
print(len([x for x in os.listdir('folder_name/']))  

普通用户可以尝试

 import os                                                                                                     
 cd Desktop/Maheep/                                                     
 print(len([x for x in os.listdir('folder_name/']))  

下面是一个简单的单行命令,我觉得很有用:

print int(os.popen("ls | wc -l").read())

简短而简单

import os
directory_path = '/home/xyz/'
No_of_files = len(os.listdir(directory_path))

一行和递归:

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

count_files('path/to/dir')