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))

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}

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。

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

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))])

我很惊讶没有人提到os.scandir:

def count_files(dir):
    return len([1 for x in list(os.scandir(dir)) if x.is_file()])