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

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

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
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

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())

我用glob。Iglob的目录结构类似于

data
└───train
│   └───subfolder1
│   |   │   file111.png
│   |   │   file112.png
│   |   │   ...
│   |
│   └───subfolder2
│       │   file121.png
│       │   file122.png
│       │   ...
└───test
    │   file221.png
    │   file222.png

以下两个选项都返回4(正如预期的那样,即不计算子文件夹本身)

len (list (glob。iglob(“数据/火车/ * / * . png ", recursive = True)) sum(我在环球公司工作过一次)

简单的一个:

import os
number_of_files = len(os.listdir(directory))
print(number_of_files)