Question

我具有以下文件结构，并希望使用python创建每个文件夹中文件数量的字典。底部的示例将翻译为以下字典：

{Employee A: {Jan : 3}, {Feb : 2}, Employee B: {Jan : 2}, {Feb : 1}}

有人知道如何使用os来遍历目录吗？

Employee A
    Jan
        File 1
        File 2
        File 3
    Feb
        File 1
        File 2
Employee B
    Jan
        File 1
        File 2
    Feb
        File 1

Answer 1

研究解析os.walk的输出

例如：

mydict = {}
for (root,dirs,files) in os.walk('testdir', topdown=False)
    if len(files)>0:
        mydict[root]=len(files)
print mydict

返回

{'testdir/EmployeeB/Jan': 2, 'testdir/EmployeeA/Feb': 2, 'testdir/EmployeeB/Feb': 1, 'testdir/EmployeeA/Jan': 3}

您可以轻松地解析这些键以生成您要查找的嵌套字典。

Answer 2

通过这种方式，您可以遍历目录中的所有文件并创建它们的列表。您可以根据需要对其进行修改：

import os
import glob
from pathlib import Path

error_log_list = []

def traverse_structure():

  try:
    root = r"C:\\Users\Whatever\Desktop\DirectoryToSearch"
    # Change working directory
    os.chdir(root)

    print("Creating master list of the directory structure...\n")

    # Traverse the folder structure
    for folder, subfolders, files in os.walk(root):

      # Pass over each file
      for file in files:

        absolute_path = os.path.join(folder,file)

        # Create a master file list
        file_paths_list.append(absolute_path)

  except Exception as e:
    error_log_list.append( "Failed to open the root directory specified "+root+"\n Error: "+str(e)+"\n" )

traverse_structure()

Answer 3

使用os库：

import os
parent = os.listdir(path) # return directory files to list
child = []
for x in parent:
    if os.path.isdir(path +'/' + x):
        child.append(os.listdir(path + '/' + x))
    else
        child.append('')
d = dict(zip(parent,child))
print(d)

这是使字典脱离目录的基本逻辑。但这支持2个级别。我将n级部分留给自己。

Answer 4

只需进行一些细微调整，就可以根据需要执行ActiveState Python配方Create a nested dictionary from os.walk：

try:
    reduce
except NameError:  # Python 3
    from functools import reduce
import os

def count_files_in_directories(rootdir):
    """ Creates a nested dictionary that represents the folder structure
        of rootdir with a count of files in the lower subdirectories.
    """
    dir = {}
    rootdir = rootdir.rstrip(os.sep)
    start = rootdir.rfind(os.sep) + 1
    for path, dirs, files in os.walk(rootdir):
        folders = path[start:].split(os.sep)
        subdir = len(files) if files else dict.fromkeys(files)
        parent = reduce(dict.get, folders[:-1], dir)
        parent[folders[-1]] = subdir

    return list(dir.values())[0]

startdir = "./sample"
res = count_files_in_directories(startdir)
print(res)  # -> {'Employee A': {'Feb': 2, 'Jan': 3}, 'Employee B': {'Feb': 1, 'Jan': 2}}

请注意./sample目录是我创建的用于测试的文件夹结构的根目录，与您的问题中显示的目录结构完全相同。

如何计算子目录中的文件数？

4 个答案: