Question

我有一个名为notes的文件夹，它们自然会被分类到文件夹中，在这些文件夹中也会有子类别的子文件夹。现在我的问题是我有一个功能，它遍历3个子目录：

def obtainFiles(path):
      list_of_files = {}
      for element in os.listdir(path):
          # if the element is an html file then..
          if element[-5:] == ".html":
              list_of_files[element] = path + "/" + element
          else: # element is a folder therefore a category
              category = os.path.join(path, element)
              # go through the category dir
              for element_2 in os.listdir(category):
                  dir_level_2 = os.path.join(path,element + "/" + element_2)
                  if element_2[-5:] == ".html":
                      print "- found file: " + element_2
                      # add the file to the list of files
                      list_of_files[element_2] = dir_level_2
                  elif os.path.isdir(element_2):
                      subcategory = dir_level_2
                      # go through the subcategory dir
                      for element_3 in os.listdir(subcategory):
                          subcategory_path = subcategory + "/" + element_3
                        if subcategory_path[-5:] == ".html":
                            print "- found file: " + element_3
                            list_of_files[element_3] = subcategory_path
                        else:
                            for element_4 in os.listdir(subcategory_path):
                                 print "- found file:" + element_4

请注意，这仍然是一项正在进行中的工作。它在我眼中非常难看...... 我要在这里尝试实现的是遍历所有文件夹和子文件夹，并将所有文件名放在名为“list_of_files”的字典中，名称为“key”，完整路径为“value”。该函数还没有完全正常工作，但是想知道如何使用os.walk函数做类似的事情？

由于

Answer 1

根据您的简短描述，这样的事情应该有效：

list_of_files = {}
for (dirpath, dirnames, filenames) in os.walk(path):
    for filename in filenames:
        if filename.endswith('.html'): 
            list_of_files[filename] = os.sep.join([dirpath, filename])

Answer 2

另一种方法是使用生成器，构建在@ ig0774的代码

上

import os
def walk_through_files(path, file_extension='.html'):
   for (dirpath, dirnames, filenames) in os.walk(path):
      for filename in filenames:
         if filename.endswith(file_extension): 
            yield os.path.join(dirpath, filename)

然后

for fname in walk_through_files():
    print(fname)

Answer 3

你可以这样做：

list_of_files = dict([ (file, os.sep.join((dir, file)))
                       for (dir,dirs,files) in os.walk(path)
                       for file in files
                       if file[-5:] == '.html' ])

Answer 4

我多次遇到过这个问题，而且没有一个答案让我满意 - 所以创建了a script for that。在走遍目录时，使用Pytohn非常麻烦。

以下是如何使用它：

import file_walker


for f in file_walker.walk("/a/path"):
     print(f.name, f.full_path) # Name is without extension
     if f.isDirectory: # Check if object is directory
         for sub_f in f.walk(): # Easily walk on new levels
             if sub_f.isFile: # Check if object is file (= !isDirectory)
                 print(sub_f.extension) # Print file extension
                 with sub_f.open("r") as open_f: # Easily open file
                     print(open_f.read())

你如何使用python浏览目录？

4 个答案: