python - 我想使用 os.walk 迭代所有子目录，但只打印链接而不介入。如何降低复杂性？

Question

我有一段我一直在研究的代码，目前看起来像这样:

  def buildManifestFile(self):
    manifestFile = self.createManifestFileForWriting(self.manifest_target)
    symlinks = []
    for subdir, dirs, files in os.walk(self.basedir, followlinks=True):
      if self.isIncluded(self.includes,self.excludes,subdir[len(self.basedir):]) and not any(link in subdir for link in symlinks):
        if not os.path.islink(subdir):                                      ## symlinks should be listed as files, not followed.
          manifestFile.write('%dir "' + subdir +  '"\n')
        else:
          manifestFile.write('"' + subdir + '"\n')
          symlinks.append(subdir)
          continue
        for fileName in files:
          fileNameWithSubdir = '"' + subdir + "/" + fileName + '"\n'
          if self.isIncluded(self.includes,self.excludes,fileNameWithSubdir):
            manifestFile.write(fileNameWithSubdir)
    self.closeManifestFile(manifestFile)

我想从本质上遍历一个目录，访问所有子目录、文件和符号链接(symbolic link)，但我不想跟随符号链接(symbolic link)，因为它会在我的输出中产生大量冗余。我实际上只是想将它们打印为一个文件。上面的代码似乎工作正常，但困扰我的部分是当我在具有许多符号链接(symbolic link)的目录上使用它时，它的复杂性增长得非常快。

Answer 1

os.walk 会默认列出符号链接(symbolic link)，但不会遵循它们。在 os.walk tail 中循环文件夹时，看看还剩下什么。

def oswalker(path):
    walked = list(os.walk(path))
    toplevel = walked[0]
    poss_folder = toplevel[1]
    folders = walked[1:]

    for folder in folders:
        # symlinks will be listed in the walk folders, but are not
        # present in the tail of the walk list
        check = [i for i,x in enumerate(poss_folder) if x in folder[0]]
        if check:
            poss_folder.pop(check[0])

    print "Symlinks %s", poss_folder


oswalker('.')