Python:无法在我的代码中找到错误。 os.walk() 返回的文件名

Question

我有以下目录结构:

DICOMs:
      Dir_1 
           /sub-dir 1
           /sub-dir 2
                   /file1.dcm
      Dir_2 
           /sub-dir 1
           /sub-dir 2
                    /file1.dcm

我编写了以下代码来读取每个子目录的第一个文件。

dire_str = '/DICOMs:/

for dirname,dirnames,filenames in os.walk(dire_str,topdown=True):
    for subdirname in dirnames:
        print(os.path.join(dirname,subdirname))
        a = 1
        for filename in filenames:
            firstfilename = os.path.join(dirname, filename)
            dcm_info = dicom.read_file(firstfilename, force=True)

如果我在 python 控制台上运行它，它会给我

dirname = dir_1 
dirnames=[subdir_1, subdir_2]
filenames = .DSstore

对于文件名数组中的此错误，我无法获取第一个文件的文件名。如果代码有错误或语法错误，有人可以帮助我吗？ 我在 subdir_2 和 subdir_1 下有 file1.dcm。但仍然显示文件是.DS_Store。

我想要实现的是:

1) go into Dir(say  dir_1)
   go inside subdir_1 
   look for first .dcm file to read header tag 
   if tag is present in first file (yes) then call a function which will excute code on this subdir.(note i want to check just first file)
   if not go out of this subdir
   in this way check every subdir

once done with one dir 
repeat these steps for dir2

非常感谢!

Answer 1

这可能更接近您的意图:

for dirpath, dirnames, filenames in os.walk(dire_str, topdown=True):
    # Do some stuff that is per directory
    for filename in filenames:
        # Do some stuff that is per file
        pass

如果您只想对以 .dcm 结尾的文件进行操作，遵循以下原则可能会很好:

for dirpath, dirnames, filenames in os.walk(dire_str, topdown=True):
    # Do some stuff that is per directory
    for filename in filenames:
        if filename.endswith('.dcm')
            # Do some stuff that is per file
            pass

为了专门解决您添加到问题中的新伪代码，我相信您会需要如下所示的内容:

for dirpath, dirnames, filenames in os.walk(dire_str, topdown=True):
    for filename in filenames:
        if filename.endswith('.dcm') and tag_present_in_dcm(
            os.path.join(dirpath, filename)
        ):
            execute_code_on_subdir(dirpath)
            break