Python 如何获取文件的父目录和子目录
我尝试列出一个特定文件所属目录的名称。下面是我的文件树的一个示例:Python 如何获取文件的父目录和子目录,python,Python,我尝试列出一个特定文件所属目录的名称。下面是我的文件树的一个示例: root_folder ├── topic_one │ ├── one-a.txt │ └── one-b.txt └── topic_two ├── subfolder_one │ └── sub-two-a.txt ├── two-a.txt └── two-b.txt 理想情况下,我希望打印出来的是: "File: file_name belongs in parent dir
root_folder
├── topic_one
│ ├── one-a.txt
│ └── one-b.txt
└── topic_two
├── subfolder_one
│ └── sub-two-a.txt
├── two-a.txt
└── two-b.txt
理想情况下,我希望打印出来的是:
"File: file_name belongs in parent directory"
"File: file_name belongs in sub directory, parent directory"
我写了这个剧本:
for root, dirs, files in os.walk(root_folder):
# removes hidden files and dirs
files = [f for f in files if not f[0] == '.']
dirs = [d for d in dirs if not d[0] == '.']
if files:
tag = os.path.relpath(root, os.path.dirname(root))
for file in files:
print file, "belongs in", tag
这给了我这个输出:
one-a.txt belongs in topic_one
one-b.txt belongs in topic_one
two-a.txt belongs in topic_two
two-b.txt belongs in topic_two
sub-two-a.txt belongs in subfolder_one
我似乎不知道如何将文件的父目录包含在子目录中。非常感谢您的任何帮助或替代方法。感谢此解决方案:
for root, dirs, files in os.walk(root_folder):
# removes hidden files and dirs
files = [f for f in files if not f[0] == '.']
dirs = [d for d in dirs if not d[0] == '.']
if files:
tag = os.path.relpath(root, root_folder)
for file in files:
tag_parent = os.path.dirname(tag)
sub_folder = os.path.basename(tag)
print "File:",file,"belongs in",tag_parent, sub_folder if sub_folder else ""
打印出:
File: one-a.txt belongs in topic_one
File: one-b.txt belongs in topic_one
File: two-a.txt belongs in topic_two
File: two-b.txt belongs in topic_two
File: sub-two-a.txt belongs in topic_two subfolder_one
表示relpath将第一个参数作为目标,第二个参数作为原点。在您的帖子中,您正在将
root
与其自身进行比较,但是您应该执行relpath(root,root\u文件夹)
并且对于打印,您可以使用tag.replace(os.path.sep,',')
谢谢!这非常接近。我得到这个sub-two-a.txt属于topic\u two/子文件夹\u one,topic\u two
。对于子目录打印输出,是否仍要删除topic\u two/
?os.path.basename()
会做一些小动作,谢谢你,@Jean Françoisfab!您的回答和@Jjpx的建议帮助我解决了问题。现在root
已经包含了从root\u文件夹到文件的相对路径。对于更深层次的文件,您最好打印“文件:”,文件“属于”,“,”。加入(root.split(os.path.sep))