我有一个目录 DATA ,其中包含几个子目录,并且在每个子目录中,都有更多的目录和文件。
这是我的代码:
for dirpath,subs,filenames in os.walk("/Users/.../DATA"):
for f in filenames:
print(os.path.abspath(os.path.join(dirpath, f)))
此代码输出的结果是绝对目录(例如“ / Users /.../ Data / SubFile / SubFile.txt”)
我想要的结果是(例如“ Data / SubFile / Subfile.txt”)
答案 0 :(得分:1)
像这样简单的事情呢?
dir_path = "/Users/.../DATA"
for dirpath,subs,filenames in os.walk("/Users/.../DATA"):
for f in filenames:
print(os.path.abspath(os.path.join(dirpath, f))[len(dir_path):])
答案 1 :(得分:0)
common_prefix = os.path.commonprefix(["/Users/.../DATA"])
for dirpath, subs, filenames in os.walk("/Users/.../DATA"):
for f in filenames:
print(os.path.relpath(os.path.join(dirpath, f), common_prefix))