在python中,假设我有一个这样的路径:
/folderA/folderB/folderC/folderD/
我怎样才能得到folderD部分?
在python中,假设我有一个这样的路径:
/folderA/folderB/folderC/folderD/
我怎样才能得到folderD部分?
当前回答
path = "/folderA/folderB/folderC/folderD/"
last = path.split('/').pop()
其他回答
你可以这样做
>>> import os
>>> os.path.basename('/folderA/folderB/folderC/folderD')
UPDATE1:如果你给它/folderA/folderB/folderC/folderD/xx.py,这种方法是有效的。这将xx.py作为基名。我猜这不是你想要的。所以你可以这样做-
>>> import os
>>> path = "/folderA/folderB/folderC/folderD"
>>> if os.path.isdir(path):
dirname = os.path.basename(path)
UPDATE2:正如lars指出的那样,进行更改以适应以'/'结尾。
>>> from os.path import normpath, basename
>>> basename(normpath('/folderA/folderB/folderC/folderD/'))
'folderD'
path = "/folderA/folderB/folderC/folderD/"
last = path.split('/').pop()
如果你使用本地python包pathlib,它真的很简单。
>>> from pathlib import Path
>>> your_path = Path("/folderA/folderB/folderC/folderD/")
>>> your_path.stem
'folderD'
假设在文件夹d中有一个文件的路径。
>>> from pathlib import Path
>>> your_path = Path("/folderA/folderB/folderC/folderD/file.txt")
>>> your_path.name
'file.txt'
>>> your_path.parent
'folderD'
在我目前的项目中,我经常将路径的后面部分传递给函数,因此使用path模块。为了得到第n部分的倒序,我使用:
from typing import Union
from pathlib import Path
def get_single_subpath_part(base_dir: Union[Path, str], n:int) -> str:
if n ==0:
return Path(base_dir).name
for _ in range(n):
base_dir = Path(base_dir).parent
return getattr(base_dir, "name")
path= "/folderA/folderB/folderC/folderD/"
# for getting the last part:
print(get_single_subpath_part(path, 0))
# yields "folderD"
# for the second last
print(get_single_subpath_part(path, 1))
#yields "folderC"
此外,为了以包含剩余路径的路径的相反顺序传递第n部分,我使用:
from typing import Union
from pathlib import Path
def get_n_last_subparts_path(base_dir: Union[Path, str], n:int) -> Path:
return Path(*Path(base_dir).parts[-n-1:])
path= "/folderA/folderB/folderC/folderD/"
# for getting the last part:
print(get_n_last_subparts_path(path, 0))
# yields a `Path` object of "folderD"
# for second last and last part together
print(get_n_last_subparts_path(path, 1))
# yields a `Path` object of "folderc/folderD"
注意,这个函数返回一个可以很容易地转换为字符串的Pathobject(例如str(path))
str = "/folderA/folderB/folderC/folderD/"
print str.split("/")[-2]