Python中获取绝对文件路径的目录路径
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/15022854/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
提示：将鼠标放在中文语句上可以显示对应的英文。显示中英文
时间：2020-08-18 13:08:00 来源:igfitidea点击:
Get the directory path of absolute file path in Python
提问by ddinchev
I want to get the directory where the file resides. For example the full path is:
我想获取文件所在的目录。例如完整路径是：
fullpath = "/absolute/path/to/file"
# something like:
os.getdir(fullpath) # if this existed and behaved like I wanted, it would return "/absolute/path/to"
I could do it like this:
我可以这样做：
dir = '/'.join(fullpath.split('/')[:-1])
But the example above relies on specific directory separator and is not really pretty. Is there a better way?
但是上面的例子依赖于特定的目录分隔符，并不是很漂亮。有没有更好的办法？
采纳答案by isedev
You are looking for this:
你正在寻找这个：
>>> import os.path
>>> fullpath = '/absolute/path/to/file'
>>> os.path.dirname(fullpath)
'/absolute/path/to'
Related functions:
相关功能：
>>> os.path.basename(fullpath)
'file'
>>> os.path.split(fullpath)
('/absolute/path/to','file')
