On Jun 7, 12:55 am, eliben <[EMAIL PROTECTED]> wrote:
> Hello,
>
> os.path.split returns the head and tail of a path, but what if I want
> to have all the components ? I could not find a portable way to do
> this in the standard library, so I've concocted the following
> function. It uses os.path.split to be portable, at the expense of
> efficiency.
>
> ----------------------------------
> def parse_path(path):
> """ Parses a path to its components.
>
> Example:
> parse_path("C:\\Python25\\lib\\site-packages\
> \zipextimporter.py")
>
> Returns:
> ['C:\\', 'Python25', 'lib', 'site-packages',
> 'zipextimporter.py']
>
> This function uses os.path.split in an attempt to be portable.
> It costs in performance.
> """
> lst = []
>
> while 1:
> head, tail = os.path.split(path)
>
> if tail == '':
> if head != '': lst.insert(0, head)
> break
> else:
> lst.insert(0, tail)
> path = head
>
> return lst
> ----------------------------------
>
> Did I miss something and there is a way to do this standardly ?
> Is this function valid, or will there be cases that will confuse it ?
>
You can just split the path on `os.sep', which contains the path
separator of the platform on which Python is running:
components = pathString.split(os.sep)
Sebastian
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