On Tue, 30 Dec 2008 at 21:30, rdmurray at bitdance.com wrote: > On Tue, 30 Dec 2008 at 17:51, Phillip J. Eby wrote: >> At 02:32 PM 12/30/2008 -0800, Scott David Daniels wrote: >> > More trouble with the "just take the dirname": >> > >> > paths = ['/a/b/c', '/a/b/d', '/a/b'] >> > os.path.dirname(os.path.commonprefix([ >> > os.path.normpath(p) for p in paths])) >> > >> > give '/a', not '/a/b'. >> >> ...because that's the correct answer. > > But not the answer that is wanted. > > So the challenge now is to write a single expression that will yield > '/a/b' when passed the above paths list, and also produce '/a/b' when > passed the following paths list: > > paths = ['/a/b/c', '/a/b/cd'] Sorry, now I see what you are saying: that in '/a/b' the 'b' is the filename. Clearly that wasn't what I intuitively expected our notional 'commonpathprefix' command to produce, for whatever that is worth :) --RDM
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