Denis McMahon writes: > On Thu, 06 Nov 2014 15:14:05 +1100, Chris Angelico wrote: > > On Thu, Nov 6, 2014 at 3:00 PM, Denis McMahon wrote: > >> def baseword(s): > >> """find shortest sequence which repeats to generate s""" > >> return s[0:["".join([s[0:x]for k in range(int(len(s)/x)+1)])[0:len > >> (s)]for x in range(1,len(s)+1)].index(s)+1] > > > > That's hardly a PEP-8 compliant line, but I can help out a bit. > > > > return s[0:[(s[0:x]*(len(s)//x+1))[0:len(s)]for x in > > range(1,len(s)+1)].index(s)+1] > > > > That's still 83 characters without indentation, but it's close now. > > l = len > r = range > > but technically then it's no longer a one liner. > > > I love the algorithm. Took me a bit of analysis (and inspection of > > partial results) to understand what your code's doing, but it's stupidly > > elegant and elegantly stupid. > > :) > > Well yes, building that list is a stupid way to solve the problem, but I > can't see another way to do it in one line. It's an implementation of my > algorithm 3 (which I think you described) but working from the other end > as it were.
70-character expression, even with the spaces: >>> x = 'tästäkötästäkötästä' >>> x[:min(k for k in range(len(x) + 1) if x == (x[:k] * len(x))[:len(x)])] 'tästäkö' -- https://mail.python.org/mailman/listinfo/python-list
