On 2018-09-18 17:05, ast wrote: > Le 18/09/2018 à 17:01, ast a écrit : >> Hello >> >> I found a smart and very concise code to >> generate all permutations of a list. >> I put it here if someone is interested to >> figure out how it works
When you say "found a [...] code" I hope you mean "wrote a function" rather than "discovered a bit of code and decided to share it without attribution or permission". >> >> >> def permut(li, prefix=[]): >> >> if len(li)==1: >> yield prefix + li >> else: >> for elt in li: >> li2 = li.copy() >> li2.remove(elt) >> yield from S(li2, prefix+[elt]) > > error: permut instead of S > >> yield from permut(li2, prefix+[elt]) Nice. If you want to see more implementations of the same thing, have a look in the standard library docs ;-) https://docs.python.org/3/library/itertools.html#itertools.permutations -- https://mail.python.org/mailman/listinfo/python-list
