Re: [Tutor] Creating lists with definite (n) items without repetitions
Hello Marcus, On Fri, Sep 04, 2015 at 04:28:10PM +0200, marcus lütolf wrote: [...] > I should probably tell you the real task are a series (maximum ~ 301) > lists in which real names of people are assigned to the items/letters > for 2 people(golfers) can be in the same list(flight) only once for an > extended period of time. The next step would be to assign compatible > and noncompatible attributes to the items/letters which will reduce > the maximum of possible lists(flights) Sorry, that description doesn't help me understand your problem. Perhaps you could show how to generate the pairs you want given (say) a small list of names. Let us call them A, B, C, D and E (five people), taken three at a time? (1) Can you write out all the possible lists for n=3 (the maximum)? Assuming order does not matter, I get ten lists: A B C A B D A B E A C D A C E A D E B C D B C E B D E C D E (2) Can you show what you mean by "compatible and noncompatible attributes", and use them to "reduce the maximum of possible lists"? How do you decide which of the ten above are allowed and which are not? -- Steve ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Creating lists with definite (n) items without repetitions
On Fri, Sep 4, 2015 at 7:28 AM, marcus lütolf wrote: > I should probably tell you the real task are a series (maximum ~ 301) > lists in which real names of people are assigned to the items/letters for > 2 people(golfers) can be in the same list(flight) only once for an > extended period of time. > The next step would be to assign compatible and noncompatible attributes > to the items/letters which will reduce the maximum of possible > lists(flights) > > I came up with this (obviously it could be made shorter, but I thought it would be clearer this way): import string, itertools def main(): validCombos = [] usedPairs = [] allCombos = itertools.combinations(string.ascii_lowercase, 3) for combo in allCombos: pair1 = (combo[0],combo[1]) pair2 = (combo[0],combo[2]) pair3 = (combo[1],combo[2]) if pair1 in usedPairs or pair2 in usedPairs or pair3 in usedPairs: next else: usedPairs.append(pair1) usedPairs.append(pair2) usedPairs.append(pair3) validCombos.append(combo) print(validCombos) print(len(validCombos)) if __name__ == '__main__': main() The resulting triplets seem to meet your criteria - but there are only 90 of them. How confident are you about the number 301? ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
