algorithm - Efficient way to make two pairs in python -

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i make 2 pairs pairs. pair consists of 2 elements, , two-pair consists of 2 pairs. here list of constraints:

  1. in pair, order of elements important: (element1, element2) != (element2, element1)
  2. in two-pair, order of pairs not important: (pair1, pair2) == (pair2, pair1)

i wrote pseudo code satisfies above constraints follows:

class pair:     def __init__(self, element1, element2):         assert isinstance(element1, element)         assert isinstance(element2, element)         self.element1 = element1         self.element2 = element2      def __eq__(self, other):         if not isinstance(other, pair):             return false         if self.element1 != other.element1:             return false         if self.element2 != other.element2:             return false         return true      def __ne__(self, other):         return not (self.__eq__(other))      def __hash__(self):         return hash(self.element1) ^ hash(self.element2)      def getfirst(self):         return self.element1      def getsecond(self):         return self.element2
class twopair:     def __init__(self, pair1, pair2):         assert isinstance(pair1, pair)         assert isinstance(pair2, pair)         self.pair1 = pair1         self.pair2 = pair2      def __eq__(self, other):         if not isinstance(other, twopair):             return false         if self.pair1 == other.pair1 , self.pair2 == other.pair2:             return true         if self.pair1 == other.pair2 , self.pair2 == other.pair1:             return true         return false      def __ne__(self, other):         return not (self.__eq__(other))      def __hash__(self):         return hash(self.pair1) ^ hash(self.pair2)      def getfirst(self):         return self.pair1      def getsecond(self):         return self.pair2
def maketwopairs(allpairs):     alltwopairs = set([])     pair1 in allpairs:         pair2 in allpairs:             if pair1 == pair2:                 continue             twopair = twopair(pair1, pair2)             if twopair in alltwopairs:                 continue             else:                 alltwopairs.add(twopair)     return alltwopairs

the function maketwopairs takes long time in code. there other representation 2 pairs? or, can above code improved?

you better off sticking standard python datastructures. tuple pair , set twopair (although might write set subclass add __hash__ method).

for example:

import operator  class twopairs(set):   def __hash__(self):     return reduce(operator.xor, map(hash, self))

regarding fact maketwopairs function takes long time execute, can rewrite :

def make_two_pairs(all_pairs):   all_two_pairs = set()   # uniqify pairs list   all_pairs = list(set(all_pairs))   in range(len(all_pairs)-1):     j in range(i+1, len(all_pairs)):       all_two_pairs.add(twopairs(all_pairs[i], all_pairs[j]))    return all_two_pairs

you produce unique twopairs, without combinatorial explosion or overhead of testing everytime before adding new pair result set.


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