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pythonlistcombinationsapriori

find 2^n -2 combinations of elements in a list


I have the following list:

list1 = ['g1','g2','g3','g4']

I want to find 2^n-2 combinations where n is the total number of items in the list. For n = 4 the result should be 2^4 -2 = 14, i.e. 14 combinations.

The combinations are:

[[['g1'],['g2','g3','g4']],[['g2'],['g1','g3','g4']], [['g3'],['g1','g2','g4']],['g4'],['g1','g2','g3']],[['g1','g2'],['g3','g4']],[['g1','g3'],['g2','g4']],[['g1','g4'],['g3','g4']],[['g2','g3'],['g1','g4']],
[['g2','g4'],['g1','g3']],[['g3','g4'],['g1','g2']],[['g1','g2','g3'],['g4']],[['g2','g3','g4'],['g1']],[['g3','g4','g1'],['g2']],[['g4','g1','g2'],['g3']]]

I know one approach: in first iteration take single element and put it into a list and other elements in second list: ['g1'],['g2','g3','g4'] in second iteration take 2 elements in a list and other elements in second list. ['g1','g2'],['g1','g4'] Is there any other approach ?? I'm writing this program in python. My approach is costly. Is there any library method to do this quickly.


Solution

  • Here's a functional approach using itertools

    import itertools as iter
    
    list1 = ['g1', 'g2', 'g3', 'g4']
    combinations = [iter.combinations(list1, n) for n in range(1, len(list1))]
    flat_combinations = iter.chain.from_iterable(combinations)
    result = map(lambda x: [list(x), list(set(list1) - set(x))], flat_combinations)
    # [[['g1'], ['g4', 'g3', 'g2']], [['g2'], ['g4', 'g3', 'g1']], [['g3'], ['g4', 'g2', 'g1']],...
    len(result)
    # 14