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Saw recipe 502194 and wondered if I could do similar. I initially came up with function comb that enumerates all combinations. I wanted a generator however and so abandoned that approach for comb2 which is a little more complex.

Python, 79 lines
 ``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79``` ```def comb(*sequences): ''' combinations of multiple sequences so you don't have to write nested for loops >>> from pprint import pprint as pp >>> pp(comb(['Guido','Larry'], ['knows','loves'], ['Phyton','Purl'])) [['Guido', 'knows', 'Phyton'], ['Guido', 'knows', 'Purl'], ['Guido', 'loves', 'Phyton'], ['Guido', 'loves', 'Purl'], ['Larry', 'knows', 'Phyton'], ['Larry', 'knows', 'Purl'], ['Larry', 'loves', 'Phyton'], ['Larry', 'loves', 'Purl']] >>> ''' combinations = [[seq] for seq in sequences[0]] for seq in sequences[1:]: combinations = [comb+[item] for comb in combinations for item in seq ] return combinations def comb2(*sequences): ''' Generator of combinations of multiple sequences so you don't have to write nested for loops Note: rightmost sequence changes the quickest as if it were the inner loop. >>> for x in comb2(['Guido','Larry'], ['knows','loves','hates'], ['Phyton','Purl','Wuby','Auk']): ... print x ... ['Guido', 'knows', 'Phyton'] ['Guido', 'knows', 'Purl'] ['Guido', 'knows', 'Wuby'] ['Guido', 'knows', 'Auk'] ['Guido', 'loves', 'Phyton'] ['Guido', 'loves', 'Purl'] ['Guido', 'loves', 'Wuby'] ['Guido', 'loves', 'Auk'] ['Guido', 'hates', 'Phyton'] ['Guido', 'hates', 'Purl'] ['Guido', 'hates', 'Wuby'] ['Guido', 'hates', 'Auk'] ['Larry', 'knows', 'Phyton'] ['Larry', 'knows', 'Purl'] ['Larry', 'knows', 'Wuby'] ['Larry', 'knows', 'Auk'] ['Larry', 'loves', 'Phyton'] ['Larry', 'loves', 'Purl'] ['Larry', 'loves', 'Wuby'] ['Larry', 'loves', 'Auk'] ['Larry', 'hates', 'Phyton'] ['Larry', 'hates', 'Purl'] ['Larry', 'hates', 'Wuby'] ['Larry', 'hates', 'Auk'] >>> The algorithm relies on noting the following and generalising: If you had three sequences of length 2, 3, and 4 left-to-right, Then the indices of the *elements* for all combinations can be generated with: for x in range(2*3*4): print ( (x/(3*4*1))%2, (x/(4*1))%3, (x/(1))%4 ) ''' import operator lengths = [len(seq) for seq in sequences] range_len_seq = range(len(sequences)) max_count = reduce(operator.mul, lengths) _tmp = lengths + [1] # append multiplicative identity dividers = [reduce(operator.mul, _tmp[-x-1:]) for x in range_len_seq][::-1] modulos = lengths for n in range(max_count): yield [sequences[r][(n/dividers[r])%modulos[r]] for r in range_len_seq] ```

Deeply nested code is usually code in need of refactoring, but if you still think you need it then comp2 will save you some levels of indentation.

• Donald 'Paddy' McCarthy.
 Created by Paddy McCarthy on Wed, 31 Jan 2007 (PSF)