I took a swing at answering a question on StackOverflow, albeit I liked the accepted answer better than mine. But I think my suggested answer should be more simple to digest and provide more control with “allEqual()” function.
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| import itertools | |
| a = [7, 3] | |
| b = [3, 1, 2] | |
| c = [4, 3, 5] | |
| def allEqual(t): | |
| same = True | |
| if len(t) == 1: | |
| return True | |
| if len(t) == 0: | |
| return False | |
| for i in range(1, len(t)): | |
| if t[i] != t[i – 1]: | |
| same = False | |
| i = len(t) – 1 | |
| else: | |
| same = same and True | |
| return same | |
| combo = list(itertools.product(a, b, c)) | |
| # print list(itertools.permutations(a,2)) | |
| # print combo | |
| # combo = [x for x in combo if x[0]==x[1]==x[2]] | |
| # print combo | |
| combo = [x for x in combo if allEqual(x)] | |
| print combo |
Aarsh 