Transforming nested key-tuples into their dictionary values

[ClassicalSoul]

Surely there has got to be a better way of doing this?

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>>> d = dict(enumerate(list('abcde'))) >>> INPUT = [[(0, 1, 2), (3, 4)], [(0, 1, 3), (2, 4)], [(0, 1, 4), (2, 3)], [(0, 2, 3), (1, 4)], [(0, 2, 4), (1, 3)], [(0, 3, 4), (1, 2)], [(1, 2, 3), (0, 4)], [(1, 2, 4), (0, 3)], [(1, 3, 4), (0, 2)], [(2, 3, 4), (0, 1)]] >>> list(map(lambda grouping: tuple(map(lambda group: tuple(map(lambda id_: d[id_], group)), grouping)), INPUT))

Output:
[(('a', 'b', 'c'), ('d', 'e')), (('a', 'b', 'd'), ('c', 'e')), (('a', 'b', 'e'), ('c', 'd')), (('a', 'c', 'd'), ('b', 'e')), (('a', 'c', 'e'), ('b', 'd')), (('a', 'd', 'e'), ('b', 'c')), (('b', 'c', 'd'), ('a', 'e')), (('b', 'c', 'e'), ('a', 'd')), (('b', 'd', 'e'), ('a', 'c')), (('c', 'd', 'e'), ('a', 'b'))]

[buran]

even without going into understanding your goal - instead of 3 lamdas - define separarte functions... oneliners are not always better coding

[bowlofred]

In what way do you want it to be "better"? Run faster, clearer code, different output, something else?

[ClassicalSoul]

My goal is to substitute the keys in the input object with their corresponding dictionary value. I want to write clearer code or faster code (or both). In particular, I'm wondering if there is no more standard way of doing what I'm doing -- for example, if there is some function I can import.

[bowlofred]

No, there's no obvious mapping function that does mapping within nested structures natively. I'd probably write it as separate loops to improve readability if the performance weren't critical.