python - How to sort a list of tuples according to another list

Question

There is a list:

a = [("ax", 1), ("ec", 3), ("bk", 5)]

another list:

b = ["ec", "ax", "bk"]

I want to sort a according to b:

sort_it(a, b)

a = [("ec", 3), ("ax", 1), ("bk", 5)]

How to do this?

Answer 1

a.sort(key=lambda x: b.index(x[0]))

This sorts a in-place using the the index in b of the first element of each tuple from a as the values it sorts on.

Another, possibly cleaner, way of writing it would be:

a.sort(key=lambda (x,y): b.index(x))

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If you had large numbers of items, it might be more efficient to do things a bit differently, because .index() can be an expensive operation on a long list, and you don't actually need to do a full sorting since you already know the order:

mapping = dict(a)
a[:] = [(x,mapping[x]) for x in b]

Note that this will only work for a list of 2-tuples. If you want it to work for arbitrary-length tuples, you'd need to modify it slightly:

mapping = dict((x[0], x[1:]) for x in a)
a[:] = [(x,) + mapping[x] for x in b]