How to sort a list of tuples according to another list

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]

Another posibility is to sort a, sort the indices of b according to b and than sort the a according to the indices

a.sort(key=lambda x: x[0])
ind = [i[0] for i in sorted(enumerate(b),key=lambda x: x[1])]
a = [i[0] for i in sorted(zip(a,ind),key=lambda x: x[1])]

since every sorting takes n*log(n) this is still scalable for bigger lists