Python, sort a list by another list [duplicate]

Answer 1

You can try passing a lambda function to the key parameter of the sorted() built-in function:

a = ['c', 'd', 'B', 'a', 'e']
b = ['a001', 'B002', 'c003', 'd004', 'e005']
c = sorted(b, key = lambda x: a.index(x[0])) # ['c003', 'd004', 'b002', 'a001', 'e005']

Answer 2

If you have a very big list, the solutions using .index will not be very efficient as the first list will be index'd for each entry in the second list. This will take O(n^2) time.

Instead, you can construct a sort mapping:

order = {v:i for i,v in enumerate(a)}
c = sorted(b, key=lambda x: order[x[0]])

Answer 3

You can do it using the key named argument of sorted():

c = sorted(b, key = lambda e: a.index(e[0]))