Sort tuple list with another list

Question

I have a tuple list to_order such as:

to_order = [(0, 1), (1, 3), (2, 2), (3,2)]

And a list which gives the order to apply to the second element of each tuple of to_order:

order = [2, 1, 3]

So I am looking for a way to get this output:

ordered_list = [(2, 2), (3,2), (0, 1), (1, 3)]

Any ideas?

Answer 1

You can provide a key that will check the index (of the second element) in order and sort based on it:

to_order = [(0, 1), (1, 3), (2, 2), (3,2)]
order = [2, 1, 3]
print(sorted(to_order, key=lambda item: order.index(item[1]))) # [(2, 2), (3, 2), (0, 1), (1, 3)]

EDIT

Since, a discussion on time complexities was start... here ya go, the following algorithm runs in O(n+m), using Eric's input example:

N = 5
to_order = [(randrange(N), randrange(N)) for _ in range(10*N)]
order = list(set(pair[1] for pair in to_order))
shuffle(order)


def eric_sort(to_order, order):
    bins = {}

    for pair in to_order:
        bins.setdefault(pair[1], []).append(pair)

    return [pair for i in order for pair in bins[i]]


def alfasin_new_sort(to_order, order):
    arr = [[] for i in range(len(order))]
    d = {k:v for v, k in enumerate(order)}
    for item in to_order:
        arr[d[item[1]]].append(item) 
    return [item for sublist in arr for item in sublist]


from timeit import timeit
print("eric_sort", timeit("eric_sort(to_order, order)", setup=setup, number=1000))
print("alfasin_new_sort", timeit("alfasin_new_sort(to_order, order)", setup=setup, number=1000))

OUTPUT:

eric_sort 59.282021682999584
alfasin_new_sort 44.28244407700004