How to sort a list of tuples according to another list

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Question :

How to sort a list of tuples according to another list

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?

Asked By: alwbtc


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))

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]
Answered By: Amber

Answer #2:

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

Answered By: fritz-johann

Answer #3:

There’s actually a way to do this in linear O(n) time, because this isn’t really a sorting operation. The existence of the list b means that the sorting is already done; all we really need to do is to rearrange the elements of a to be in the same order. This can be done efficiently thanks to dictionaries.

from collections import defaultdict

def sorted_by(seq_to_sort, desired_order, key=None):
    if key is None:
        key = lambda x: x

    # group the elements by their key
    grouped_items = defaultdict(list)
    for item in seq_to_sort:
        k = key(item)

    # flatten the dict of groups to a list
    return [item for key in desired_order for item in grouped_items[key]]


a = [("ax", 1), ("ec", 3), ("bk", 5)]
b = ["ec", "ax", "bk"]
result = sorted_by(a, b, lambda tup: tup[0])
print(result)  # output: [("ec", 3), ("ax", 1), ("bk", 5)]


  • This is a stable sort; if two list items have the same key, their order will be preserved. Example:

    >>> sorted_by([1, 2, 3], [5], key=lambda x: 5)
    [1, 2, 3]
  • If any list elements are mapped to keys that don’t exist in desired_order, those elements are silently discarded. For example:

    >>> sorted_by([1, 2, 3], [1, 2, 3], key=lambda x: 5)

See also:

Answered By: Aran-Fey

Answer #4:

Traditional sorting may not be needed.

[tup for lbl in b for tup in a if tup[0] == lbl]
# [('ec', 3), ('ax', 1), ('bk', 5)]
Answered By: pylang

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