Python：根据一个列表的索引对字典列表进行排序

Question

So I have dict like the following:

dict1 = {1: [-1, 1, 2, 4, 3], 2: [11, 10, 9, 8, 7]}

How can I sort the lists in dict1 based on the indices that would sort one of the lists. Such as the sorted indices of list

[11, 10, 9, 8, 7]

which is

[4, 3, 2, 1, 0]

So the new dict1 will be:

dict1 = {1: [3, 4, 2, 1, -1], 2: [7, 8, 9, 10, 11]}

I have tried the code like this:

sorted(dict1.items(), key=lambda t: t[1])

But still cannot figure it out? I can use a for loop to do this work but I believe there exist a more pythonic way. Any idea?

Answer 1

One way would be to compute the index list and apply it to all lists:

>>> dict1 = {1: [-1, 1, 2, 4, 3], 2: [11, 10, 9, 8, 7]}
>>> basis = dict1[2]
>>> indexes = sorted(range(len(basis)), key=basis.__getitem__)
>>> for lst in dict1.values():
        lst[:] = [lst[i] for i in indexes]

>>> dict1
{1: [3, 4, 2, 1, -1], 2: [7, 8, 9, 10, 11]}

If you're ok with "the first" list determining the order (you only said "based on [...] one of the lists" ), you could also build columns and sort those:

>>> dict1 = {1: [-1, 1, 2, 4, 3], 2: [11, 10, 9, 8, 7]}
>>> dict(zip(dict1.keys(), zip(*sorted(zip(*dict1.values())))))
{1: (-1, 1, 2, 3, 4), 2: (11, 10, 9, 7, 8)}

Answer 2

Sort using enumerate to get the sorted indexes and just use the indexes to sort the other lists:

from operator import itemgetter
dict1 = {1: [-1, 1, 2, 4, 3], 2: [11, 10, 9, 8, 7]}

l = [i for i, _ in sorted(enumerate(dict1[2]), key=itemgetter(1))]


for v in dict1.values():
    v[:] = (v[i] for i in l)
print(dict1)

Or if you want a new dict:

l = [i for i, _ in sorted(enumerate(dict1[2]), key=itemgetter(1))]
new = {k: [v[i] for i in l] for k, v in dict1.items()}

Or if you don't mind tuples:

new = {k: itemgetter(*l)(v) for k, v in dict1.items()})

Answer 3

Using itemgetter in a dictionary comprehension

>>> from operator import itemgetter
>>> order = [4, 3, 2, 1, 0]
>>> key = itemgetter(*order)
>>> {k: key(dict1[k]) for k in dict1}
{1: (3, 4, 2, 1, -1), 2: (7, 8, 9, 10, 11)}