I know that this question has been asked a hundred times, but the answer always seems to be “use numpy’s argsort”. But either I am misinterpreting what most people are asking, or the answers are not correct for the question. Whatever be the case, I wish to get indices of a list’s ascending order. The phrasing is confusing, so as an example, given a list [4, 2, 1, 3] I expect to get a list back [3, 1, 0, 2]. The smallest item is 1, so it gets index 0, the largest one is 4 so it gets index 3. It seems to me that argsort is often suggested, but it just doesn’t seem to do that.
from numpy import argsort l = [4, 2, 1, 3] print(argsort(l)) # [2, 1, 3, 0] # Expected [3, 1, 0, 2]
Clearly argsort is doing something else, so what is it actually doing and how is it similar to the expected behaviour so that it is so often (wrongly) suggested? And, more importantly, how can I get the desired output?
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Answer
The argsort() is basically converting your list to a sorted list of indices.
l = [4, 2, 1, 3]
First it gets index of each element in the list so new list becomes:
indexed=[0, 1, 2, 3]
Then it sorts the indexed list according to the items in the original list. As 4:0 , 2:1 , 1:2 and 3:3 where : means “corresponds to”.
Sorting the original list we get
l=[1, 2, 3, 4]
And placing values of each corresponding index of old list
new=[2,1,3,0]
So basically it sorts the indices of a list according to the original list.