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pythonarrayslistsorting
Izalion
Izalion

Reputation: 758

How can I get the position from smaller number to biggest number of list?

I know my question seems confusing, but let's say I have the following list:

List1 = [15,16,1,2]

So, the position of [ 15, 16, 1, 2 ] is [ 0, 1, 2, 3 ].

What I am looking for is a result list containing the position from the smallest number of list1 to biggest.

Result List = [2,3,0,1]

I don't exactly know the most efficient way to do it in Python.

Upvotes: 1

Views: 399

Answers (2)

janluke
janluke

Reputation: 1677

What you want is called argsort in some libraries. In mathematical terms, you want a "permutation" (look it up on wikipedia) that when applied to the input list gives a sorted list.

The best way to do it in Python is the following:

def argsort(lst):
    return sorted(range(len(lst)), key=lst.__getitem__)

lst = [15,16,1,2]
print(argsort(lst))
# print: [2, 3, 0, 1]

This is how it works. Instead of sorting the values in lst, you sort the indexes (range(0, len(lst))) using the corresponding lst values as "keys". Keep in mind that lst.__getitem__(i) is just equivalent to lst[i].

EDIT: for people new to Python and unfamiliar with the "key" argument and the magic methods of Python (those written like __this__), this may be intimidating. A more friendly versione could be:

def argsort(lst):
    indexes = range(len(lst))
    return sorted(indexes, key=lambda i: lst[i])

Upvotes: 2

Patrick Artner
Patrick Artner

Reputation: 51643

Use enumerate and sorted with a custom key using a lambda expression:

List1 = [15,16,1,2]

v  = [i for i,_ in sorted(enumerate(List1), key=lambda x:x[1])]

print(v)

Output:

[2,3,0,1]

enumerate(List1) produces tuples of (index, value) that are sorted based on the value and then only the index is used to create the resulting list.

Upvotes: 3

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