I have a list like this:
myList = [10,30,40,20,50]
Now I use numpy's argsort
function to get the indices for the sorted list:
import numpy as np
so = np.argsort(myList)
which gives me the output:
array([0, 3, 1, 2, 4])
When I want to sort an array using so
it works fine:
myArray = np.array([1,2,3,4,5])
myArray[so]
array([1, 4, 2, 3, 5])
But when I apply it to another list, it does not work but throws an error
myList2 = [1,2,3,4,5]
myList2[so]
TypeError: only integer arrays with one element can be converted to an index
How can I now use so
to sort another list without using a for-loop and without converting this list to an array first?
You can't. You have to convert it to an array then back.
myListSorted = list(np.array(myList)[so])
Edit: I ran some benchmarks comparing the NumPy way to the list comprehension. NumPy is ~27x faster
>>> from timeit import timeit
>>> import numpy as np
>>> myList = list(np.random.rand(100))
>>> so = np.argsort(myList) #converts list to NumPy internally
>>> timeit(lambda: [myList[i] for i in so])
12.29590070003178
>>> myArray = np.random.rand(100)
>>> so = np.argsort(myArray)
>>> timeit(lambda: myArray[so])
0.42915570305194706
myList2
is a normal python list, and it does not support that kind of indexing.
You would either need to convert that to a numpy.array
, Example -
In [8]: np.array(myList2)[so]
Out[8]: array([1, 4, 2, 3, 5])
Or you can use list comprehension -
In [7]: [myList2[i] for i in so]
Out[7]: [1, 4, 2, 3, 5]
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