Hi I have an array with X amount of values in it I would like to locate the indexs of the ten smallest values. In this link they calculated the maximum effectively, How to get i
Since this question was posted, numpy has updated to include a faster way of selecting the smallest elements from an array using argpartition. It was first included in Numpy 1.8.
Using snarly's answer as inspiration, we can quickly find the k=3
smallest elements:
In [1]: import numpy as np
In [2]: arr = np.array([1, 3, 2, 4, 5])
In [3]: k = 3
In [4]: ind = np.argpartition(arr, k)[:k]
In [5]: ind
Out[5]: array([0, 2, 1])
In [6]: arr[ind]
Out[6]: array([1, 2, 3])
This will run in O(n) time because it does not need to do a full sort. If you need your answers sorted (Note: in this case the output array was in sorted order but that is not guaranteed) you can sort the output:
In [7]: sorted(arr[ind])
Out[7]: array([1, 2, 3])
This runs on O(n + k log k) because the sorting takes place on the smaller output list.
If you call
arr.argsort()[:3]
It will give you the indices of the 3 smallest elements.
array([0, 2, 1], dtype=int64)
So, for n
, you should call
arr.argsort()[:n]
Just don't reverse the sort results.
In [164]: a = numpy.random.random(20)
In [165]: a
Out[165]:
array([ 0.63261763, 0.01718228, 0.42679479, 0.04449562, 0.19160089,
0.29653725, 0.93946388, 0.39915215, 0.56751034, 0.33210873,
0.17521395, 0.49573607, 0.84587652, 0.73638224, 0.36303797,
0.2150837 , 0.51665416, 0.47111993, 0.79984964, 0.89231776])
Sorted:
In [166]: a.argsort()
Out[166]:
array([ 1, 3, 10, 4, 15, 5, 9, 14, 7, 2, 17, 11, 16, 8, 0, 13, 18,
12, 19, 6])
First ten:
In [168]: a.argsort()[:10]
Out[168]: array([ 1, 3, 10, 4, 15, 5, 9, 14, 7, 2])
I don't guarantee that this will be faster, but a better algorithm would rely on heapq
.
import heapq
indices = heapq.nsmallest(10,np.nditer(arr),key=arr.__getitem__)
This should work in approximately O(N)
operations whereas using argsort
would take O(NlogN)
operations. However, the other is pushed into highly optimized C, so it might still perform better. To know for sure, you'd need to run some tests on your actual data.
This code save 20 index of maximum element of split_list
in Twenty_Maximum
:
Twenty_Maximum = split_list.argsort()[-20:]
against this code save 20 index of minimum element of split_list
in Twenty_Minimum
:
Twenty_Minimum = split_list.argsort()[:20]