CODEWITHSUNDEEP
pythonarraysnumpyminimum
astrochris
astrochris

Reputation: 1866

I have need the N minimum (index) values in a numpy array

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 indices of N maximum values in a numpy array? however I cant comment on links yet so I'm having to repost the question.

I'm not sure which indices i need to change to achieve the minimum and not the maximum values. This is their code 

In [1]: import numpy as np

In [2]: arr = np.array([1, 3, 2, 4, 5])

In [3]: arr.argsort()[-3:][::-1]
Out[3]: array([4, 3, 1])

Upvotes: 36

Views: 66209

Answers (5)

mohammadali68
mohammadali68

Reputation: 31

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]

Upvotes: 2

Alex
Alex

Reputation: 824

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.

Upvotes: 31

Mike Müller
Mike Müller

Reputation: 85442

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])

Upvotes: 4

mgilson
mgilson

Reputation: 309929

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.

Upvotes: 8

petrichor
petrichor

Reputation: 6569

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]

Upvotes: 56

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