CODEWITHSUNDEEP
pythonnumpy
Alex McLean
Alex McLean

Reputation: 2764

Quickest way to find the nth largest value in a numpy Matrix

There are lots of solutions to do this for a single array, but what about a matrix, such as:

>>> k
array([[ 35,  48,  63],
       [ 60,  77,  96],
       [ 91, 112, 135]])

You can use k.max(), but of course this only returns the highest value, 135. What if I want the second or third?

Upvotes: 39

Views: 94911

Answers (7)

SiyuK
SiyuK

Reputation: 1

To obtain the index of the 2nd (or nth) largest number of the array (my_array), you may use,

index_2 = np.argsort(my_array)[-2]

or, more generally,

index_n = np.argsort(my_array)[-n]

Now, to obtain the value for the nth largest element,

nth_largest_val = my_array[index_n]

Upvotes: 0

Alb
Alb

Reputation: 1320

Using the 'unique' function is a very clean way to do it, but likely not the fastest:

k = array([[ 35,  48,  63],
           [ 60,  77,  96],
           [ 91, 112, 135]])
i = numpy.unique(k)[-2]

for the second largest

Upvotes: 7

Luka Tikaradze
Luka Tikaradze

Reputation: 67

nums = [[ 35,  48,  63],
        [ 60,  77,  96],
        [ 91, 112, 135]]

highs = [max(lst) for lst in nums]
highs[nth]

Upvotes: 0

drsealks
drsealks

Reputation: 2354

Another way of doing this when repeating elements are presented in the array at hand. If we have something like 

a = np.array([[1,1],[3,4]])

then the second largest element will be 3, not 1.

Alternatively, one could use the following snippet:

second_largest = sorted(list(set(a.flatten().tolist())))[-2]

First, flatten matrix, then only leave unique elements, then back to the mutable list, sort it and take the second element. This should return the second largest element from the end even if there are repeating elements in the array.

Upvotes: 0

Lakshay Chawla
Lakshay Chawla

Reputation: 9

import numpy as np
a=np.array([[1,2,3],[4,5,6]])
a=a.reshape((a.shape[0])*(a.shape[1]))  # n is the nth largest taken by us
print(a[np.argsort()[-n]])

Upvotes: 0

serv-inc
serv-inc

Reputation: 38147

As said, np.partition should be faster (at most O(n) running time):

np.partition(k.flatten(), -2)[-2]

should return the 2nd largest element. (partition guarantees that the numbered element is in position, all elements before are smaller, and all behind are bigger).

Upvotes: 70

rofls
rofls

Reputation: 5115

You can flatten the matrix and then sort it:

>>> k = np.array([[ 35,  48,  63],
...        [ 60,  77,  96],
...        [ 91, 112, 135]])
>>> flat=k.flatten()
>>> flat.sort()
>>> flat
array([ 35,  48,  60,  63,  77,  91,  96, 112, 135])
>>> flat[-2]
112
>>> flat[-3]
96

Upvotes: 29

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