Finding minimum value of 2d numpy array based on second value of row

Question

I have a NumPy array similar to as shown below and I want to find the minimum element based on the second item of the row. I can achieve this using the sorted function, is there a built-in NumPy function to achieve the same (probably the one-liner function)?

import numpy as np
a = np.array([[1,20], [2,3], [3,3], [4,3], [5,4],[1,2], [2,3], [3,3], [4,3], [5,4],[1,2], [2,3], [3,3], [4,3], [5,4], [1,2], [2,3], [3,3], [4,3], [5,4], [1,2], [2,3]])

min_ = sorted(a, key=lambda row: row[1])[0]

Thanks a lot in advance.

Does this work? numpy.org/doc/stable/reference/generated/… — whege
– whege, Commented May 10, 2022 at 15:47

Giovanni Tardini · Accepted Answer · 2022-05-10 15:49:01Z

1

You can also have it like

a[np.argmin(a[:, 1])]

answered May 10, 2022 at 15:49

Giovanni Tardini

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not_speshal · Accepted Answer · 2022-05-10 16:10:52Z

1

Try with argsort:

>>> a[a[:, 1].argsort()][0]
array([1, 2])

edited May 10, 2022 at 16:10

answered May 10, 2022 at 15:48

not_speshal

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3 Comments

stochastic learner Over a year ago

np.sort(a, axis=0) gives [[ 1, 2], [ 1, 2], [ 1, 2], [ 1, 2], [ 1, 3], [ 2, 3], [ 2, 3], [ 2, 3], [ 2, 3], [ 2, 3], [ 3, 3], [ 3, 3], [ 3, 3], [ 3, 3], [ 4, 3], [ 4, 3], [ 4, 3], [ 4, 4], [ 5, 4], [ 5, 4], [ 5, 4], [ 5, 20]]. Looks like it doesn't preserve the order.

stochastic learner Over a year ago

My point is. Your solution just works for just this example. Change a=np.array([[1,20], [2,3], [3,3], [4,3], [5,1],[1,2], [2,3], [3,3], [4,3], [5,4],[1,2], [2,3], [3,3], [4,3], [5,4], [1,2], [2,3], [3,3], [4,3], [5,4], [1,2], [2,3]]), which your np.sort(a, axis=0)[0] gives [1, 1] which is not even in a, is it?

not_speshal Over a year ago

@stochasticlearner - Aah, I see what you mean. Apologies, edited

Mahdi F. · Accepted Answer · 2022-05-10 16:51:34Z

1

You can use numpy.lexsort to sort multiple axes like the below:

>>> a[np.lexsort((a[:,1], a[:,0]))][0]
array([ 1,  2])

In python, sorts are guaranteed to be stable. If you have a value similar to the other values in the second column, maybe you get the wrong answer.

Ref:

Sort Stability and Complex Sorts

Sorts are guaranteed to be stable. That means that when multiple records have the same key, their original order is preserved.

>>> a = np.array([[2,1], [1,1]])
>>> a[np.argmin(a[:, 1])]
array([ 2,  1])

>>> a[a[:, 1].argsort()][0]
array([ 2,  1])

>>> a[np.lexsort((a[:,1], a[:,0]))][0]
array([ 1,  1])

edited May 10, 2022 at 16:51

answered May 10, 2022 at 16:07

Mahdi F.

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2 Comments

stochastic learner Over a year ago

since you are slicing twice, will it hurt the performance?

Mahdi F. Over a year ago

@stochasticlearner edited answer. It's better you consider the first column too.

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