Python: finding index of an array under several conditions

Question

I have the following problem. There are two n-dimensional arrays of integers and I need to determine the index of an item that fulfills several conditions.

The index should have a negative element in "array1".
Of this subset with negative elements, it should have the smallest value in "array2".
In case of a tie, select the value that has the smallest value in "array1" (or the first otherwise)

So suppose we have:

array1 = np.array([1,-1,-2])
array2 = np.array([0,1,1])

Then it should return index 2 (the third number). I'm trying to program this as follows:

import numpy as np
n = 3
array1 = np.array([1,-1,-2])
array2 = np.array([0,1,1])
indices = [i for i in range(n) if array1[i]<0] 
indices2 = [i for i in indices if array2[i] == min(array2[indices])] 
index = [i for i in indices2 if array1[i] == min(array1[indices2])][0] #[0] breaks the tie.

This seems to work, however, I don't find it very elegant. To me it seems like you should be able to do this in one or two lines and with defining less new variables. Anyone got a suggestion for improvement? Thanks in advance.

your arrays seems to be 1-dimensional and of size n, not n-dimensional — njzk2
– njzk2, Commented May 16, 2013 at 15:06
I usually use the terms vector and array interchangeably, which is probably the wrong thing to do :). I meant n-dimensional vectors. — Forzaa
– Forzaa, Commented May 16, 2013 at 16:19

Hensing · Accepted Answer · 2013-05-16 15:39:22Z

6

you can get the indices of all negative elements from array1 with:

np.where(array1 < 0)

then you can access the subset via:

subset = array2[array1 < 0]

to get the index of the smallest (negative) value of array1, you can use array1.argmin()

value = array2[array1.argmin()]

putting all together gives you:

value = array2[array1 < 0][array1[array1 < 0].argmin()]

but you have to catch ValueErrors, if array1 has only positive values.

edited May 16, 2013 at 15:39

answered May 16, 2013 at 15:08

Hensing

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2 Comments

Forzaa Over a year ago

Yea, this is closest to what I had and quite easy to understand. I will only execute this part when there is a negative element, btw, so don't worry. The np.where function is very useful... I hadn't found out yet. Thanks a lot! I did accept the other answers as best since it is more intricate.

Hensing Over a year ago

slicing an array with bool-masks is faster than np.where array1[array1 < 0] is much faster than array1[np.where(array1 < 0)] so try not to use np.where, when you don't want the index!

njzk2 · Accepted Answer · 2013-05-16 15:18:04Z

3

I don't know much about numpy (though apparently i should really look into it), so here is a plain python solution

This

sorted([(y, x, index) for (index, (x, y)) in enumerate(zip(array1, array2)) if x < 0])

will give you the tripley of elements from array2, array1, index sorted by value in array2 and value in array1 in case of tie, index in case of tie

The first element is what you seek. This gives the following result :

[(1, -2, 2), (1, -1, 1)]

The index is therefore 2, and is obtained by [0][2]

answered May 16, 2013 at 15:18

njzk2

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1 Comment

Forzaa Over a year ago

This almost looks like magic to me! Very elegant. I don't completely understand why this works yet, though, since I'm not 100% familiar with python yet. I will find out though :)

elyase · Accepted Answer · 2013-05-16 15:43:01Z

2

The index should have a negative element in "array1":

indices_1 = np.argwhere(array1 < 0)

Of this subset with negative elements, it should have the smallest value in "array2".

indices_2 = array2[indices_1].argmin()

In case of a tie, select the value that has the smallest value in "array1" (or the first otherwise)

array1[indices_2].argmin()

edited May 16, 2013 at 15:43

answered May 16, 2013 at 15:36

elyase

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1 Comment

Forzaa Over a year ago

argmin() only returns one element for me... which is why I refrained from using it.

Sukrit Kalra · Accepted Answer · 2013-05-16 15:36:44Z

1

I came up with this. Just devised one more test case and seemed to work. See if it suits your needs.

sorted([(array2[index], element, index) for index, element in enumerate(array1) if element<0])[0][2]

Or through using the min function.

min([(array2[index], element, index) for index, element in enumerate(array1) if element<0])[2]

edited May 16, 2013 at 15:36

answered May 16, 2013 at 15:27

Sukrit Kalra

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Comments

unutbu · Accepted Answer · 2013-05-16 15:36:46Z

The following will work with numpy arrays of any dimension. It uses numpy.lexsort to order the indices.

numpy.lexsort(Y,X) sorts the items in X in ascending order, and breaks ties according to the values in Y. It returns the indices in order (not the values of X or Y.) This is the only function I know of in numpy which "breaks ties" for you.

import numpy as np
def find_index(array1, array2):
    indices = np.where(array1 < 0)
    i = np.lexsort((array1[indices],array2[indices]))[0]
    return [idx[i] for idx in indices]

array1 = np.array([1,-1,-2])
array2 = np.array([0,1,1])

array3 = np.array([(1, 2),
                   (-1, -2),
                   (-2, 0)])
array4 = np.array([(0, 2),
                   (1, 1),
                   (3, 0)])

print(find_index(array1, array2))
# [2]

print(find_index(array3, array4))
# [1, 1]

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