How to compare two numpy arrays with multiple condition

Question

I have 2 NumPy arrays like the below

array_1 = np.array([1.2, 2.3, -1.0, -0.5]) 
array_2 = np.array([-0.5, 1.3, 2.5, -0.9])

We can do the element-wise simple arithmetic calculation (addition, subtraction, division etc) easily using different np functions

array_sum = np.add(array_1, array_2)
print(array_sum) # [ 0.7  3.6  3.5 -0.4]  

array_sign = np.sign(array_1 * array_2)
print(array_sign) # [-1.  1.  1. -1.]

However, I need to check element-wise multiple conditions for 2 arrays and want to save them in 2 new arrays (say X and Y).

For example, if both elements contain different sign (e.g.: 1st and 3rd element pairs of the given example)) then, X will contain 0 and Y will be the sum of the poitive element and abs(negative element)

X = [0]
Y = [1.7]

When both elements are positive (e.g.: 2nd element pair of the given example) then, X will contain the lower value and Y will contain the greater value

X = [1.3]
Y = [2.3]

If both elements are negative, then, X will be 0 and Y will be the sum of the abs(negative element) and abs(negative element)

So, the final X and Y will be something like

X = [0, 1.3, 0, 0]
Y = [1.7, 2.3, 3.5, 1.4]

I have gone through some posts (this, and this) that described, the comparison procedures between 2 arrays, but not getting idea for multiple conditions. Here, 2 arrays are very small but, my real arrays are very large (e.g.: contains 2097152 element per array).

Any ideas are highly appreciated.

To simplify a bit, if EITHER element is negative, then X is 0 and Y is the sum of the abs.. Otherwise, X is min and Y is max. Right? — Tim Roberts
– Tim Roberts, Commented Mar 10, 2022 at 19:59

not_speshal · Accepted Answer · 2022-03-10 20:07:33Z

1

Try with numpy.select:

conditions = [(array_1>0)&(array_2>0), (array_1<0)&(array_2<0)]
choiceX = [np.minimum(array_1, array_2), np.zeros(len(array_1))]
choiceY = [np.maximum(array_1, array_2), -np.add(array_1,array_2)]

X = np.select(conditions, choiceX)
Y = np.select(conditions, choiceY, np.add(np.abs(array_1), np.abs(array_2)))

>>> X
array([0. , 1.3, 0. , 0. ])

>>> Y
array([1.7, 2.3, 3.5, 1.4])

answered Mar 10, 2022 at 20:07

not_speshal

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2 Comments

Opps_0 Over a year ago

would you mind giving some hints to get -min(abs(array_1), abs(array_2)) instead of 0for the X?

Opps_0 Over a year ago

I have used this -np.minimum(np.abs(array_1), np.abs(array_2)) in choiceX instead of np.zeros(len(array_1)) and as an additional result in the X and working for the demo data. If I need to change or I am doing wrong, please, let me know.

Tim Roberts · Accepted Answer · 2022-03-10 20:05:38Z

1

This will do it. It does require vertically stacking the two arrays. I'm sure someone will pipe up if there is a more efficient solution.

import numpy as np

array_1 = np.array([1.2, 2.3, -1.0, -0.5]) 
array_2 = np.array([-0.5, 1.3, 2.5, -0.9]) 

def pick(t):
    if t[0] < 0 or t[1] < 0:
        return (0,abs(t[0])+abs(t[1]))
    return (t.min(), t.max())

print( np.apply_along_axis( pick, 0, np.vstack((array_1,array_2))))

Output:

[[0.  1.3 0.  0. ]
 [1.7 2.3 3.5 1.4]]

The second line of the function can also be written:

        return (0,np.abs(t).sum())

But since these will only be two-element arrays, I doubt that saves anything at all.

answered Mar 10, 2022 at 20:05

Tim Roberts

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1 Comment

Tim Roberts Over a year ago

The select answer is almost certainly better, because it spends less time in Python code.

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