How to conditionally combine two numpy arrays of the same shape

Question

This sounds simple, and I think I'm overcomplicating this in my mind.

I want to make an array whose elements are generated from two source arrays of the same shape, depending on which element in the source arrays is greater.

to illustrate:

import numpy as np
array1 = np.array((2,3,0))
array2 = np.array((1,5,0))

array3 = (insert magic)
>> array([2, 5, 0))

I can't work out how to produce an array3 that combines the elements of array1 and array2 to produce an array where only the greater of the two array1/array2 element values is taken.

Any help would be much appreciated. Thanks.

Answer 1

We could use NumPy built-in np.maximum, made exactly for that purpose -

np.maximum(array1, array2)

Another way would be to use the NumPy ufunc np.max on a 2D stacked array and max-reduce along the first axis (axis=0) -

np.max([array1,array2],axis=0)

Timings on 1 million datasets -

In [271]: array1 = np.random.randint(0,9,(1000000))

In [272]: array2 = np.random.randint(0,9,(1000000))

In [274]: %timeit np.maximum(array1, array2)
1000 loops, best of 3: 1.25 ms per loop

In [275]: %timeit np.max([array1, array2],axis=0)
100 loops, best of 3: 3.31 ms per loop

# @Eric Duminil's soln1
In [276]: %timeit np.where( array1 > array2, array1, array2)
100 loops, best of 3: 5.15 ms per loop

# @Eric Duminil's soln2
In [277]: magic = lambda x,y : np.where(x > y , x, y)

In [278]: %timeit magic(array1, array2)
100 loops, best of 3: 5.13 ms per loop

---

Extending to other supporting ufuncs

Similarly, there's np.minimum for finding element-wise minimum values between two arrays of same or broadcastable shapes. So, to find element-wise minimum between array1 and array2, we would have :

np.minimum(array1, array2)

For a complete list of ufuncs that support this feature, please refer to the docs and look for the keyword : element-wise. Grep-ing for those, I got the following ufuncs :

add, subtract, multiply, divide, logaddexp, logaddexp2, true_divide, floor_divide, power, remainder, mod, fmod, divmod, heaviside, gcd, lcm, arctan2, hypot, bitwise_and, bitwise_or, bitwise_xor, left_shift, right_shift, greater, greater_equal, less, less_equal, not_equal, equal, logical_and, logical_or, logical_xor, maximum, minimum, fmax, fmin, copysign, nextafter, ldexp, fmod

Answer 2

If your condition ever becomes more complex, you could use np.where:

import numpy as np
array1 = np.array((2,3,0))
array2 = np.array((1,5,0))
array3 = np.where( array1 > array2, array1, array2)
# array([2, 5, 0])

You could replace array1 > array2 with any condition. If all you want is the maximum, go with @Divakar's answer.

And just for fun :

magic = lambda x,y : np.where(x > y , x, y)
magic(array1, array2)
# array([2, 5, 0])