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possibly this has been asked before, but I have a hard time finding a corresponding solution, since I can't find the right keywords to search for it.

One huge advantage of numpy arrays that I like is that they are transparent for many operations. However in my code I have a function that has a conditional statement in the form (minimal working example): import numpy as np

arr1 = np.array([1, 2, 3])
arr2 = np.array([1, 1, 3])

def func(x, y):
    if x > y:
        z = 1
    else:
        z = 2
    return z

func(arr1, arr2) obviously results in an error message: ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

I do understand what the problem here is and why it can't work like this. What I would like to do here is that x > y is evaluated for each element and then an array z is returned with the corresponding result for each comparison. (Needs to ensure of course that the arrays are of equal length, but that's not a problem here) I know that I could do this by changing func such that it iterates over the elements, but since I'm trying to improve my numpy skills: is there a way to do this without explicit iteration?

Berniyh
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  • `np.where(arr1>arr2, 1, 2)` – Sayandip Dutta Jul 28 '20 at 11:14
  • I think that `np.equal(arr1, arr2)` is what you're looking for here. It'll return `array([ True, False, True])` in this example. Here's the documentation: https://numpy.org/doc/stable/reference/generated/numpy.equal.html – amunnelly Jul 28 '20 at 11:17
  • Does this answer your question? [ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()](https://stackoverflow.com/questions/10062954/valueerror-the-truth-value-of-an-array-with-more-than-one-element-is-ambiguous) – Trenton McKinney Aug 11 '20 at 18:04
  • No, not really. The problem here is not the element-wise comparison (which is the topic of the linked question), but rather using this comparison to set different output values for each element. For this numpy.where was the appropriate answer as pointed out below. – Berniyh Aug 12 '20 at 08:44

1 Answers1

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arr1 > arr2 does exactly what you'd hope it does: compare each element of the two arrays and build an array with the result. The result can be used to index in any of the two arrays, should you need to. The equivalent of your function, however, can be done with np.where:

res = np.where(arr1 > arr2, 1, 2)

or equivalently (but slightly less efficiently), using the boolean array directly:

res = np.ones(arr1.shape)
res[arr1 <= arr2] = 2 # note that I have to invert the condition to select the cells where you want the 2
GPhilo
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  • np.where does indeed sound interesting. 1 and 2 are only placeholders, the actual statement is a bit more complicated, but that's irrelevant here, it works. :) Thanks! – Berniyh Jul 28 '20 at 11:30
  • Yes, I guessed the actual statement would be more elaborate, but the principle is the same :) – GPhilo Jul 28 '20 at 11:30