如何正确比较3个不同的numpy数组的元素？

Question

I'm trying to compare elements at the same index from 3 different arrays. When I try if arr1[i] == arr2[i] I get the The truth value of an array with more than one element is ambiguous. Use a.any() or a.all() The truth value of an array with more than one element is ambiguous. Use a.any() or a.all() . Here's the whole function:

def tmr(arr1, arr2, arr3):
arr4 = arr1
for i in range(arr4.size):
    if arr1[i] == arr2[i]:
        arr4[i] = arr1[i]
    else:
        if arr2[i] == arr3[i]:
            arr4[i] = arr3[i]
return arr4

I'm more familiar with C++ than Python and I really can't see why this doesn't exactly work. I also tried using zip like this:

for w, x, y, z in zip(arr4, arr1, arr2, arr3):
    if x == y == z:
        w = x

Answer 1

Interpreting this as for the intended result in your last example using zip , try:

arr4 = arr1[np.equal(arr1, arr2) & np.equal(arr2, arr3)]

For the interpretation in your first code block you can use list comprehension:

list4 = [arr1[i] if arr1[i] == arr2[i] else arr3[i] if arr2[i] == arr3[i] else None for i in range(len(arr1))]
arr4 = np.array(list4)

Based on your example I'm not sure the default values for arr4 if neither arr1[i] == arr2[i] nor arr2[i] == arr3[i] , so have left them as None above.

The two approaches provide different answers, but if I'm interpreting correctly, the first is the desired behaviour.

Answer 2

numpy array functions np.equal , np.logical_and , np.where - vectorized/broadcasting

import numpy as np
arr1 = np.array([0,1,2,3,7])
arr2 = np.array([0,1,0,2,7])
arr3 = np.array([0,0,2,1,7])

arr4 = np.ones(5)*10


eq_idx = np.where(np.logical_and(np.equal(arr1, arr3), np.equal(arr2, arr3)))

arr4[eq_idx] = arr1[eq_idx]

arr4
Out[28]: array([ 0., 10., 10., 10.,  7.])