如何计算两个 numpy arrays 之间匹配零元素的数量？

Question

I have two numpy arrays of equal size. They contain the values 1 , 0 , and -1 . I can count the number of matching ones and negative ones, but I'm not sure how to count the matching elements that have the same index and value of zero.

I'm a little confused on how to proceed here.

Here is some code:

print(actual_direction.shape)
print(predicted_direction.shape)
act = actual_direction
pre = predicted_direction
part1 = act[pre == 1]
part2 = part1[part1 == 1]
result1 = part2.sum()
part3 = act[pre == -1]
part4 = part3[part3 == -1]
result2 = part4.sum() * -1
non_zeros = result1 + result2
zeros = len(act) - non_zeros
print(f'zeros : {zeros}\n')
print(f'non_zeros : {non_zeros}\n')
final_result = non_zeros + zeros
print(f'result1 : {result1}\n')
print(f'result2 : {result2}\n')
print(f'final_result : {final_result}\n')

Here is the printout:

(11279,)
(11279,)
zeros : 5745.0
non_zeros : 5534.0
result1 : 2217.0
result2 : 3317.0
final_result : 11279.0

So what I've done here is simply subtract the summation of the ones and negative ones from the total length of the array. I can't assume that the difference (zeros: 5745) contains ALL matching elements that contain zeros can I?

Answer 1

You could try this:

import numpy as np

a=np.array([1,0,0,1,-1,-1,0,0])
b=np.array([1,0,0,1,-1,-1,0,1])
summ = np.sum((a==0) & (b==0))
print(summ)

Output:

3

Answer 2

You can use numpy.ravel() to flatten out the array, then use zip() to compare each element side by side:

import numpy as np

ar1 = np.array([[1, 0, 0],
                [0, 1, 1],
                [0, 1, 0]])

ar2 = np.array([[0, 0, 0],
                [1, 0, 1],
                [0, 1, 0]])

count = 0
for e1, e2 in zip(ar1.ravel(), ar2.ravel()):
    if e1 == e2:
        count += 1

print(count)

Output:

6

---

You can also do this to list all the matches found, as well as print out the amount:

dup = [e1 for e1, e2 in zip(ar1.ravel(), ar2.ravel()) if e1 == e2]
print(dup)
print(len(dup))

Output:

[0, 0, 1, 0, 1, 0]
6

Answer 3

You have two arrays and want to count the positions where both of these are 0, right?

You can check where the array meets your required condition (a == 0) , and then use the 'and' operator & to check where both arrays meet your requirement:

import numpy as np
a = np.array([1, 0, -1, 0, -1, 1, 1, 1, 1])
b = np.array([1, 0, -1, 1, 0, -1, 1, 0, 1])

both_zero = (a == 0) & (b == 0)  # [False,  True, False, False, False, False]

both_zero.sum()  # 1

---

In your updated question you appear to be interested in the similarities and differences between actual values and predictions. For this, a confusion matrix is ideally suited.

from sklearn.metrics import confusion_matrix
confusion_matrix(a, b, labels=[-1, 0, 1])

will give you a confusion matrix as output telling you how many -1s were predicted as -1, 0 and 1, and the same for 0 and +1:

[[1 1 0]   # -1s predicted as -1, 0 and 1
 [0 1 1]   # 0s predicted as -1, 0 and 1
 [1 1 3]]  # 1s predicted as -1, 0 and 1