How to get the percentage of non Matching at particular Element value in an two arrays-CodePudding

I have an array like: import numpy as np

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

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

Requirement:

  where the elementvalue of 1 in an array 'a' is equal to element value of 1 in array 'b'.
  Need to access the percentage of matching only with element value 1,not at zero

I tried with:

  matching_error=(np.mean(a!=b)

Output:

   0.25
   # because out of 24 elements in an array 18 are matching(here 0,1 both the values are 
   in matching action), I need matching action at only element value 1

Required Output:

   0.83
   #because out of 24 elements, the element value is matching at 4 points

CodePudding user response：

Following your previous logic, you can do:

1-np.mean((a==b) & (a==1))  # 1 - mean(values are equal AND equal to 1)

Alternative:

np.mean((a!=b) | (a!=1))    # mean(values are different OR not equal to 1)

output: 0.8333333333333334

CodePudding user response：

Since your data is (implicitly) binary, i.e. only consisting of 0s and 1s, one could use

1 - np.mean(a b==2)
>0.8333333333333334

CodePudding user response：

Given the specified result, I believe what the question is attempting to ask is equivalent to this:

For what percentage of array positions is the item in at least one of the arrays equal to zero.

This can be calculated like this:

np.mean( (a==0) | (b==0) )

Output:

0.8333333333333334