My possible values are:
0: [0 0 0 0]
1: [1 0 0 0]
2: [1 1 0 0]
3: [1 1 1 0]
4: [1 1 1 1]
I have some values:
[[0.9539342 0.84090066 0.46451256 0.09715253],
[0.9923432 0.01231235 0.19491441 0.09715253]
....
I want to figure out which of my possible values this is the closest to my new values. Ideally I want to avoid doing a for loop and wonder if there's some sort of vectorized way to search for the minimum mean squared error?
I want it to return an array that looks like: [2, 1 ....
You can use np.argmin to get the lowest index of the rmse value which can be calculated using np.linalg.norm
import numpy as np
a = np.array([[0, 0, 0, 0], [1, 0, 0, 0], [1, 1, 0, 0],[1, 1, 1, 0], [1, 1, 1, 1]])
b = np.array([0.9539342, 0.84090066, 0.46451256, 0.09715253])
np.argmin(np.linalg.norm(a-b, axis=1))
#outputs 2 which corresponds to the value [1, 1, 0, 0]
As mentioned in the edit, b can have multiple rows. The op wants to avoid for loop, but I can't seem to find a way to avoid the for loop. Here is a list comp way, but there could be a better way
[np.argmin(np.linalg.norm(a-i, axis=1)) for i in b]
#Outputs [2, 1]