I have 2 large arrays with the exact same ammount of elements.
Array1=[[1,2,3][1,1,2]]
Array2=[[0,2,0][3,1,3]]
if Element in Array1="1", Replace "1" with whatever is in the same place as Array2
Output=[[0,2,3][3,1,2]]
Should be easy, but this late on a friday has my brains scrambled.
import numpy as np
Array1 = np.array([[1,2,3], [1,1,2]])
Array2 = np.array([[0,2,0], [3,1,3]])
b = np.where(Array1 == 1)
Array1[b] = Array2[b]
Result:
>>> Array1
array([[0, 2, 3],
[3, 1, 2]])
As jorgeca pointed out the above solution can be reduced to:
b = Array1 == 1
Array1[b] = Array2[b]
np.where (since Array1 == 1 returns a boolean mask that's already an index) or maybe use its three arguments' form.This one's based on Akaval's solution, but in one line. It takes advantage of other features of np.where():
import numpy as np
Array1 = np.array([[1,2,3], [1,1,2]])
Array2 = np.array([[0,2,0], [3,1,3]])
Output = np.where(Array1 == 1, Array2, Array1)
np.where() until your answer sent me to the documentation. This is actually pretty close to one of the examples on that page.