Given an array of dimension N how do I divide all values in the array by the first value from a selected dimension?
Example code:
import numpy as np
A = np.random.randint(1, 10, size=(3,3,3))
B = A[:,:,0]
C = np.divide(A,B)
A
print()
B
print()
C
print()
C[:,:,0]
Output:
array([[[1, 8, 5],
[3, 6, 5],
[5, 4, 2]],
[[6, 2, 9],
[4, 2, 2],
[5, 6, 8]],
[[3, 3, 1],
[2, 7, 7],
[6, 4, 6]]])
array([[1, 3, 5],
[6, 4, 5],
[3, 2, 6]])
array([[[1. , 2.66666667, 1. ],
[0.5 , 1.5 , 1. ],
[1.66666667, 2. , 0.33333333]],
[[6. , 0.66666667, 1.8 ],
[0.66666667, 0.5 , 0.4 ],
[1.66666667, 3. , 1.33333333]],
[[3. , 1. , 0.2 ],
[0.33333333, 1.75 , 1.4 ],
[2. , 2. , 1. ]]])
array([[1. , 0.5 , 1.66666667],
[6. , 0.66666667, 1.66666667],
[3. , 0.33333333, 2. ]])
I was expecting the final output from C[:,:,0] to be all 1's. I guess it has to do with the broadcasting of B but I don't think I understand why it isn't broadcasting B into a shape (3,3,3) where it is replicated along dimension 2.
To get your expected results you could reshape your B
array to:
B = A[:,:,0].reshape(3,-1, 1)
Then when you divide you will get a result like:
array([[[1. , 0.11111111, 0.11111111],
[1. , 0.25 , 0.5 ],
[1. , 0.88888889, 0.44444444]],
[[1. , 0.88888889, 1. ],
[1. , 1.8 , 1.6 ],
[1. , 4.5 , 0.5 ]],
[[1. , 0.66666667, 0.5 ],
[1. , 1.125 , 0.75 ],
[1. , 0.5 , 2.25 ]]])
You could also maintain the proper dimension for broadcasting by taking B
as:
B = A[:,:,0:1]