Difference between all elements in a row across two 2d NumPy arrays?

Question

I'm relatively new to programming/Python/NumPy, so I apologize if my ignorance is showing...

Say I have two 2d NumPy arrays:

import numpy as np

a = np.array([[1,   5,              20],
              [2,   15,             float("NaN")],
              [4,   float("NaN"),   float("NaN")]])
b = np.array([[4,   float("NaN")],
              [1,   13],
              [5,   float("NaN")]])

I would like to find the differences between all elements of the first row of each matrix, differences between all elements of the second row of each matrix, and so on. An example of a desired result given the above matrices is

[[  3.  nan  -1.  nan -16.  nan]  # <-- differences between all elements b[0] and a[0]
 [ -1.  11. -14.  -2.  nan  nan]  # <-- differences between all elements b[1] and a[1]
 [  1.  nan  nan  nan  nan  nan]] # <-- differences between all elements b[2] and a[2]

A way of obtaining the above result while using a loop is

outcome = []
for a_row, b_row in zip(a, b):
    outcome.append((b_row - a_row[:,None]).flatten())
outcome = np.array(outcome)

However, is there a way about this without looping through the rows that would be faster?

Unless your matrix elements follow some pattern you didn't disclose you will have to iterate through it — mont_
– mont_, Commented Oct 19, 2022 at 13:00

Guy · Accepted Answer · 2022-10-19 13:18:04Z

1

You can transpose the array to avoid the shape differences when substructing, and then ravel and reshape the results

arr = b.T[:, None] - a.T
np.ravel([arr[0], arr[1]], 'F').reshape((a.shape[0], a.shape[1] * b.shape[1]))

Output

[[  3.  nan  -1.  nan -16.  nan]
 [ -1.  11. -14.  -2.  nan  nan]
 [  1.  nan  nan  nan  nan  nan]]

answered Oct 19, 2022 at 13:18

Guy

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1 Comment

Vincent Cialdella Over a year ago

Thanks SO much! This is exactly what I was hoping for, but I clearly couldn't find my to this solution. I appreciate your time :)

learner · Accepted Answer · 2022-10-19 12:56:27Z

1

It is still using a loop, but in a list comprehension fashion:

np.array([(b_row - a_row[:,None]).flatten() for a_row, b_row in zip(a, b)])

answered Oct 19, 2022 at 12:56

learner

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