Adding a smaller array to a larger array at a specified location (using variables)

Question

Suppose I have a 6x6 matrix I want to add into a 9v9 matrix, but I also want to add it at a specified location and not necessarily in a 6x6 block.

The below code summarizes what I want to accomplish, the only difference is that I want to use variables instead of the rows 0:6 and 3:9.

import numpy as np

a = np.zeros((9,9)) 
b = np.ones((6,6))

a[0:6,3:9] += b  #Inserts the 6x6 ones matrix into the top right corner of the 9x9 zeros

Now using variables:

rows = np.array([0,1,2,3,4,5])
cols = np.array([3,4,5,6,7,8])

a[rows,3:9] += b #This works fine
a[0:6,cols] += b #This also works fine
a[rows,cols += b #But this gives me the following error: ValueError: shape mismatch: value array of shape (6,6) could not be broadcast to indexing result of shape (6,)

I have spent hours reading through forums and trying different solutions but nothing has ever worked. The reason I need to use variables is because these are input by the user and could be any combination of rows and columns. This notation worked perfectly in MatLab, where I could add b into a with any combination of rows and columns.

a[rows[:,None], cols] += b, to index the rows with rows. 1-D arrays are different from matrices in MatLab. — Michael Szczesny
– Michael Szczesny, Commented Feb 17, 2022 at 9:36

Green 绿色 · Accepted Answer · 2022-02-17 09:32:02Z

1

Explanation:

rows = np.array([0,1,2,3,4,5])
cols = np.array([3,4,5,6,7,8])

a[rows,cols] += b

You could translate the last line to the following code:

for x, y, z in zip(rows, cols, b):
  a[x, y] = z

That means: rows contains the x-coordinate, cols the y-coordinate of the field you want to manipulate. Both arrays contain 6 values, so you effectively manipulate 6 values, and b must thus also contain exactly 6 values. But your b contains 6x6 values. Therefore this is a "shape mismatch". This site should contain all you need about indexing of np.arrays.

answered Feb 17, 2022 at 9:32

Green 绿色

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2 Comments

Ned Taylor Over a year ago

Thanks for the answer. Although I couldn't get the above solution to work. I read through the link and was able to get what I wanted with the np.ix_() function

Green 绿色 Over a year ago

Normally, @Michael Szczensny's answer should work for you: a[rows[:,None], cols] += b. At least, it does on your example.

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