Iterating through rows in python

Question

I have a (68x2) matrix named shape and I am trying to iterate through all the 68 rows by placing column 0 and column 1 of shape in array B. This is then multiplied by a (3x3) transformation matrix A. Then my intent was to create a single array (which is why I used np.append) but actually all I am getting are 68 singular 2 dimensional matrices and I do not know why.

Here is my code:

import numpy as np

for row in shape:
    B = np.array([[row[0]],[row[1]],[1]])
    result = np.matmul(A,B)
    result = np.append(result[0], result[1], axis = 0)
    print(result)

Anyone know how I can fix my problem?

If you need to iterate through rows, collect the results in a list, and turn that into an array once at the end. np.append is EVIL. — hpaulj
– hpaulj, Commented Mar 10, 2021 at 21:02
I created an empty list result = [] and at the end I added result.append(result) is that what you meant because I'm getting: AttributeError: 'numpy.ndarray' object has no attribute 'append' — Justin
– Justin, Commented Mar 10, 2021 at 21:10

Blckknght · Accepted Answer · 2021-03-10 23:06:48Z

2

You can concatenate a new column onto your shape array and then multiply all your rows by the transform matrix at once using a single matrix multiplication.

result = (np.concatenate((shape, np.ones((68, 1))), axis=1) @ A)[:,:2]

It's possible you need to multiply by the transpose of the transformation matrix, A.T, rather than by A itself.

answered Mar 10, 2021 at 23:06

Blckknght

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1 Comment

Justin Over a year ago

Thank you this worked wonders. For my case I did not need to use the transpose of A

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