How to assign rows of np array according to rows of another array?

Question

I want to build an array of points where each row depends on the corresponding row of a timestep array as in:

steps = 5
output = np.empty((steps, 3))
timesteps = np.linspace(0, 1, steps)

for i, step in enumerate(timesteps):
    output[i] = [sin(step), cos(step), 0]

With the expected output:

[[0.         1.         0.        ]
 [0.24740396 0.96891242 0.        ]
 [0.47942554 0.87758256 0.        ]
 [0.68163876 0.73168887 0.        ]
 [0.84147098 0.54030231 0.        ]]

How can I vectorize this operation?

I could assign each column independently like so:

out[:, 0] = np.sin(timesteps)
out[:, 1] = np.cos(timesteps)
out[:, 2] = 0

But I'd like to maintain a row-based approach for readability, if possible.

I was thinking of something like

output[:] = [np.sin(timesteps[:]), np.cos(timestep[:]), 0]

but obviously that doesn't work due to setting an array element with a sequence.

The column assignment is the right way to do this. I don't see how that makes things any less readable. — Tim Roberts
– Tim Roberts, Commented Jun 8, 2021 at 18:09
@TimRoberts I suppose it hides the point=row relation a bit, but yeah, not a big issue. — JS Lavertu
– JS Lavertu, Commented Jun 8, 2021 at 18:18

Fonzie · Accepted Answer · 2021-06-08 18:41:19Z

you can create a numpy array in the following way:
np.array([np.sin(timesteps), np.cos(timesteps), [0]*len(timesteps)])

However this sets sines as the first line, cosines as second and so on. To change them from being lines to being columns you can use the transpose() method also accesible via the .T attribute. Finally you'd get:

output = np.array([np.sin(timesteps), 
                   np.cos(timesteps), 
                   [0]*len(timesteps)]).transpose()

or

output = np.array([np.sin(timesteps), np.cos(timesteps), [0]*len(timesteps)]).T

I find this close enough to what you tried and guess this is ok about the "readability" concern. To check that this is indeed what you want:

import numpy as np

steps = 5
output = np.empty((steps, 3))
timesteps = np.linspace(0, 1, steps)

for i, step in enumerate(timesteps):
    output[i] = [np.sin(step), np.cos(step), 0]

output -= np.array([np.sin(timesteps), np.cos(timesteps), [0]*len(timesteps)]).transpose()

print(output)

> [[0. 0. 0.]
  [0. 0. 0.]
  [0. 0. 0.]
  [0. 0. 0.]
  [0. 0. 0.]]

NB: [0]*len(timesteps) create a list of size len(timesteps) populated only with zeros

Yogesh Chander · Accepted Answer · 2021-06-08 19:04:41Z

1

I have something looks like this, can you try that ?

output[:] = [np.sin(timesteps[:]), np.cos(timestep[:]), 0]

Here is my solution:

import numpy as np
steps = 5
timesteps = np.linspace(0, 1, steps)

out=np.concatenate(([np.sin(timesteps)],[np.cos(timesteps)],[np.zeros(steps)]))

out=out.transpose()

out
array([[0.        , 1.        , 0.        ],
       [0.24740396, 0.96891242, 0.        ],
       [0.47942554, 0.87758256, 0.        ],
       [0.68163876, 0.73168887, 0.        ],
       [0.84147098, 0.54030231, 0.        ]])

I think this would be helpful, let me know any issue persists.

answered Jun 8, 2021 at 19:04

Yogesh Chander

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