How can I separate a vector into basis vectors?

Question

I have the basis:

>>> m = 6
>>> basis = np.zeros((m, m))
>>> np.fill_diagonal(basis, 1)
array([[1. 0. 0. 0. 0. 0.]
       [0. 1. 0. 0. 0. 0.]
       [0. 0. 1. 0. 0. 0.]
       [0. 0. 0. 1. 0. 0.]
       [0. 0. 0. 0. 1. 0.]
       [0. 0. 0. 0. 0. 1.]]

How can I access the superposition of a vector? Let's say from:

vec = [0, 1, 1, 0, 1, 0]

So I want the output:

array([[0. 1. 0. 0. 0. 0.]
       [0. 0. 1. 0. 0. 0.]
       [0. 0. 0. 0. 1. 0.]]

So for each 1 you want a new row with `1` at the position that is the position in the `vec`? — Willem Van Onsem, Jul 31 '21 at 20:07

Willem Van Onsem · Accepted Answer · 2021-07-31T20:22:57.077

1

We can construct a diagonal with the number of elements of vec, and then filter the rows, so:

>>> vec = np.array([0,1,1,0,1,0])
>>> np.identity(6)[vec.astype(bool)]
array([[0., 1., 0., 0., 0., 0.],
       [0., 0., 1., 0., 0., 0.],
       [0., 0., 0., 0., 1., 0.]])

This is however not very efficient if the number of items in the vec is large, since constructing an identity matrix takes quadratic time in the length of vec.

We can also work with .diag(…) as @AlexAlex says:

>>> vec = np.array([0,1,1,0,1,0])
>>> np.diag(vec)[vec.astype(bool)]
array([[0, 1, 0, 0, 0, 0],
       [0, 0, 1, 0, 0, 0],
       [0, 0, 0, 0, 1, 0]])

Here we can specify different values, for example:

>>> vec = np.array([0,1,2,0,3,0])
>>> np.diag(vec)[vec.astype(bool)]
array([[0, 1, 0, 0, 0, 0],
       [0, 0, 2, 0, 0, 0],
       [0, 0, 0, 0, 3, 0]])

edited Jul 31 '21 at 20:22

answered Jul 31 '21 at 20:10

Willem Van Onsem

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or this: np.diag(vec) – Alex Alex Jul 31 '21 at 20:15
I think the np.diag(vec) would be more efficient but it gives me an m x m matrix with rows which are just 0. Can i somehow get rid of them? – Luxamba Jul 31 '21 at 20:20
@Luxamba: yes, by filtering with the `[vec.astype(bool)]` :) – Willem Van Onsem Jul 31 '21 at 20:21

How can I separate a vector into basis vectors?

1 Answers1