I have a 2 x 2
numpy.array()
matrix
, and an array N x 2
X
, containing N
2-dimensional vectors.
I want to multiply each vector in X
by the 2 x 2
matrix. Below I use a for loop, but I am sure there is a faster way. Please, could someone show me what it is? I assume there is a way using a numpy
function.
# the matrix I want to multiply X by
matrix = np.array([[0, 1], [-1, 0]])
# initialize empty solution
Y = np.empty((N, 2))
# loop over each vector in X and create a new vector Y with the result
for i in range(0, N):
Y[i] = np.dot(matrix, X[i])
For example, these arrays:
matrix = np.array([
[0, 1],
[0, -1]
])
X = np.array([
[0, 0],
[1, 1],
[2, 2]
])
Should result in:
Y = np.array([
[0, 0],
[1, -1],
[2, -2]
])
One-liner is (matrix @ X.T).T
Just transpose your X, to get your vectors in columns. Then matrix @ X.T
or (np.dot(matrix, X.T)
if you prefer this solution, but now that @
notation exists, why not using it) is a matrix made of columns of matrix times X[i]
. Just transpose back the result if you need Y to be made of lines of results
matrix = np.array([[0, 1], [-1, 0]])
X = np.array([[1,2],[3,4],[5,6]])
Y = (matrix @ X.T).T
Y is
array([[ 2, -1],
[ 4, -3],
[ 6, -5]])
As expected, I guess.
In detail:
X
is
array([[1, 2],
[3, 4],
[5, 6]])
so X.T
is
array([[1, 3, 5],
[2, 4, 6]])
So, you can multiply your 2x2 matrix by this 2x3 matrix, and the result will be a 2x3 matrix whose columns are the result of multiplication of matrix
by the column of this. matrix @ X.T
is
array([[ 2, 4, 6],
[-1, -3, -5]])
And transposing back this gives the already given result.
So, tl;dr: one-liner answer is (matrix @ X.T).T