Hi I'd like to create a 10 x 5 matrix with the first column filled with 1 to 10 and the subsequent columns filled with 2, 3, 4 and 5 times the values of the first column.
I've made things work with the following code, but is there a shorter way to do this?
import numpy as np
mat = np.zeros([10,5])
mat[:,0] = np.arange(1,11)
mat[:,1] = np.dot(mat[:,0],2)
mat[:,2] = np.dot(mat[:,0],3)
mat[:,3] = np.dot(mat[:,0],4)
mat[:,4] = np.dot(mat[:,0],5)
I think you can achieve this by outer product.
Try:
import numpy as np
a = np.arange(1, 11).reshape(-1, 1) # column vector (1,2,3,...,10)
b = np.arange(1, 6).reshape(1, -1) # row vector (1,2,3,...,5)
np.matmul(a, b) # matrix of entries of multiplication of the indices (1-based indices)
or the one-liner:
np.arange(1, 11).reshape(-1, 1) * np.arange(1, 6).reshape(1, -1)
This works for me:
>>> np.array([np.array([1,2,3,4,5]) * i for i in range(1,11)])
array([[ 1, 2, 3, 4, 5],
[ 2, 4, 6, 8, 10],
[ 3, 6, 9, 12, 15],
[ 4, 8, 12, 16, 20],
[ 5, 10, 15, 20, 25],
[ 6, 12, 18, 24, 30],
[ 7, 14, 21, 28, 35],
[ 8, 16, 24, 32, 40],
[ 9, 18, 27, 36, 45],
[10, 20, 30, 40, 50]])
This is exactly what builtin numpy outer does:
>>> np.outer(np.arange(1, 11), np.arange(1, 6))
array([[ 1, 2, 3, 4, 5],
[ 2, 4, 6, 8, 10],
[ 3, 6, 9, 12, 15],
[ 4, 8, 12, 16, 20],
[ 5, 10, 15, 20, 25],
[ 6, 12, 18, 24, 30],
[ 7, 14, 21, 28, 35],
[ 8, 16, 24, 32, 40],
[ 9, 18, 27, 36, 45],
[10, 20, 30, 40, 50]])
np.arange(1,11)[:,None]*np.arange(1,6).