CODEWITHSUNDEEP
pythonmatrixnumpy
Sha
Sha

Reputation: 105

Filling a 2D Array (a Matrix)

Consider, for instance, the following formula:

$$f_{nm}(x) = \sin(n \ m \ x^2/\pi)$$

where $n$ and $m$ both are integers $\in [1,4]$. Forming a matrix with this information would result in:

$$\mathbf{F} = \begin{pmatrix}\ f_{11} & f_{12} & f_{13} & f_{14} \\ f_{21} & f_{22} & f_{23} & f_{24} \\ f_{31} & f_{32} & f_{33} & f_{34} \\ f_{41} &f_{42}&f_{43}&f_{44}\end{pmatrix}$$

How to create such a $4\times4$ matrix in Python and fill it with the provided formula for a given $x$ please. I know this question may be off-topic here, but the $LaTeX$ style of the equations did not work for me in the StackOverflow, that is why posted it here.

Upvotes: 0

Views: 89

Answers (2)

rdbisme
rdbisme

Reputation: 877

A faster implementation avoiding Python loops (if I understood your question correctly) that assumes that also x is an array from 0 to 1 and exploits Numpy Broadcasting

import numpy as np

n = np.arange(1, 5)
m = n.reshape(4, 1)
x = np.linspace(0, 1, 100).reshape(100, 1, 1)

result = np.sin(n * m * x**2 / np.pi)

print(result.shape)

This gives a result array of dimensions:

(100, 4, 4)

where the first dimensions indexes the x value. That is:

result[0, ...]

is the 4x4 matrix corresponding to x = x[0] = 0

Upvotes: 0

Zebx
Zebx

Reputation: 26

For a fixed $x$ you should be able to do this

import numpy as np

f = np.zeros((4,4))
for n in range(4):
    for m in range(4):
        f[n][m] = np.sin(n*m*(x**2)/np.pi)

Upvotes: 1

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