Matlab匿名函数如何评估矩阵？

Question

Given the following matlab code:

A = reshape (1:9 ,[3 3]);
f = @(x) x^2;
f(A)

the output is:

A =                                                                                                                                                                                                                                                    

   1   4   7                                                                                                                                                                                                                                           
   2   5   8                                                                                                                                                                                                                                           
   3   6   9  

ans =                                                                                                                                                                                                                                                  

    30    66   102                                                                                                                                                                                                                                     
    36    81   126                                                                                                                                                                                                                                     
    42    96   150 

Can you please explain to me how this output is calculated? how the anonymous function is invoked?

I want to write the same program in python.

Is there any compatible function in python which lets me implement this?

Answer 1

An anonymous function with a matrix as an input behaves just as the contents normally would if executed from the command line. It does nothing special.

The issue is that you're using ^ which is the matrix power ( mpower ) and I believe you are looking for the element-wise power ( power ) which is .^ which in your case would square each element individually.

f = @(x)x.^2;
B = f(A);

If you want to do the same in Python, you can do exactly this with numpy

import numpy as np

A = np.arange(1, 10).reshape(3, 3).T
B = np.power(A, 2)

If you want to encaptulate this operation, it's best pratice to put it into it's own function

def f(x):
    return np.power(x, 2)

B = f(A)

If you named lambda (anonymous) function ( this isn't recommended ) you can do:

f = lambda x: np.power(x, 2)
B = f(A)