substitute each element of a vector into a matrix without using loop
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Hi I want to substitute each element of vector1 into 'x' in matrix1 and store each matrix in an array without using a loop. Please tell me how.
vector1=[1:1:10];
matrix1=[4*x 5*x ; 4*x 2*x];
Thanks in advance.
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Più risposte (3)
Sean de Wolski
il 19 Nov 2013
So:
vector1=[1:1:10];
x = vector1;
matrix1=[4*x 5*x ; 4*x 2*x];
Or is x symbolic?
clear x;
syms x
matrix1=[4*x 5*x ; 4*x 2*x];
matrix1 = subs(matrix1,x,vector1)
5 Commenti
AllKindsofMath AllKinds
il 19 Nov 2013
Sean de Wolski
il 19 Nov 2013
Modificato: Sean de Wolski
il 19 Nov 2013
i.e. the second half of what I have...
AllKindsofMath AllKinds
il 19 Nov 2013
Sean de Wolski
il 19 Nov 2013
subs(matrix1,x,vector1(4))
AllKindsofMath AllKinds
il 19 Nov 2013
Jan
il 19 Nov 2013
Maybe something like the following?
matrix = [4, 5; 4, 2];
[p, q] = size( matrix );
vector = 1:1:10;
matrix = repmat( matrix(:), 1, numel( vector ) );
matrix = matrix .* repmat( vector, p*q, 1 );
matrix = reshape( matrix, p, q, numel( vector ) );
This gives you a 3d matrix, where each layer contains the specified matrix, mulitplied by one entry in vector
Alfonso Nieto-Castanon
il 20 Nov 2013
perhaps something like:
f = @(x)[4*x 5 ; 4 2*x]; % Matrix in functional form
vector = 1:10; % Your vector of values for 'x'
matrix = arrayfun(f,vector,'uni',0); % A cell array of matrices
values = cellfun(@det,matrix); % Determinant of each of those matrices
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