substitute each element of a vector into a matrix without using loop
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AllKindsofMath AllKinds
2013 年 11 月 19 日
回答済み: Alfonso Nieto-Castanon
2013 年 11 月 20 日
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.
0 件のコメント
採用された回答
Jan
2013 年 11 月 19 日
Like this?
vector = 1:1:10;
vector = reshape( vector, [1, 1, numel(vector)] );
one_vector = ones( 1, 1, numel(vector) );
matrix = [4 * vector, 5 * one_vector; 4 * one_vector, 2 * vector ];
5 件のコメント
Jan
2013 年 11 月 20 日
Not as far as I can see. You will probably have to iterate over the third dimension.
その他の回答 (3 件)
Sean de Wolski
2013 年 11 月 19 日
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 件のコメント
Jan
2013 年 11 月 19 日
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
2013 年 11 月 20 日
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
0 件のコメント
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