(Answers Dev) Restored edit
the description below .. for loop and matrices
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Matthew Worker
am 26 Jun. 2021
Kommentiert: Rena Berman
am 16 Dez. 2021
% when o=1 I want to multiply the second row of H with the second column of W and sum it with the multiplication of the third row of H with the third column of W and the multiplication of the fourth row of H with the fourth column of W.
% when o=2 I want to multiply the first row of H with the first column of W and sum it with the multiplication of the third row of H with the third column of W and the multiplication of the fourth row of H with the fourth column of W.
% when o=3 I want to multiply the first row of H with the first column of W and sum it with the multiplication of the second row of H with the second column of W and the multiplication of the fourth row of H with the fourth column of W.
% and so on .....
clc;
clear;
H=[1 2 2 1 ; 3 1 1 2;1 3 2 4;2 1 3 5]
W=[4 1 2 1;1 3 2 1;2 1 1 3;2 1 1 4]
for o=1:4
p = abs(H(o,:)*W(:,o))
end
Akzeptierte Antwort
Walter Roberson
am 26 Jun. 2021
H=[1 2 2 1 ; 3 1 1 2;1 3 2 4;2 1 3 5]
W=[4 1 2 1;1 3 2 1;2 1 1 3;2 1 1 4]
locs = [2 2 3 3 4 4;
1 1 3 3 4 4;
1 1 2 2 4 4];
for o = 1 : size(locs,1)
p(o) = H(locs(o,1),:)*W(:,locs(o,2)) + H(locs(o,3),:)*W(:,locs(o,4)) + H(locs(o,5),:)*W(:,locs(o,6));
end
p
6 Kommentare
Walter Roberson
am 27 Jun. 2021
Suppose for discussion that H is 4 x 3, and W is 5 x 4
H=[1 2 2; 3 1 1;1 3 2;2 1 3]
W=[4 1 2 1;1 3 2 1;2 1 1 3;2 1 1 4; 1 2 4 5]
Now for the first iteration, you define
o = 1
p = abs(H(o,:)*W(:,o))
H(o,:) is 1 x 3 and W(:,o) is 5 x 1. But you cannot use * between a 1 x 3 and a 5 x 1.
Your definitions cannot work unless you define that W has the same number of rows that H has columns -- which you do not do.
If you had defined
p = abs(W(:,o)*H(o,:))
then that would work, giving a 5 x 3 result.
Which do you want? That each entry should be size(W,1) x size(H,2) ? Or that each entry should be 1 x 1, in which case you would have to define H as having the same number of columns as W has rows.
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