There is a loop for working with a vector, can it be made to work with a matrix?
z=zeros(100,900); %
x=z(:,1);
y=x;
j=1;
i=1;
u=1;
c=rand(1,900);
for i=1:100
j=j+1;
x(j)=x(i)+0.4*cos(20)/0.226*pi+c(1);
% in the c(1) loop this is because the loop is for the first column,
% in the matrix loop it should be c(num)
u=u+1
y(u)=y(i)+0.64*sin(78)/x(i)*pi+c(1);
end
I will be glad to any advice

4 Kommentare

Arif Hoq
Arif Hoq am 22 Feb. 2022
Bearbeitet: Arif Hoq am 22 Feb. 2022
what is value of "a" ? if i choose the variable "a" as a random number, then
a=randi(10,1,100);
z=zeros(100,900); %
x=z(:,1);
y=x;
j=1;
i=1;
u=1;
c=rand(1,900);
for i=1:100
j=j+1;
x(j)=x(i)+0.4*cos(20)/0.226*pi+c(1);
% in the c(1) loop this is because the loop is for the first column,
% in the matrix loop it should be c(num)
u=u+1;
y(u)=y(i)+0.64*sin(78)/a(i)*pi+c(1);
end
Lev Mihailov
Lev Mihailov am 22 Feb. 2022
the value of "a" is the value of "x", in future versions of the code, the variable "a" had to be abandoned
Dyuman Joshi
Dyuman Joshi am 22 Feb. 2022
Which variable(s) is/are supposed to be a matrix?
KSSV
KSSV am 22 Feb. 2022
It looks like you need not to use a loop here....tell us your problem, you can vectorize the code striaght away.

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DGM
DGM am 22 Feb. 2022
Bearbeitet: DGM am 22 Feb. 2022

0 Stimmen

You mean like this?
c = rand(1,900);
x = zeros(100,900);
y = x;
for k = 1:99
x(k+1,:) = x(k,:) + 0.4*cos(20)/0.226*pi + c;
y(k+1,:) = y(k,:) + 0.64*sin(78)./x(k,:)*pi + c;
end
... but I should ask how you say it's working when it's clearly dividing by zero and creating an entire array of Infs

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DGM
am 22 Feb. 2022

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