using for loop for manipulating matrices

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0 个投票

Hello there, can anyone please help me out with this

Suppose, I have

x = [1,0.5,1,0.5,5]'
for i=1:2
for j=1:2
h(i,j) = x(5)/(2*pi*0.7^2) * exp(- ( (i-x(1))^2 + 
j-x(3))^2  )  /  (2*0.7^2)   );
end
end

That will create a 2x2 matrix called 'h'.

Now I have

x = [1,0.5,1,0.5,5 ; 2,0.5,1,0.5,10]'

and I need to calculate h for each column of x, so h would be a 2x(2xN) matrix, where N is the number of columns in 'x'

Can any one help me how to get this?

I tried

for k=1:N
for i=1:2
for j=1:2
h(i,j) = x(5,k)/(2*pi*0.7^2) * exp(- ( (i-x(1,k))^2 + 
j-x(3,k))^2  )  /  (2*0.7^2)   );
end
end
end

but it fails!

I tried

for k=1:N
for i=1:2
for j=1:2
h(i,j) = x(5,k)/(2*pi*0.7^2) * exp(- ( (i-x(1,k))^2 + 
j-x(3,k))^2  )  /  (2*0.7^2)   );
end
end
h1 = horzcat(h)
end

and it fails!

Please help mates!

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Image Analyst 2011-10-13

Here's some help:

http://www.mathworks.com/matlabcentral/answers/13205-tutorial-how-to-format-your-question-with-markup

Jan 2011-10-13

The code examples are not running due to a missing parenthesis.

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Answer 1

Jan 2011-10-13

在 MATLAB Online 中打开

0 个投票

What is a "2x(2xN) matrix"? I assume, you want an [2 x 2 x N] array.

a = 2*pi*0.7^2;
b = 2*0.7^2;
h = zeros(2, 2, N);   % Pre-allocate!
for k=1:N
  for i=1:2
    for j=1:2
      h(i,j,k) = x(5,k)/a * exp(-((i-x(1,k))^2 + j-x(3,k))^2) / b ???)???;
    end
  end
end

Your code exampled contains a not matching parenthesis. I've included it in question marks.

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PChoppala 2011-10-13

Here you go, the correct statement for 'h'

h(i,j) = x(5,k)/(2*pi*0.7^2) * exp(- ( (i-x(1,k))^2 + (j-x(3,k))^2 ) / (2*0.7^2) )

Matching parenthesis there! The last statement was not copied enough.

PChoppala 2011-10-13

h = nan(2,(2*N))

x = [1,0.5,1,0.5,5 ; 2,0.5,1,0.5,10]'

for k=1:N

k

for i=1:2

i

for j=1:2

j

h(i,j) = x(5,k)/(2*pi*0.7^2) * exp(- ( (i-x(1,k))^2 + (j-x(3,k))^2 ) / (2*0.7^2) )

end

%h1 = horzcat(h)

end

that's the entire code I tried, but fails to get the desired result!

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Answer 2

bym 2011-10-13

在 MATLAB Online 中打开

0 个投票

to what Jan has provided add:

h = reshape(h,2,[]);

after the for loops

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PChoppala 2011-10-13

h = nan(2,(2*N))

x = [1,0.5,1,0.5,5 ; 2,0.5,1,0.5,10]'

for k=1:N

k

for i=1:2

i

for j=1:2

j

h(i,j,k) = x(5,k)/(2*pi*0.7^2) * exp(- ( (i-x(1,k))^2 + (j-x(3,k))^2 ) / (2*0.7^2) )

end

%h1 = horzcat(h)

end

that's the entire code I tried, but fails to get the desired result!

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Answer 3

bym 2011-10-14

在 MATLAB Online 中打开

0 个投票

is this what you are after?

clc;clear
x = [1,0.5,1,0.5,5 ; 2,0.5,1,0.5,10]';
h = zeros(2,size(x,2));
for i=1:2 
    for j=1:2*size(x,2) 
        h(i,j) = x(5)/(2*pi*0.7^2)...
                 * exp(- ( (i-x(1))^2 + j-x(3))^2 )...
                 / (2*0.7^2);
    end
end
disp(h)

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PChoppala 2011-10-14

Hi, I think I am almost there, but new doubts crept in when trying your code.

Using

x = [1,0.5,1,0.5,5 ; 2,0.5,1,0.5,10]'

hArr = []

for k=1:N

k

for i=1:2

i

for j=1:2

j

h(i,j) = x(5,k)/(2*pi*0.7^2) * exp(- ( (i-x(1,k))^2 + (j-x(3,k))^2 ) / (2*0.7^2) )

end

hArr = [hArr h]

end

gives

hArr =

1.6240 0.5854 1.1708 0.4220

0.5854 0.2110 3.2481 1.1708

and...using

clc;

clear

x = [1,0.5,1,0.5,5 ; 2,0.5,1,0.5,10]';

h = zeros(2,size(x,2));

for i=1:2

for j=1:2*size(x,2)

% h(i,j) = x(5)/(2*pi*0.7^2)...

% * exp(- ( (i-x(1))^2 + j-x(3))^2 )...

% / (2*0.7^2);

h(i,j) = (x(5)/(2*pi*0.7^2)) * exp(- (((i-x(1))^2) + ((j-x(3))^2)) / (2*0.7^2) )

end

disp(h)

gives

h=

1.6240 0.5854 0.0274 0.0002

0.5854 0.2110 0.0099 0.0001

You may as well try for each column of 'x' separately to see the correct answer.

Please tell me if my method, the first one is correct or not!

bym 2011-10-14

well, I think you would be in the best position to judge whether it is correct or not, since I have only a partial idea of what you want to accomplish. If you want the 2x2 output of a column to be concatenated horizontally, then I would say you have achieved this.

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