Manipulating Matrix elements based on their index without using a for loop.

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Noah Chrein
Noah Chrein 2015-6-18
编辑： Tim 2015-6-18
Can I do simple manipulations on each matrix element based on its index without having to run a for loop?
for example, say I want to take the inscribed circular region of diameter n in an n by n matrix A, and set everything else to zero, I would then do
for i = 1:n
for j = 1:n
if ((i-n/2)^2+(j-n/2)^2>n) A(i,j) = 0; end;
end
end
for large n, this becomes a very high memory problem.
the same thing with a square can be done easily, say the matrix A is some arbitrary size and I want the first m by n matrix, I could just do this
A(1:m,1:n)
I am thinking there is some syntax like
A([logical operation based on index])
I know for just value, this can be done with a logical operator directly on the matrix
A<1
returns all the positions of elements less than 1

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采纳的回答

Azzi Abdelmalek
Azzi Abdelmalek 2015-6-18
[jj,ii]=meshgrid(1:n,1:n);
A((ii-n/2).^2+(jj-n/2).^2>n)=0
  1 个评论
Tim
Tim 2015-6-18
This doesn't quite work, I think you meant something like this:
n=100
[jj,ii]=meshgrid(1:n,1:n);
A = (sqrt((ii-n/2).^2 + (jj-n/2).^2)< n/2)*100
image(A)

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更多回答（1 个）

Tim
Tim 2015-6-18
编辑：Tim 2015-6-18
Try this.
clear all;
%We want to inscribe a circle of diameter n on an nxn matrix A.
%More generally we want an index based method for assinging
%values to a matrix without having to run a calculation on each element.
%Presumably we want the circle center in the middle. My plan of attack is
%to generate a list of index pairs first and then assign it to the matrix.
%Matrix row/column and coincidentally circle diameter
n=95;
%For efficiency, instantiate the matrix A
A = zeros(n+1);
%Use a paramterized function to generate the i and j indices to appropriate
%resolution(This will change based on how big n is; higher n, more
%resolution required in this step.)
%We must use linear indexing to assign scattered values so calculate a new
%indices vector which gives a numeric value for the element number of each
%index pair i.e. A(1,1) = 1 and A(n,n) would be like 10,000 for n=100.
%luckily we have the sub2ind function which does this for us.
t=0:.01:2*pi;
idx = sub2ind(size(A), int16((n/2).*sin(t))+n/2, int16((n/2).*cos(t))+n/2);
%assign values to your generated indices in A
A(idx)=100;
%Check your work
image(A);
This method should work for any shape you could parameterize or define in a vector form. See my comment on Azzi's answer for the way to generate a filled in shape.

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