profile of circular image - more efficient way to do it? - matlab

Question

I need to get a 1-D profile of a circular image, for example 256x256 sin(R) image
I've written a matlab function for the task but it turns out to be very un-efficient.
the function averages over radius intervals of the original images.
matlab profiler reveals that the first line in the for-loop [indxs=find(...)]
takes ~86% of the running time.
i need to run the function on a some thousands of simulated images (some larger then 256x256) and it takes very long time to complete.
does anyone knows how can i make this code run faster?
maybe someone has another, more efficient way to do the task??
i also tried to convert to function into C++ & mex file using matlab coder
but it took longer (x3) to perform the task, might be because the sub-function- "findC"
uses some 2D-ffts to find the center of the image.
Thanks you All,
Dudas
My Matlab function:
function [sig R_axis Center]= Im2Polar (imR,ch,Center_Nblock)
% Converts Circular image to 1-D sig
% based on true image values w/o interpolation
% Input -
% imR - circular sinuns image
% ch - number of data-points in output signal (sig)
% Center_Nblock - a varible related to the image center finding method
% Output -
% sig - 1D vector of the circular image profile
% R_axis - axis data-points for sig
% Center - image center in pixels
[Mr Nr] = size(imR); % size of rectangular image
[Center]=findC(imR,Center_Nblock);
Xc=Center(1);
Yc=Center(2);
rMax=sqrt((Mr/2)^2 + (Nr/2)^2);
x=[0:1:Mr-1]-Xc+1;
y=[0:1:Nr-1]-Yc+1;
[X,Y]=meshgrid(x,y);
[TH,R] = cart2pol(X,Y);
% Assembling 1-D signal
sig=single([]);
ii=1;
dr=floor(rMax)/ch;
V=dr:dr:floor(rMax);
for v=V
indxs=find((v-dr)<=R & R<v);**
sig(ii)=mean(imR(indxs));
Nvals(ii)=length(indxs);
ii=ii+1;
end %for v
R_axis=V-dr/2;
end % of function

Answer 1

Following from the comments here's an example of something I might try. Let's work with a 9x9 example. Suppose you have the following annulus.
A =
0 0 0 0 0 0 0 0 0
0 0 1 1 1 1 1 0 0
0 1 1 1 0 1 1 1 0
0 1 1 0 0 0 1 1 0
0 1 0 0 0 0 0 1 0
0 1 1 0 0 0 1 1 0
0 1 1 1 0 1 1 1 0
0 0 1 1 1 1 1 0 0
0 0 0 0 0 0 0 0 0
Then the indices of your sort of mask are, lets say [k n]
>> [k n]
ans =
3 2
4 2
5 2
6 2
7 2
2 3
3 3
4 3
6 3
7 3
8 3
2 4
3 4
7 4
8 4
2 5
8 5
2 6
3 6
7 6
8 6
2 7
3 7
4 7
6 7
7 7
8 7
3 8
4 8
5 8
6 8
7 8
Now have a 9x9 matrix of zeroes on hand called B, we can shift the whole thing over to the left by one pixel as follows using the formula (i+9*(j-1)) to convert double index to a single index.
>> B=zeros(9,9);
>> B((k)+9*(n-2))=1
B =
0 0 0 0 0 0 0 0 0
0 1 1 1 1 1 0 0 0
1 1 1 0 1 1 1 0 0
1 1 0 0 0 1 1 0 0
1 0 0 0 0 0 1 0 0
1 1 0 0 0 1 1 0 0
1 1 1 0 1 1 1 0 0
0 1 1 1 1 1 0 0 0
0 0 0 0 0 0 0 0 0
Or move down and to the right as follows
>> B=zeros(9,9);
>> B((k+1)+9*(n-0))=1
B =
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 1 1 1 1 1 0
0 0 1 1 1 0 1 1 1
0 0 1 1 0 0 0 1 1
0 0 1 0 0 0 0 0 1
0 0 1 1 0 0 0 1 1
0 0 1 1 1 0 1 1 1
0 0 0 1 1 1 1 1 0
As long as it doesn't go out of bounds you should be able to shift a single annular mask around with a simple addition to put the center at the image center.