Thread Subject: locating colored pixels in an image

I have an image (3000x2000 pixels). I want to scan each
column at a time and look for the red pixels (R>200). Most
pixels are just black. When it finds a red pixel, I want
it to search forward and find how many consecutive red
pixels (R>200) there are. Any ideas?

In article <fa4tcl$69q$1@fred.mathworks.com>, Craig <hillx154@umn.edu> wrote:
>I have an image (3000x2000 pixels). I want to scan each
>column at a time and look for the red pixels (R>200). Most
>pixels are just black. When it finds a red pixel, I want
>it to search forward and find how many consecutive red
>pixels (R>200) there are. Any ideas?

There are many ways of doing this. This is just a small variant
on finding the boundaries of "runs", which in one form or another
ends up getting asked about every 10 days.

*One* possible method:

If the original image is A, then construct

B = [zeros(1,size(A,2)), A(:,:,1)>200, zeros(1,size(A,2))];

This would be 3002 x 2000 (x1)

Then,

C = B(2:end,:) xor B(1:end-1,:)

C would be 3001 x 2000 .

For any particular column C(:,K) the first 1 in the column marks
the beginning of the first run of red at that same offset into
A(:,K) -- e.g., if the first 1 in C(:,K) was in row 5, then A(5,K)
is the first red pixel. The second 1 in the column C(:,K) marks
the beginning of the first run of *non*-red, so if the second 1
in the column C(:,K) were at (say) position 17, then the last
red element in the run was at A(16,K). The pattern continues
with the first 1 of each pair marking the beginning of a red run,
and the second 1 of each pair marking one past the last red in
the run. If C(3001,K) is set then A(3000,K) was a red pixel.
If you encounter a pair C(N,K) and C(N+1,K) that are both 1s,
then A(N,K) was a single red pixel -- you can see then why it is
important that the end-marker be one past the end of the run,
so you can find the "runs" of length 1.

To find the boundaries, you can iterate over each column K and
use find(C(:,K)). Finding the boundaries in a vectorized
way is left as an exercise for the reader ;-)
--
  All is vanity. -- Ecclesiastes

"Craig " <hillx154@umn.edu> wrote in message
<fa4tcl$69q$1@fred.mathworks.com>...
> I have an image (3000x2000 pixels). I want to scan each
> column at a time and look for the red pixels (R>200). Most
> pixels are just black. When it finds a red pixel, I want
> it to search forward and find how many consecutive red
> pixels (R>200) there are. Any ideas?

Except that R >= 200 is not a good criterion for a color
being red.

patch(rand(3,1),rand(3,1),[201 0 0]/255)
patch(rand(3,1),rand(3,1),[201 255 255]/255)
patch(rand(3,1),rand(3,1),[150 0 0]/255)
patch(rand(3,1),rand(3,1),[201 150 0]/255)
patch(rand(3,1),rand(3,1),[201 0 150]/255)

http://www.mathworks.com/matlabcentral/fileexchange/loadFile.do?
objectId=12326&objectType=FILE

John

Craig:
<SNIP sees red...

> I want to scan each column at a time and look for the red
pixels (R>200)...

one of the many solutions

% the data
% - assume rgb is a nxmx3 mat
% - extract the <red> plane
% m=rgb(:,:,1);
% - eg,
     m=[
0 0 203 0 0 0
201 0 0 0 207 210
202 0 0 204 0 211
0 0 0 205 208 212
0 0 0 206 209 0
     ];
% the engine
     [mf,mf]=histc(m,[200,inf]);
     mm=[0,mf(:).',0];
     ib=strfind(mm,[0,1]);
     ie=strfind(mm,[1,0]);
     il=ie-ib;
     rowx=mod(ib,size(m,1));
     colx=fix(ib/size(m,1))+1;
     r=[colx.',rowx.',il.'];
% the result
     m
     disp(sprintf('%% col/row %2d/%2d lenth %2d\n',r.'));

-but- carefully look at what <john d'errico> says...
us