Thread Subject: How to check pixel value in bmp image is of red color?

I have an image of a circle red in color. Its boundary os
black in color. I have to find out whether a particular
pixel value is red in color. How can i do this?

If I remember correctly, a bitmap (bmp) is stored in a mxnx3
matrix, where mxn is the image size and the third dimension
is the color information (rgb). For red, a the pixel data
in an image im would look like this:

im(16,35,1:3)

ans(:,:,1) =

  255


ans(:,:,2) =

    0


ans(:,:,3) =

    0



And for black, all values should be zero.

"Matt Fig" <spamanon@yahoo.com> wrote in message <g3j719
$qvn$1@fred.mathworks.com>...
> If I remember correctly, a bitmap (bmp) is stored in a
mxnx3
> matrix, where mxn is the image size and the third
dimension
> is the color information (rgb). For red, a the pixel data
> in an image im would look like this:
>
> im(16,35,1:3)
>
> ans(:,:,1) =
>
> 255
>
>
> ans(:,:,2) =
>
> 0
>
>
> ans(:,:,3) =
>
> 0
>
>
>
> And for black, all values should be zero.

thanks a lot!!!!
Actually I have to find the coordinates of this image. How
can I do that???

In article <g3j6eh$mpo$1@fred.mathworks.com>,
Rinachi Garg <rinachi_garg@yahoo.co.in> wrote:
>I have an image of a circle red in color. Its boundary os
>black in color. I have to find out whether a particular
>pixel value is red in color. How can i do this?

When does a colour stop being red and start being a different
colour? in RGB-255, we agree that [255 0 0] is red, but
is [255 0 1] "not red"? Is [128 0 0] red? Is [32 0 0] red? Is
[1 0 0] red?

--
Current spam load: 750-800 messages per day (March 4, 2008)

roberson@ibd.nrc-cnrc.gc.ca (Walter Roberson) wrote in message
<g3jnpb$p03$1@canopus.cc.umanitoba.ca>...
> In article <g3j6eh$mpo$1@fred.mathworks.com>,
> Rinachi Garg <rinachi_garg@yahoo.co.in> wrote:
> >I have an image of a circle red in color. Its boundary os
> >black in color. I have to find out whether a particular
> >pixel value is red in color. How can i do this?
>
> When does a colour stop being red and start being a different
> colour? in RGB-255, we agree that [255 0 0] is red, but
> is [255 0 1] "not red"? Is [128 0 0] red? Is [32 0 0] red? Is
> [1 0 0] red?

This is why I wrote fuzzycolor. It uses a lookup
table that identifies most colors in the range
you show as being red. A quick test shows that
it accepts anything down to roughly [.18 0 0]
(0-1 RGB units) as red. Too far down to zero
in the red channel, and the color is better
described as black.

The tables themselves were defined by a visual
test that I did, on my own monitor, but a user
can redefine those tables.

fuzzycolor([.18 0 0],'red')
ans =
     1

fuzzycolor([.17 0 0],'red')
ans =
         0.67

fuzzycolor([.16 0 0],'red')
ans =
         0.16

fuzzycolor([.15 0 0],'red')
ans =
     0

The problem is that the definition of "red"
is highly subjective. As well, the set of
colors that any one person will choose to
define that way is probably not even a
convex set in RGB space.

John

"Matt Fig" <spamanon@yahoo.com> wrote in message <g3j719
$qvn$1@fred.mathworks.com>...
> If I remember correctly, a bitmap (bmp) is stored in a
mxnx3
> matrix, where mxn is the image size and the third
dimension
> is the color information (rgb). For red, a the pixel data
> in an image im would look like this:
>
> im(16,35,1:3)
>
> ans(:,:,1) =
>
> 255
>
>
> ans(:,:,2) =
>
> 0
>
>
> ans(:,:,3) =
>
> 0
>
>
>
> And for black, all values should be zero.


hiii
I have written the following code and tried to find out the
pixel values which store red color by inserting them in
another array. I have tried to find the position of column
and row seperately.
But i am not getting desired result.
>> I = imread('circle.bmp');
>> J = rgb2gray(I);
>> a;b;
>> [m,n] = size(j);
>> for i = 1:1:m
      for j = 1:1:n
         if (I(i,j,1) == 255)
           if(I(i,j,2) == 0)
             if(I(i,j,3) == 0)
                  a(i)=i;
                  b(j)=j;
              end;
           end;
          end;
       end;
     end;
I think I am doing some mistake in declaring arrays a and
b . And then assigning the values of positions os row and
columns 2 a and b.
How can I correct it?

"Rinachi Garg" <rinachi_garg@yahoo.co.in> wrote in message
<g3l5nj$a2c$1@fred.mathworks.com>...
> "Matt Fig" <spamanon@yahoo.com> wrote in message <g3j719
> $qvn$1@fred.mathworks.com>...
> > If I remember correctly, a bitmap (bmp) is stored in a
> mxnx3
> > matrix, where mxn is the image size and the third
> dimension
> > is the color information (rgb). For red, a the pixel
data
> > in an image im would look like this:
> >
> > im(16,35,1:3)
> >
> > ans(:,:,1) =
> >
> > 255
> >
> >
> > ans(:,:,2) =
> >
> > 0
> >
> >
> > ans(:,:,3) =
> >
> > 0
> >
> >
> >
> > And for black, all values should be zero.
>
>
> hiii
> I have written the following code and tried to find out
the
> pixel values which store red color by inserting them in
> another array. I have tried to find the position of
column
> and row seperately.
> But i am not getting desired result.
> >> I = imread('circle.bmp');
> >> J = rgb2gray(I);
> >> a;b;
> >> [m,n] = size(j);
> >> for i = 1:1:m
> for j = 1:1:n
> if (I(i,j,1) == 255)
> if(I(i,j,2) == 0)
> if(I(i,j,3) == 0)
> a(i)=i;
> b(j)=j;
> end;
> end;
> end;
> end;
> end;
> I think I am doing some mistake in declaring arrays a and
> b . And then assigning the values of positions os row and
> columns 2 a and b.
> How can I correct it?

I have solved this problem.
By declaring a = zeros(1,m);
b = zeros(1,n);

Rinachi

"Rinachi Garg" <rinachi_garg@yahoo.co.in> wrote in message
<g3l5nj$a2c$1@fred.mathworks.com>...
> "Matt Fig" <spamanon@yahoo.com> wrote in message <g3j719
> $qvn$1@fred.mathworks.com>...
> > If I remember correctly, a bitmap (bmp) is stored in a
> mxnx3
> > matrix, where mxn is the image size and the third
> dimension
> > is the color information (rgb). For red, a the pixel
data
> > in an image im would look like this:
> >
> > im(16,35,1:3)
> >
> > ans(:,:,1) =
> >
> > 255
> >
> >
> > ans(:,:,2) =
> >
> > 0
> >
> >
> > ans(:,:,3) =
> >
> > 0
> >
> >
> >
> > And for black, all values should be zero.
>
>
> hiii
> I have written the following code and tried to find out
the
> pixel values which store red color by inserting them in
> another array. I have tried to find the position of
column
> and row seperately.
> But i am not getting desired result.
> >> I = imread('circle.bmp');
> >> J = rgb2gray(I);
> >> a;b;
> >> [m,n] = size(j);
> >> for i = 1:1:m
> for j = 1:1:n
> if (I(i,j,1) == 255)
> if(I(i,j,2) == 0)
> if(I(i,j,3) == 0)
> a(i)=i;
> b(j)=j;
> end;
> end;
> end;
> end;
> end;
> I think I am doing some mistake in declaring arrays a and
> b . And then assigning the values of positions os row and
> columns 2 a and b.
> How can I correct it?
----------------------------------------------------------
Rinachi:
You're not yet familiar with MATLAB, right? You could do
this faster with just 3 lines to extract out the individual
color bands from the color image and two lines to get the
coordinates of the pure red pixels. Why not try this:

clc; % Clear command window.
% Make up some arbitrary data for this demo.
redBand = uint8(255*rand(6,6));
greenBand = uint8(255*rand(6,6));
blueBand = uint8(255*rand(6,6));
redPixels = rand(6,6) > 0.75
redBand(redPixels(:)) = 255;
greenBand(redPixels(:)) = 0;
blueBand(redPixels(:)) = 0;
colorImage = cat(3, redBand, greenBand, blueBand);
imshow(colorImage); % Display it.
% Done making up demo image.

% Now, here's the crucial part that Rinachi needs:
% Extract the 3 color bands out of the color image.
redBand = colorImage(:,:, 1)
greenBand = colorImage(:,:, 2)
blueBand = colorImage(:,:, 3)

% Find x,y of pixels that are pure red.
reds = (redBand == 255) & (greenBand == 0) & (blueBand == 0)
[redRows redColumns] = find(reds)
% Check out the command window for results.

And like Walter pointed out, checking for 255,0,0 will only
find pure, brightest reds as would be created from computer
graphics or annotation, not necessarily for other arbitrary
reds that you might find in real world images snapped with,
say, a digital camera. For that, there's other methods.
Let us know if that applies to you.
Regards,
ImageAnalyst
---------------------------------------------------

"Image Analyst" <imageanalyst@mailinator.com> wrote in
message <g3lodn$c90$1@fred.mathworks.com>...
> "Rinachi Garg" <rinachi_garg@yahoo.co.in> wrote in
message
> <g3l5nj$a2c$1@fred.mathworks.com>...
> > "Matt Fig" <spamanon@yahoo.com> wrote in message <g3j719
> > $qvn$1@fred.mathworks.com>...
> > > If I remember correctly, a bitmap (bmp) is stored in
a
> > mxnx3
> > > matrix, where mxn is the image size and the third
> > dimension
> > > is the color information (rgb). For red, a the pixel
> data
> > > in an image im would look like this:
> > >
> > > im(16,35,1:3)
> > >
> > > ans(:,:,1) =
> > >
> > > 255
> > >
> > >
> > > ans(:,:,2) =
> > >
> > > 0
> > >
> > >
> > > ans(:,:,3) =
> > >
> > > 0
> > >
> > >
> > >
> > > And for black, all values should be zero.
> >
> >
> > hiii
> > I have written the following code and tried to find out
> the
> > pixel values which store red color by inserting them in
> > another array. I have tried to find the position of
> column
> > and row seperately.
> > But i am not getting desired result.
> > >> I = imread('circle.bmp');
> > >> J = rgb2gray(I);
> > >> a;b;
> > >> [m,n] = size(j);
> > >> for i = 1:1:m
> > for j = 1:1:n
> > if (I(i,j,1) == 255)
> > if(I(i,j,2) == 0)
> > if(I(i,j,3) == 0)
> > a(i)=i;
> > b(j)=j;
> > end;
> > end;
> > end;
> > end;
> > end;
> > I think I am doing some mistake in declaring arrays a
and
> > b . And then assigning the values of positions os row
and
> > columns 2 a and b.
> > How can I correct it?
> ----------------------------------------------------------
> Rinachi:
> You're not yet familiar with MATLAB, right? You could do
> this faster with just 3 lines to extract out the
individual
> color bands from the color image and two lines to get the
> coordinates of the pure red pixels. Why not try this:
>
> clc; % Clear command window.
> % Make up some arbitrary data for this demo.
> redBand = uint8(255*rand(6,6));
> greenBand = uint8(255*rand(6,6));
> blueBand = uint8(255*rand(6,6));
> redPixels = rand(6,6) > 0.75
> redBand(redPixels(:)) = 255;
> greenBand(redPixels(:)) = 0;
> blueBand(redPixels(:)) = 0;
> colorImage = cat(3, redBand, greenBand, blueBand);
> imshow(colorImage); % Display it.
> % Done making up demo image.
>
> % Now, here's the crucial part that Rinachi needs:
> % Extract the 3 color bands out of the color image.
> redBand = colorImage(:,:, 1)
> greenBand = colorImage(:,:, 2)
> blueBand = colorImage(:,:, 3)
>
> % Find x,y of pixels that are pure red.
> reds = (redBand == 255) & (greenBand == 0) & (blueBand ==
0)
> [redRows redColumns] = find(reds)
> % Check out the command window for results.
>
> And like Walter pointed out, checking for 255,0,0 will
only
> find pure, brightest reds as would be created from
computer
> graphics or annotation, not necessarily for other
arbitrary
> reds that you might find in real world images snapped
with,
> say, a digital camera. For that, there's other methods.
> Let us know if that applies to you.
> Regards,
> ImageAnalyst
> ---------------------------------------------------

I have a bitmap image which contain red, blue and green
coloured filled contous. I have to find the coordinates of
the boundary pixels of these three contours.
How to do this?

On Jun 22, 11:57=A0am, "Rinachi Garg" <rinachi_g...@yahoo.co.in> wrote:
> "Image Analyst" <imageanal...@mailinator.com> wrote in
> message <g3lodn$c9...@fred.mathworks.com>...
>
>
>
>
>
> > "Rinachi Garg" <rinachi_g...@yahoo.co.in> wrote in
> message
> > <g3l5nj$a2...@fred.mathworks.com>...
> > > "Matt Fig" <spama...@yahoo.com> wrote in message <g3j719
> > > $qv...@fred.mathworks.com>...
> > > > If I remember correctly, a bitmap (bmp) is stored in
> a
> > > mxnx3
> > > > matrix, where mxn is the image size and the third
> > > dimension
> > > > is the color information (rgb). =A0For red, a the pixel
> > data
> > > > in an image im would look like this:
>
> > > > im(16,35,1:3)
>
> > > > ans(:,:,1) =3D
>
> > > > =A0 255
>
> > > > ans(:,:,2) =3D
>
> > > > =A0 =A0 0
>
> > > > ans(:,:,3) =3D
>
> > > > =A0 =A0 0
>
> > > > And for black, all values should be zero. =A0
>
> > > hiii
> > > I have written the following code and tried to find out
> > the
> > > pixel values which store red color by inserting them in
> > > another array. I have tried to find the position of
> > column
> > > and row seperately.
> > > But i am not getting desired result.
> > > >> I =3D imread('circle.bmp');
> > > >> J =3D rgb2gray(I);
> > > >> a;b;
> > > >> [m,n] =3D size(j);
> > > >> for i =3D 1:1:m
> > > =A0 =A0 =A0 for j =3D 1:1:n
> > > =A0 =A0 =A0 =A0 =A0if (I(i,j,1) =3D=3D 255)
> > > =A0 =A0 =A0 =A0 =A0 =A0if(I(i,j,2) =3D=3D 0)
> > > =A0 =A0 =A0 =A0 =A0 =A0 =A0if(I(i,j,3) =3D=3D 0)
> > > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 a(i)=3Di;
> > > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 b(j)=3Dj;
> > > =A0 =A0 =A0 =A0 =A0 =A0 =A0 end;
> > > =A0 =A0 =A0 =A0 =A0 =A0end;
> > > =A0 =A0 =A0 =A0 =A0 end;
> > > =A0 =A0 =A0 =A0end;
> > > =A0 =A0 =A0end;
> > > I think I am doing some mistake in declaring arrays a
> and
> > > b . And then assigning the values of positions os row
> and
> > > columns 2 a and b.
> > > How can I correct it?
> > ----------------------------------------------------------
> > Rinachi:
> > You're not yet familiar with MATLAB, right? =A0You could do
> > this faster with just 3 lines to extract out the
> individual
> > color bands from the color image and two lines to get the
> > coordinates of the pure red pixels. =A0Why not try this:
>
> > clc; % Clear command window.
> > % Make up some arbitrary data for this demo.
> > redBand =3D uint8(255*rand(6,6));
> > greenBand =3D uint8(255*rand(6,6));
> > blueBand =3D uint8(255*rand(6,6));
> > redPixels =3D rand(6,6) > 0.75
> > redBand(redPixels(:)) =3D 255;
> > greenBand(redPixels(:)) =3D 0;
> > blueBand(redPixels(:)) =3D 0;
> > colorImage =3D cat(3, redBand, greenBand, blueBand);
> > imshow(colorImage); % Display it.
> > % Done making up demo image.
>
> > % Now, here's the crucial part that Rinachi needs:
> > % Extract the 3 color bands out of the color image.
> > redBand =3D colorImage(:,:, 1)
> > greenBand =3D colorImage(:,:, 2)
> > blueBand =3D colorImage(:,:, 3)
>
> > % Find x,y of pixels that are pure red.
> > reds =3D (redBand =3D=3D 255) & (greenBand =3D=3D 0) & (blueBand =3D=3D
> 0)
> > [redRows redColumns] =3D find(reds)
> > % Check out the command window for results.
>
> > And like Walter pointed out, checking for 255,0,0 will
> only
> > find pure, brightest reds as would be created from
> computer
> > graphics or annotation, not necessarily for other
> arbitrary
> > reds that you might find in real world images snapped
> with,
> > say, a digital camera. =A0For that, there's other methods. =A0
> > Let us know if that applies to you.
> > Regards,
> > ImageAnalyst
> > ---------------------------------------------------
>
> I have a bitmap image which contain red, blue and green
> coloured filled contous. I have to find the coordinates of
> the boundary pixels of these three contours.
> How to do this?- Hide quoted text -
>
> - Show quoted text -

-------------------------------------------------------------------
Rinachi:
Well now, you've totally changed the situation. This is what you
should have said in the first place, because just getting the
coordinates of the red pixels may have seemed like what you should
need to do but it's not. If you have filled contours, basically solid
blobs of uniform color, then what you need to do is call bwboundaries
in the image processing toolikt (which hopefully you already have
since you're dealing with images). (It finds the boundaries, like if
you had eroded the binary image and subtracted that from the original
binary image.) If you have nested contours (colored blobs inside of
other blobs) then it gets a bit tricker but you can still do it.
Good luck,
ImageAnalyst