how to calculate intensity of an image

i wanted to calculate intensity of an image... please could someone help me.....

[vipul utsav]

if you want single value of intensity you can take mean(total image(:)) otherwise every element of image represents intensity

Answer 1

[Image Analyst]

There is nothing to do. The image is already the intensity.

[Elysi Cochin]

sir actually i wanted to compute the histogram for an intensity equal to "intensity_value".... to determine "intensity_value" i need to extract the intensity of largest area and then add a tolerance of ±10..... how to do that..... please do reply sir...

[Image Analyst]

This is kind of confusing: " the histogram for an intensity equal to "intensity_value" If you just want the number of pixels with that intensity, you don't need to call imhist(), you just need to do:

binaryImage = (yourImage == intensity_value);
numPixels = sum(binaryImage(:));

If you want to find all the areas of all the connected blobs in that binary image, you can do

measurements = regionprops(binaryImage, 'area');
allAreas = [measurements.Area];

To find the largest area:

[sortedAreas sortIndexes] = sort(allAreas, 'descend');
largestArea = sortedArea(1);

I don't know what you mean by adding a tolerance to the largest area, but maybe it's this:

largestAreaWithTolerance = 1.1 * largestArea;

You should really study the image segmentation tutorial in my File Exchange.

[Elysi Cochin]

This is kind of confusing: " the histogram for an intensity equal to "intensity_value"...... sir please refer this link.... this is what i want to compute..... i'm also confused sir..... please do reply sir.....

http://img841.imageshack.us/img841/9431/50272323.png

[Walter Roberson]

Use the code Image Analyst showed earlier:

binaryImage = (yourImage == intensity_value);
numPixels = sum(binaryImage(:));

Or, more compactly,

numPixels = sum(yourImage == intensity_value);

Or, as we were discussing in a previous one of your questions on this same topic, use histc() or hist() or imhist(), after having figured out what value to pass in for the boundaries. The advantage of doing this is that it does the counting all at once instead of having to do counting for each intensity.

[Elysi Cochin]

sir i'm getting error when i do so....

    ??? Undefined function or variable 'intensity_value'.

    Error in ==> Untitled at 9
    binaryImage = (yourImage == intensity_value);

but when i did as below no error... is what i did correct??

    [filename pathname]=uigetfile('*.png','Select An Image');
    yourImage=imread([pathname filename]);

    grayImage=rgb2gray(yourImage);
    binaryImage = im2bw(grayImage);

    numPixels = sum(binaryImage(:))

    measurements = regionprops(binaryImage, 'area');
    allAreas = [measurements.Area];

    % To find the largest area:

    [sortedAreas sortIndexes] = sort(allAreas, 'descend');
    largestArea = sortedAreas(1)

    largestAreaWithTolerance = 1.1 * largestArea

[Image Analyst]

That paper says you have to take the whole histogram of a 10 by 10 chunk of your image, which I guess slides along the image. Then, for that histogram of the 10x10 block, it wants you to find the tallest bin - the max of the histogram. Then it asks you to sum the histogram bins from that bin minus 10 to that bin + 10 gray levels, but excluding the tallest bin itself. This was not well explained by your original wording. Even the papers wording is not optimal. For example using "area" to describe a sum of bins. Yes, I guess it is an area but those pixels represented by those bins may be scattered all over the image and might not even touch each other so it's not like the area of some contiguous blob in the image, which is what I thought from your description.

So, scan the image, say with blockproc(). Then get the histogram in a 10x10 block with imhist() or hist(). Then use max() to find the tallest bin location. Then use sum() to sum between that bin-10 and that bin+10 and subtract the value of the tallest bin. I don't really know what this means, but that's what they ask for. Give it a shot. I told you step by step the functions to use. Search for a blockproc demo that I've posted in Answers if you need help with blockproc.

[Elysi Cochin]

thank u sir..