How to calculate the maximum intensity projection of a stack of images?

I know we can use the A= max(image,[],3) to get the maximum intensity. But I wonder to know how I can calculate the 3D volume by using the maximum intensity projection. And then I can rotate the 3D volume. Thank you.

Answer 1

[Image Analyst]

Hopefully "image" is not the actual name of your image, but you can just use sum:

volume = sum(image(:));

[Penny]

Thanks,the name of the image is Img, so A= max(Img,[],3). So how to show the volume?

[Image Analyst]

volume = sum(Img(:));

[Penny]

Thanks, but it doesn't work. Do I still use imshow or some other command?

[Penny]

This method only returns a value, not a 3D array.

[Image Analyst]

You already have the 3-D array. You called it "Img". Since it is probably a grayscale image, I should have asked to to identify "foreground" pixels by some method such as thresholding. You said you want the volume of the object in there so let's find all pixels that are above the threshold and then sum them to get a single value.

binaryImage = Img > someThreshold;
volume = sum(binaryImage(:));

This will count the number of pixels above the threshold.

[Penny]

Thank you Image Analyst. But I want to show the volume by using the MIP algorithm. Do you know how to implement?

[Image Analyst]

Scan each pixel column in the image extracting the Z vector and sending it into max().

[rows, columns, numSlices] = size(Img);
outputImage = zeros(rows, columns, class(Img)); % Or whatever class you want.
for col = 1 : columns
    for row = 1 : rows
        thisZVector = Img(row, col, :);
        maxValue = max(thisZVector);
        outputImage(row, col) = maxValue;
    end
end