How can I make the image processing faster for large number of images?

Hi,

I am trying to obtain the position of the light stripe at each row. Below is an example for an image. However, identification of the light spots are performed for more than 2500 images, which takes about 10 hours. What would be the most efficient way to increase the speed of this calculation? I would be happy if you could guide me.

I = imread('C:\Users\img1.png');
img = I;
[height, width, dim] = size(I);

% Some image processing in between 
% ...

img(img~=0)=1;
intline=nan;

p =1;
tic
for j=1:width
        rows=[];        
        rows=find(img(:,j)==1);
        if length(rows)>3 % spots detectable?
            int=[];
            for k=rows(1):rows(end)
                int(k)=sum(double(I(k,j,1:3)));
            end
            y_px=find(int>=0.90*max(int));
            if length(y_px)>2
                cuts=[];
                cuts=find(y_px(2:length(y_px))-y_px(1:length(y_px)-1)>10);
                if cuts
                    cuts=cuts(end);
                    y_px=round(mean(y_px(cuts+1:end)));
                else
                    y_px=round(mean(y_px));
                end
            else   
                y_px=y_px(end);
            end
            intline(j,1:2)=[j y_px];
            p = p+1;
        end 
    end
toc

[Matt Gaidica]

Can you attach the actual image? More comments in your code would be nice too.

Answer 1

[Walter Roberson]

preallocate int and intline. You know the exact size of int at the for loop bounds. You do not know the exact size intline will turn out as but you know the maximum size so allocate that and keep track of what you use and truncate down to the used part afterward .

Do not assign [] to variables you then completely overwrite as it is a waste of time.

[Walter Roberson]

some of your operations can be vectorized . For example sum() along the third dimension and store in a matrix and access the result as needed .

Sometimes it is much more efficient to calculate over an entire matrix and ignore parts that you do not need rather than calculating only on the sections you need.

Get rid of unneeded operations . You can use

img = logical(I)

instead of explicit test for nonzero. then do not find(img(:,j)==1) just find(img(:,j)) Or skip img and just find(I(:,j)) since the test is automatically for nonzero .

[tethys]

Thanks a lot for the tips! Actually, for example, initializing the variable "int" helped, but with a very little reduction (<%5) in computational time. I have found out that the computational time is increasing over time (e.g. from 1 s to 2.5s after 20 images) for the same loop above. With the

profile on

70% of the time turned out to be spent on:

for j=1:width
    rows=find(img(:,j)); 
    for k=rows(1):rows(end)
                int(k)=sum(double(I(k,j,1:3)));
    end  
end

I have tried "accumarray", but mostly with an error:

Second input VAL must be a vector with one element for each row in SUBS, or a scalar.

How can I use "accumarray" correctly, or how can I get the most speed improvement here?

Thanks again!

[Walter Roberson]

accumarray is not relevant here . summing over the third dimension of all of I and accessing appropriate parts would improve performance .

You can do better still .

int3 = sum(I .* repmat(img(:,:,1), 3)

and now any column jj that at has at least 3 dots in, int = int3(:,jj) will act nearly the same as your current int but possibly better . Your current intline records locations calculated relative to the first nonzero pixel in the column where my modification would most naturally calculate relative to the beginning of the column . The meaning of your calculation is not documented so it is difficult to say which is better .

question for you: if one column had a particular pattern of pixels and the next column had the same pattern but 100 pixels further down then are you expecting the same y_px result for both (only pattern is important) or a difference (location of pattern affects output )? Your current code would produce the same output .

[Image Analyst]

Why not sum the image before any of the for loops, then just access it

sumImage = sum(double(I), 3);
for.....
         int(k)=s sumImage(k,j);
     

And please attach an image so we can see if there are built-in functions to do all of this.

And add comments to your code because it looks like you're "position of the light stripe at each column" rather than "position of the light stripe at each row." so it would help if you added comments to each line or block of code. Professional programmers (like me) do that - not everyone is as clairvoyant as Walter.

[tethys]

Thanks again! For me, the important point is to obtain the position of maximum intensity (or center of mass) of the light. Therefore, each row is scanned to obtain the position of the laser peak. Ideally, there must be only one peak, and its shift gives me the position of the object. But

int3 = sum(I .* repmat(img(:,:,1), 3)

does not seem to work.

I = imread('C:\Users\img1.png');
img = I;
[height, width, dim] = size(I);
mint=230; img(img(:,:,2)<mint)=0;
img=img(:,:,2)-img(:,:,1);
img(img~=0)=1;
intline=nan;

p =1; %counter

tic
for j=1:width
        rows=find(img(:,j));
        if length(rows)>3 
           for k=rows(1):rows(end)
                int(k)=sum(double(I(k,j,1:3)));
           end
            y_px=find(int>=0.90*max(int));
            if length(y_px)>2
                
                cuts=find(y_px(2:length(y_px))-y_px(1:length(y_px)-1)>10);
                if cuts
                    cuts=cuts(end);
                    y_px=round(mean(y_px(cuts+1:end)));
                else
                    y_px=round(mean(y_px));
                end
            else   
                y_px=y_px(end);
            end
            intline(j,1:2)=[j y_px];
            p = p+1;
        end 
    end 
toc
figure;  hold on; imshow(I); hold on; scatter(intline(~isnan(intline(:,1)),1),intline(~isnan(intline(:,1)),2))

Pixel values are saturated in the following image, but gives an idea about what I am doing.

[Image Analyst]

This is easily solved if you don't saturate your image and use regionprops() and ask for 'WeightedCentroid'.

[Walter Roberson]

It is not obvious to me that you want to scan by rows?? For example at about 1/3 of the way down, you have the streak at roughly 1/16 of the way across, but you have not yet reached the rows that have brightness at the far right, so the center of mass for that row would be the streak. But slightly further down where you also get the brightness at the far right, the center of mass of the row would be... ummm, about 60% of the way to the right or so? And slightly below that, the far right has dimmed but a second streak has started so the center of mass would be on the order of 1/10th of the way to the right. But this is all pretty vague and it is not obvious those lights are meaningful compared to the rest of the image, but your tests are relative to the maximum for that row, not relative to the maximum for the image. On the other hand, your logical mask is based on the red channel only, which would have only weak intensity in the green, and it gets tricky to predict visually what would be picked out.

It would seem to make more sense to be scanning column by column, which is what your current code does. img(:,j) is one column and you are looking for non-zero values in there and comparing per-column maxima that are going to include the fully-saturated white light in the middle rows. Visually the expected result would be close to a straight line across.

But you did not answer my question about whether the position information you are looking for is position relative to the first non-zero entries in the red channel in a given column, or if you are looking for absolute positions that can be easily compared column by column ? Your current code gives relative positions, so if your image had a huge shear that displaced the detectable light downwards by hundreds of pixels in part of your image, your code would not notice. The modification I talked about, int3 = sum(I .* repmat(img(:,:,1), 3) would be for working with absolute locations.

[Image Analyst]

I agree with Walter. Please explain in words what "the position of the light stripe at each row" means, because if you want something like the centerline of that bright region, or the mean or center line (row) number of the line, I'd do something different. It might mean scanning across the image column by column or it could mean doing something else. It would help if you say what the next step would be assuming you had whatever it is you want here. Like "Once I have that, I will use it to..............."