Bound an object within an image and fill that object with white

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Casey Ricks 2019-9-24

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/481810-bound-an-object-within-an-image-and-fill-that-object-with-white

评论： Casey Ricks 2019-9-24

I'm attempting to fill an image that has red particles with very 'jagged' edges. I also have images of white particles on a black background and black particles on a white background, so I'm trying to find something general. I plan to later measure the width of the 'whited out' cylinder.

Here's the image:

Initially, I was using the following line of code for black particles on a white background:

    A = imread(filename);
    
    C = imcomplement(A);
    C = im2bw(C, 150/255);
    
    SE = strel('disk',10); 
    B =  imclose(C, SE);

which produced:

This is just ok, but the boundaries aren't very precise.

I then tried the following, keeping in mind that I had red particles on a white background.

A = imread(filename);
C = imcomplement(A);
blueChannel = C(:, :, 3);
C = im2bw(blueChannel, 0.3);
B = imfill(C,'holes');

This produces the following image:

The edges are more precise, but this method is unfavorable because:

A) it is not general

B) There are are gaps left from the string

Now, I could use the strel command and fill these, but again, that creates an imprecise boundary along the edge of the column. I would prefer to fill it in such a way that connects the two closest white pixels along the outer boundary of a gap.

I've been looking into the edges function, as well as the different edge methods, but these end up just tracing lines around my object, which hasn't really helped me so far.

I feel like the imcomplement images are actually pretty close, from the code:

A = imread(filename);
C = imcomplement(A);

If only I could edit my measuring code (outlined below) to say something along the lines,

"Does the current line have a pixel that is not dark grey or black? Ff so, measure the width of the pixels that are not dark grey or black"

This still might not work though, due to the fact that the string is complemented to black.

From the image using red particles:

From the image using the Black particles:

Just as a heads up, I use the following lines of code to find the width of the cylinders im my images:

%{
"x1(idx) = (find(B(j,:),1,'last')+1)-find(B(j,:),1,'first')"  
finds the width of the 1's within the matrix. "(find(B(j,:),1,'last')+1)" 
finds the first instance of a non-zero value, where the first instance
is indicated by the 1, and 'last' tells the function to move from right
to left. We add the 1 here (+1) because the find function returns the 
column index of the first non-zero value and, without the +1, we would actually
be finding the width-1.
%}
    info = imfinfo(filename);
    x1 = zeros(0, info.Height);
    y1 = zeros(0, info.Height);
    idx = 1;
    for j = 1:info.Height
        if any(B(j,:))==1;
            x1(idx) = (find(B(j,:),1,'last')+1)-find(B(j,:),1,'first');
            y1(idx) = j;
            idx = idx+1;
        end
    end
    
    FinalStruct(i).width = x1;
    FinalStruct(i).indexwidth = y1;

A big, huge thanks in advance to anyone who even manages to read all of this!

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KALYAN ACHARJYA 2019-9-24

Your objective to find the width (horizantally) of silver strip cylinder from the given image, right?

Is it Euclidean distance?

Casey Ricks 2019-9-24

Yes, it is Euclidean distance.

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Answer 1

Image Analyst 2019-9-24

2
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/481810-bound-an-object-within-an-image-and-fill-that-object-with-white#answer_393344

testRGBImage.m

How about using the Color Thresholder App to get the thresholds in HSV color space? See attached m-file for a full demo. I did a color segmentation to get the background, then inverted the mask, filled it, and then took the largest blob.

There is still a little bit of the metal cap getting in because it seems to be an aluminum ring that is taking on some of the pink color. I suggest you cover that top ring with black electrical tape. Other than that it seems to be pretty accurate to me.