How can i condense a binary matrix?

The image attatched is what I'm working with (I've added the grids and numbers to help me explain).

I'm hoping to use MATLAB to help me achieve a table that logs every time the category changes, and at what value. e.g at 1 the category is 3, at 1.6 the category goes to 5, at 2 the category goes to 8.

The way i have gone about this so far is to export the binary image so i have the pixel matrix.

I've then extracted 20 columns out of the pixel matrix to correspond to the 20 categories:

BW1 is my binary image. 25, 74 etc have come from the pixel coordinate in the centre of each category column.

>>BW1(:,[25 74 123 172 221 270 319 368 417 466 515 564 613 662 711 760 809 858 907 956])

So now I'm left with a table which has the columns i need, and is as deep as the pixel count.I then need to add a column in this table which uses a conversion factor to convert the pixels to my scale (6/886px).

However, I'm struggling to find a way to condense my table down to just the tops now (basically whenever there is a change in category).

I was wondering if i could somehow set an increment and then get a return only at these increments?

Can anybody help me, or point me in the right direction of some further reading.

Alternatively, if there is a more efficient way of doing this id be pleased to hear it, like i say im new to this, and its a little overwhelming trying to find a solution because there seem to be so many ways to do things!

Many thanks for your time.

[Image Analyst]

What image above? You didn't attach one. Please use the green and brown frame icon to insert your image and then I'll finish reading the rest of your question and try to help you.

[Dan]

Sorry, it didnt attach properly, here is a link to the example and the source image

https://s17.postimg.org/cr2yhz6yn/Example.jpg

https://s14.postimg.org/rkydvvx01/cropped.jpg

[Jan]

@Dan: I simply used the "Image" button to let your images appear directly in your question. Alternatively include the image's URL in << and >> .

[Dan]

Thanks Jan

Answer 1

[Image Analyst]

I'm not exactly sure what you need. I'd just call imclearborder to get rid of white stuff touching the border. Then I'd call bwareafilt() or bwareaopen() to extract just the large blobs. Then you can get an average vertical profile along each column (band of column pixels)

verticalProfile = mean(binaryImage(:, col1:col2), 2) > 128;

then use strfind() to look for patterns of [0, 1] and [1, 0] to find out where the block(s) start and stop in the vertical direction.

[Dan]

Thanks for the reply.

I followed your instructions, but i got stuck at the vertical profile. I cant work out what it does?

So this is where I'm at now:

To explain a little easier I've added the red dots, they will not be in the image im processing.

Essentially, i need the X and Y locations of those red dots.

Then i need to translate those value to fit my grid which is Y 0 to 6 and X 0 to 20.

Hope this make more sense, and thanks again.

[Image Analyst]

You can fill the blobs to make sure they're solid, then take a profile though it and look for the top and bottom row. for example let's say you're looking at column "col"

binaryImage = imfill(binaryImage, 'holes');
thisColumn = binaryImage(:, col);
topLine = find(thisColumn, 1, 'first');
bottomLine = find(thisColumn, 1, 'last');

To calibrate you need to get a line through the tick marks and figure out where 0 and 6 are. For example if the tick marks are in column 100

thisColumn = binaryImage(:, 100) == 0;
topLine = find(thisColumn, 1, 'first');
bottomLine = find(thisColumn, 1, 'last');
coeffs = polyfit([topLine, bottomLine], [0,6], 1);

To get any number, pass in the line number to polyval

y = polyval(coeffs, lineNumber);

[Dan]

That's great, it's really helping, thank you.

The only problem occurs when there is more than one blob in the same column. In the example above, col 4, it takes the top of the first blob and the base of the second. I need to know the top of the second blob as well.

[Image Analyst]

If you're going to have more than one blob per column, then you're going to have to use strfind():

>> a=[0 1 1 1 0 1 1 0 1 1]
a =
     0     1     1     1     0     1     1     0     1     1
>> strfind(a, [0, 1])
ans =
     1     5     8