What are the properties in a BoundingBox from regionprops?

I'm trying to figure out what coordinates are being stored from a BoundingBox. For instance

L=logical(bw);

box=regionprops(L,'Area', 'BoundingBox'); 
box(2)

And the output

ans = 

             Area: 127
      BoundingBox: [10.5000 11.5000 10 19]

What are those 4 values? Is it the [top left, top right, bottom left, bottom right] coordinates?

[abdelrahim hashem]

Please Sir, why the values of the two components x and y is real values? Also, the last two values width and heigh are integer values?

[Walter Roberson]

There is the question of whether coordinates represent pixel centers or pixel edges.

Answer 1

[Image Analyst]

It's [left, top, width, height]. Just be aware that the left and top are 0.5 pixels to the left and above, respectively, than the actual first column and first row of the binary image. It's because the Bounding box is defined such that it "contains" the blob, and if it ran right through the center of the top-most and left-most pixel, then that definition becomes somewhat ambiguous.

[Recap]

Thank you for the answer. But i dont understand what you mean by left and top?

[Recap]

Is that the row and column from where it starts?

[Walter Roberson]

Left is column. Top is row. The order is column followed by row, which is x followed by y, because x corresponds to columns and y corresponds to row.

[Recap]

@Walter Roberson thanks for clarifying that. I greatly appreciate it.

[Maham Khan]

Hi I want to save these 4 values of Bounding box for a set of images how can i do that .

[Image Analyst]

Try

props = regionprops(L, 'Area', 'BoundingBox'); 
t = [props.BoundingBox];
allX = t(1:4:end);
allY = t(2:4:end);
allWidths = t(3:4:end);
allHeights = t(4:4:end);

[Maham Khan]

Thankyou !!!!

[Ali]

how can i get greater boundingbox rectangle out of all detected boundingbox rectangles?

[Image Analyst]

See the attached function.

[Shuvo Sarker]

That was very good explanation!

[Muhammad Rakeh Saleem]

Hello,

I want to have the values of bounding box like this [xmin, ymin, xmax, ymax].

From the command i get like this [x y width heigth]. And one thing more i want to save the result as attached in the image. I want Origin to be shift from top left to bottom left that we usually tthink of on an axis.

[Image Analyst]

You get BoundingBox = xLeft, yTop, width, height] so to get the coordinates you want, do

xMin = ceil(BoundingBox(1))
xMax = xMin + BoundingBox(3) - 1
yMin = ceil(BoundingBox(2))
yMax = yMin + BoundingBox(4) - 1

To flip the image upside down, use

flippedImage = flipud(originalImage);

Note though that the first line will always be the first line, it will just contain what used to be the last/bottom line of the image.

[cpaniaguam]

@Image Analyst Could `floor` be used instead? That way the subtraction wouldn't be needed.

[Walter Roberson]

@cpaniaguam

Suppose BoundingBox(1) is 1 and BoundingBox(3) is 5 -- width 5 starting at 1.

IA's code would give xMin = ceil(1) --> 1, xMax = 1 + 5 - 1 --> 5. Using this as subscripts, YourImage(:,1:5) would indeed select 5 elements starting from position 1.

Your proposal is xMin = floor(1) --> 1, xMax = 1 + 5 --> 6. Using this as subscripts, YourImage(:,1:6) would be 6 elements, not 5.

Okay, what about if BoundingBox(1) is 0.9 and BoundingBox(3) is 5. Then IA's could would give xMin = ceil(0.9) --> 1, xMax = 1 + 5 - 1 --> 5, subscripts would be YourImage(:,1:5) which would be valid. Your proposal would be xMin = floor(0.9) --> 0, xMax = 0 + 5 --> 5, subscripts YourImage(:,0:5) which would be invalid because of the 0 subscript. If BoundingBox(1) had been (say) 2.9 instead then floor(2.9) --> 2, xMax = 2 + 5 --> 7, subscripts YourImage(:, 2:7) which would be 6 columns -- columns 2, 3, 4, 5, 6, 7

We see from here that no matter whether you use floor() or ceil() you still need to subtract 1 from xMax.

[cpaniaguam]

@Walter Roberson Thanks :)

[Image Analyst]

@cpaniaguam the bounding box is [xLeft, yTop, width, height], and xLeft and yTop are half a pixel outside of the left most and uppermost white pixel. xLeft and yTop always end in .5 since it's outisde the blob, not on the blob, and this is just the convention they use. So if you want the actual left column and top row of the blob, you must call ceil to round them up.

And the width and height are the width and height of the bounding box, which is larger than the actual blob by half a pixel all the way around. But the width and height are of the actual blob, not the bounding box of the blob. So if you want xRight and yBottom, you must add the width-1 and height -1 to the xLeft and yTop. See this example:

img = false(5,7);
img(2:4, 3:6) = true

img = 5×7 logical array
   0   0   0   0   0   0   0
   0   0   1   1   1   1   0
   0   0   1   1   1   1   0
   0   0   1   1   1   1   0
   0   0   0   0   0   0   0

props = regionprops(img, 'BoundingBox');
boundingBox = props.BoundingBox

boundingBox = 1×4

2.5000 1.5000 4.0000 3.0000

xMin = ceil(boundingBox(1))

xMin = 3

xMax = xMin + boundingBox(3) - 1

xMax = 6

yMin = ceil(boundingBox(2))

yMin = 2

yMax = yMin + boundingBox(4) - 1

yMax = 4

I hope that explains it better.

[cpaniaguam]

It certainly does, thanks @Image Analyst!

Answer 2

[elvis okacha]

I am trying to find length and width.is this ok?:

I=Imread(filename)
Info=imfinfo(filename)
Thres=graythresh(I)
I2=~(im2bw(I,thresh))
cmp=bwconncomp(I2)
s=regionprops(cmp,'BoundingBox')
x=s.BoundingBox(3)
y=s.BoundingBox(4)
res=info.ResolutionUnit
resX=info.XResolutionUnit
resY=info.YResolutionUnit

[elvis okacha]

Error 'field reference for multiple structure elements that is followed by more reference blocks is an error

[Walter Roberson]

Your image has multiple regions, so s is returning a nonscalar structure.

bb = vertcat(s.BoundingBox);
x  = bb(:,1);
y = bb(:,2);
width = bb(:,3);
height = bb(:,4);

These will each be vectors, reflecting the fact that you have multiple regions.

[elvis okacha]

Thanks

Answer 3

[SOURABH BHATTACHARYA]

The first two are the top left (x,y) coordinates while the later ones are for max width and height respectively along the respectives axes.