Code covered by the BSD License

### Highlights from Region Growing (2D/3D grayscale)

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# Region Growing (2D/3D grayscale)

by

### Daniel (view profile)

14 Aug 2011 (Updated )

Recursive region growing algorithm for 2D/3D grayscale images with polygon and binary mask output

File Information
Description

A recursive region growing algorithm for 2D and 3D grayscale image sets with polygon and binary mask output. The main purpose of this function lies on clean and highly documented code.

Usage:
[P, J] = regionGrowing(cIM, initPos, thresVal, maxDist, tfMean, tfFillHoles, tfSimplify)

Inputs:
- cIM: 2D/3D grayscale matrix
- initPos: Coordinates for initial seed position
- thresVal: Absolute threshold level to be included
- maxDist: Maximum distance to the initial position in [px]
- tfMean: Updates the initial value to the region mean (slow)
- tfFillHoles: Fills enclosed holes in the binary mask
- tfSimplify: Reduces the number of vertices with line simplification

Outputs:
- P: VxN array (with V number of vertices, N number of dimensions). P is the enclosing polygon for all associated pixel/voxel
- J: Binary mask (with the same size as the input image) indicating 1 (true) for associated pixel/voxel and 0 (false) for being outside

Acknowledgements

Region Growing and Line Simplification inspired this file.

Required Products Image Processing Toolbox
MATLAB release MATLAB 7.12 (R2011a)
Other requirements Optional: Line Simplifcation by Wolfgang Schwanghart
04 Nov 2015 Nassir H. Salman

### Nassir H. Salman (view profile)

a lot of thanks Mr Daniel
, the algorithm of 2D/3D gray scale image is excellent , i tried it , it works correctly for segmentation, boundary, no. of pixel...but i tried to display the segmented area only with white background, there is some problem .so could you pls help me to do that.
thank u, Nassir

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04 Nov 2015 Nassir H. Salman

### Nassir H. Salman (view profile)

Matlab is great software and thanks to Mathworks

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08 Jun 2015 jack nn

### jack nn (view profile)

hi thanks.
I want to test this code.
I use this:
>> th = 0.35 * max(max(I));
>> [P, J] = regionGrowing(I, [30,50], th, 300, 'true', 'true', 'false');

but the result is this message:

RegionGrowing Opening: Initial position (30|50|1) with 0 as initial pixel value!
RegionGrowing Ending: Found 10390 pixels within the threshold range (418 polygon vertices)!

can you help me? Is there any mistake?

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20 Oct 2014 Lena Veis

### Lena Veis (view profile)

Nice work..

26 Sep 2014 Ander Biguri

### Ander Biguri (view profile)

Hi, I tried to run the code and the segmentation I get is in a circle shape and not exactly the shape of the segment.
Is this the way the code works? because according the picture it looks like a free shape

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10 May 2013 Daniel

### Daniel (view profile)

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25 Mar 2013 Isabel

10 Jan 2013 gang

### gang (view profile)

thks very much good job!

05 Dec 2012 Hanbo Chen

03 Aug 2012 Yang

### Yang (view profile)

13 May 2012 Omer Demirkaya

### Omer Demirkaya (view profile)

The following two statements should be replaced.
sumSQR = meanI*meanI; % sum of squares

stdI = sqrt((sumSQR - N*meanI*meanI)/(N-1)); % update standard deviation

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13 May 2012 Omer Demirkaya

### Omer Demirkaya (view profile)

Daniel good work.
I have a suggestion about an efficient mean and standard deviation calculations within your while loop.

Here is the inintilization of the parameters outside the loop.

meanI = regVal; % Mean intensity
sumSQR = 0; % sum of squares
stdI = 0; % Standard deviation
N = 1; % number of pixels

And then within your while loop after the statement,

J(xv+i, yv+j, zv+k) = true;

one can add the following lines of code to progessively calculate mean and standard deviation
gI = cIM(xv+i, yv+j, zv+k);
sumSQR = sumSQR + gI*gI;
meanI = (N*meanI + gI)/(N+1); % update mean
N = N+1; % increment number of pixels
stdI = (sumSQR - N*meanI*meanI)/(N-1); % update standard deviation

Omer.

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04 May 2012 Daniel

### Daniel (view profile)

Hello Michael. Unfortunately, that's true. I think the easiest way is to give over a handle to the desired axis as an additional input argument.

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04 May 2012 Michael

### Michael (view profile)

Hey Daniel, running the 2d example with no initial seed produces an error if multiple figures are open, because get(himage,'XData') returns a cell array, and is then used in axes2pix() which accepts only numerical arrays. I'm not sure of the best way to deal with this, aside from including a 'close all' in the code.

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03 Apr 2012 maram

### maram (view profile)

i want to segment a mammographic image(from mias database. after choosing the seed the result is like a curve!

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02 Apr 2012 Daniel

### Daniel (view profile)

Hello moram, could you please provide some further informations (e.g. what exactly are you trying to do)?

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02 Apr 2012 maram

### maram (view profile)

it works very well with the example associated!!but once i tried on another image i failed!

03 Mar 2012 Daniel

### Daniel (view profile)

The selection of a seeding position currently works only for 2D input images, since the function does not know, in which slice you may set the initial point. One (easy) way is to check whether the input image is 3D and show up the medial (or a random) slice.

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27 Feb 2012 Gomathi C

### Gomathi C (view profile)

when i try this code for a 3d image setting the initpos value as [] while calling, it gets the input seed point from me. But, unfortunately I get the following error.

??? Attempted to access initPos(3); index out of bounds because
numel(initPos)=2.

Kindly help me to figure out what I have to do to make the program run.

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01 Dec 2011 Ramachandran M.G.

### Ramachandran M.G. (view profile)

good work

01 Dec 2011 Ramachandran M.G.

### Ramachandran M.G. (view profile)

19 Nov 2011 Daniel

### Daniel (view profile)

Hello marwa. I'm sorry, i am not sure what you mean with "the same characteristics". Do you mean within the same intensity-range? You are welcome to contact me via mail. With regards, Daniel

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this is a good work, but i need to ask you, if there is a possibility to make this code search on all the image about all the separated regions in the image that have the same charactristics.

29 Aug 2011 Jerod Rasmussen

### Jerod Rasmussen (view profile)

This is great! Worked intuitively right out of the box! Could use an academic reference if available.