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How to identify contiguous regions between two thresholds?

조회 수: 9(최근 30일)
K E 6 Feb 2013
If I have a variable z at mapped locations x and y, is there way to obtain the x-y locations of the contour(s) enclosing any continuous region in which z is between two thresholds minZ and maxZ? I do not have the Image Processing toolbox but this task may be similar to image segmentation. Below I use contourf to obtain the x,y locations of the region's outer edge, but perhaps there is a better approach.
% Make a fake dataset
z = peaks;
z = z(1:25, :); % Let's make different numbers of x and y elements for clarity
y = 1:size(z, 1);
x = 1:size(z, 2);
% Desired thresholds for the region
minZ = 2;
maxZ = 3;
[cc, hh] = contourf('v6', x, y, z, [minZ maxZ]); % Must use version 6
for iContour = 1:length(hh)
% Obtain the x and y location of the region's outer edges
% Does not account for cutouts inside the region
if get(hh(iContour), 'cdata') == minZ
xOutline = get(hh(iContour), 'xdata');
yOutline = get(hh(iContour), 'ydata');
% Find the points inside the region
isInside = inpolygon(x, y, xOutline, yOutline);
delete(hh); % Get rid of contours

채택된 답변

ChristianW 6 Feb 2013
편집: ChristianW 6 Feb 2013
[C,h] = contour(...)
The ContourMatrix C contains directly the infomation you are looking for. Just look up ContourMatrix in the Matlab help.
Here is an example plot for the first line:
Z = peaks;
[C,h] = contour(Z,[2 3]); title('contour'); ax_c = gca;
hold on; title('proof')
plot( C(1,(1:C(2,1))+1) , C(2,(1:C(2,1))+1) )
  댓글 수: 1
K E 7 Feb 2013
Great, now I not dependent on the v6 version of contour. Thanks!

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추가 답변(1개)

Image Analyst
Image Analyst 6 Feb 2013
Explain what "identify" means. You can of course just make a binary image of where that criteria is true like this:
zIsInRange = (z >= minZ) & (z <= maxZ);
But this binary image could have several blobs on it. That's why I ask you what "identify" means to you. Normally if you had the Image Processing Toolbox, you'd call bwlabel() or bwconncomp().
  댓글 수: 3
K E 6 Feb 2013
Yes, I can do what I want with the 'v6' version of contourf. But that doesn't seem so stable!

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