Clearing the upper section of a 3-dimensional matrix
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Given a stack of 2D-images, such that they form an x-by-y-by-z volume data set, I would like to delete any data that lies above (column-wise) a set of predetermined column indexes.
For example if:
A(:,:,1) = [1 1 1 A(:,:,2) = [1 1 1 A(:,:,3) = [1 1 1
1 1 1 1 1 1 1 1 1
1 1 1] 1 1 1] 1 1 1]
And the column indexes are defined such as (Where rows correspond to different images):
B = [3 2 1
1 2 3
3 3 3]
The result should be:
C(:,:,1) = [0 0 0 C(:,:,2) = [0 0 0 C(:,:,3) = [0 0 0
0 0 1 1 0 0 0 0 0
0 1 1] 1 1 0] 0 0 0]
My current method involves the use of for-loops, looping through each image and x-position individually, but this proves to be very slow when dealing with large sets of image data. My second thought was to use linear indexing, by converting the column indexes into an array of linear indexes I can easily generate a 3D matrix with 0's at the required column index, but the issue remains of removing the data above those points.
Is there a less computationally intensive method of tackling this problem that I'm missing?
Thanks for any assistance with the matter! :)
3 Comments
Image Analyst
on 22 Feb 2014
Well if B(1,1) = 3, then any elements in A(1,1) above plane 3 should be set to zero. But there are none, since A is only 3 planes high - there are no planes above 3 for A. So C(,1,) should be the same as A for all planes. So C(1,1,:) should equal 1,1,1, which is what A is in the (1,1) column. Yet your C does not show that so I guess I misunderstood or you didn't explain it correctly.
Accepted Answer
David Young
on 21 Feb 2014
Edited: David Young
on 22 Feb 2014
Try
Bt = permute(B, [3 2 1]);
mask = bsxfun(@ge, Bt, (1:size(A,1)).');
C = A;
C(mask) = 0;
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