Extract element from column vector with condition

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I have column vector
P=[ 0; 212; 111; 190; 0; 223; 123; 200; 189; 219; 0; 190; 175; 202; 167; 181
....and so on]
I want to extract the 2nd and 3rd elements from the left 0 element IF LENGTH BETWEEN TWO 0 (the left 0 and right 0) is GREATER THAN 3. Save 2nd position elements in P2 column vector and 3rd position elements in P3 column vector. The result should be
After extraction, the remaining elements should be placed into P1 such as
P1 = [ 0; 212; 111; 190; 0; 223; 189; 219; 0; 190; 167; 181 ....and so on]
P2 = [123; 175]
P3 = [200; 202]
  7 Comments
Sagar Dhage
Sagar Dhage on 8 Aug 2014
Oh sorry, extract 2nd and 3rd position element if IF LENGTH BETWEEN two 0 is GREATER THAN 3.
Image Analyst
Image Analyst on 8 Aug 2014
Edited: Image Analyst on 8 Aug 2014
It might be clearer if you said "is 4 or more" rather than is "greater than 3". So the left run has 121, 111, 190 and that's 3, but not more than 3 (not 4 or greater) so those are left alone and get transferred to P1.
It's fairly easy if you have the Image Processing Toolbox. Do you?

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Accepted Answer

Nir Rattner
Nir Rattner on 8 Aug 2014
Edited: Nir Rattner on 8 Aug 2014
You can use the “find” function to find the indices of your zeros and then subtract each index by the one before to make sure the distance between them is not too large.
P=[ 0; 212; 111; 190; 0; 223; 123; 200; 189; 219; 0; 190; 175; 202; 167; 181; 0;];
Pzeros = find(P == 0);
violators = Pzeros(find(Pzeros(2 : end) - Pzeros(1: end - 1) > 4));
P1 = P;
P1([violators + 2; violators + 3]) = [];
P2 = P(violators + 2);
P3 = P(violators + 3);

More Answers (2)

Azzi Abdelmalek
Azzi Abdelmalek on 8 Aug 2014
Edited: Azzi Abdelmalek on 8 Aug 2014
Edit
P=[ 0; 212; 111; 190; 0; 223; 123; 200; 189; 219; 0; 190; 175; 202; 167; 181 ]
ii=strfind(P'~=0,[0 1])
idx=ii([diff([ii numel(P)])]>4)
P1=P'
jj=[idx+2 idx+3]
P1(jj)=[]
P2=P(idx+2)'
P3=P(idx+3)'

Image Analyst
Image Analyst on 8 Aug 2014
Method using the Image Processing Toolbox:
P=[ 0; 212; 111; 190; 0; 223; 123; 200; 189; 219; 0; 190; 175; 202; 167; 181]
P2 = [];
P3 = [];
measurements = regionprops(P~=0, P, 'PixelValues', 'Area')
for k = 1 : length(measurements)
% Show length of this stretch in the command window:
fprintf('Area of stretch #%d = %d\n', k, measurements(k).Area);
% Tack onto P2 and P3 if the length >= 4.
if measurements(k).Area >= 4
P2 = [P2, measurements(k).PixelValues(2)];
P3 = [P3, measurements(k).PixelValues(3)];
end
end
% Show in command window:
P2
P3
  3 Comments
Image Analyst
Image Analyst on 12 Aug 2014
I'm leaving on a week long trip in an hour. It looks like you've accepted an answer so then just go with that.
Sagar Dhage
Sagar Dhage on 22 Aug 2014
I have modified your programme.. nw its gives wat I want..Thank you.
% measurementsX = regionprops(Px~=0, Px, 'PixelValues', 'Area') % for k = 1 : length(measurementsX)
for k = 1 : length(measurementsX) % Show length of this stretch in the command window: %fprintf('Area of stretch #%d = %d\n', k, measurementsX(k).Area); % Tack onto P2 and P3 if the length >= 4. if measurementsX(k).Area == 1 P1X_1 = [P1X_1, measurementsX(k).PixelValues(1)]; end end
for k = 1 : length(measurementsX) % Show length of this stretch in the command window: %fprintf('Area of stretch #%d = %d\n', k, measurementsX(k).Area); % Tack onto P2 and P3 if the length >= 4. if measurementsX(k).Area == 3 P1X_3 = [P1X_3, measurementsX(k).PixelValues(1)]; P2X_3 = [P2X_3, measurementsX(k).PixelValues(2)]; P3X_3 = [P3X_3, measurementsX(k).PixelValues(3)]; end end
for k = 1 : length(measurementsX) % Show length of this stretch in the command window: %fprintf('Area of stretch #%d = %d\n', k, measurementsX(k).Area); % Tack onto P2 and P3 if the length >= 4. if measurementsX(k).Area ==5 P1X_5 = [P1X_5, measurementsX(k).PixelValues(1)]; P4X_5 = [P4X_5, measurementsX(k).PixelValues(2)]; P5X_5 = [P5X_5, measurementsX(k).PixelValues(3)]; P2X_5 = [P2X_5, measurementsX(k).PixelValues(4)]; P3X_5 = [P3X_5, measurementsX(k).PixelValues(5)]; end end

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