Extract element from column vector with condition
3 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
Sagar Dhage
el 8 de Ag. de 2014
Comentada: Sagar Dhage
el 22 de Ag. de 2014
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 comentarios
Image Analyst
el 8 de Ag. de 2014
Editada: Image Analyst
el 8 de Ag. de 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?
Respuesta aceptada
Nir Rattner
el 8 de Ag. de 2014
Editada: Nir Rattner
el 8 de Ag. de 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);
Más respuestas (2)
Azzi Abdelmalek
el 8 de Ag. de 2014
Editada: Azzi Abdelmalek
el 8 de Ag. de 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
el 8 de Ag. de 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 comentarios
Image Analyst
el 12 de Ag. de 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.
Ver también
Categorías
Más información sobre Octave en Help Center y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!