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Sorting sub matrices based on specific column value

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Raj Raj
Raj Raj on 20 Feb 2016
Commented: Raj Raj on 21 Feb 2016
Hello All, I try to optimize the following code. I have this example fib_dist_ang matrix. First column numbers are not relevant to discussion. second column values are sorted while third column are not sorted.
I wanted to sort the third column values. This sorting is based on same value in the second column and place the value closest to the theta_ran (e.g. theta_ran =7.0) value. During this sorting based on the third row, In this example, lets confine discussion to last three rows we have to change the all values in the row (like how SORTROWS() command does). Output I requested is as follows:
Notes: 1) I had used UNIQUE to find the first and last rows of the repetitions based on the second column value. 2) To find the closest value, I had used histc.
fib_dist_ang=%required output
2 0.008 5.2
22 0.008 6.2
42 0.014 4.7
43 0.016 7.2
3 0.016 2.2
23 0.016 6.7
fib_dist_ang= %input
22 0.008 6.2
2 0.008 5.2
42 0.014 4.7
23 0.016 6.7
43 0.016 7.2
3 0.016 2.2
%FINDING THE INDICES OF THE REPETATIONS
[~,uniq_first,~] = unique(fib_dist_ang(:,2), 'first');
[~,uniq_last,~] = unique(fib_dist_ang(:,2), 'last');
tic_sortang=tic;
for i2=1:size(uniq_first,1)
%SORTING BASED ON THE THRID COLUMN VALUES IN THE SUB MATRIX
ithrow=sortrows(fib_dist_ang(uniq_first(i2):uniq_last(i2),:),3);
%FINDING CLOSEST VALUE, IF AVAILABLE
val=theta_ran; f=ithrow(:,3);
if val > f(1) && val < f(end) % it can find the closest angle
[N,bin]=histc(val,f);%if val < f(1), then bin=0.
index=bin+1;
if abs(val-f(bin))<abs(val-f(bin+1))
fclosest=f(bin); index=bin;
else
fclosest=f(index);
end
% PLACING THE CLOSEST VALE TO THETA_RAN, IF AVALABLE
ithrow=[ithrow(index,:);ithrow(1:index-1,:);ithrow(index+1:end,:) ];
end
%COPY BACK THE SUB MATRIX INTO THE WHOLE MATRIX.
fib_dist_ang(uniq_first(i2):uniq_last(i2),:)=ithrow;
end
Only this loop is taking almost half of my total program time. Hence, I am looking for any possibility to optimize this program.

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Raj Raj
Raj Raj on 20 Feb 2016
Once we sort the third column, I am gonna search for a closest value to the theta_ran (in this example i defined theta_ran = 7, which is closest to 7.2). So I had placed 7.2 first.
For finding the closest value, I had used histc.
Azzi Abdelmalek
Azzi Abdelmalek on 20 Feb 2016
Ok, the closest is value is 7.2, after 7.2 it's 6.7, why in your result, you placed 2.2 in the second position?
Raj Raj
Raj Raj on 20 Feb 2016
Only first nearest value servers the purpose. I can place the 6.7 also after 7.2 and 2.2. But it is gonna take additional burden to code. So I am satisfied with 7.2, 2.2, 6.7.

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Accepted Answer

Guillaume
Guillaume on 20 Feb 2016
I've not really tried to understand your code, nor have I run it, but if all you want to do is sort a column according to its distance to a certain value, then you simply need to sort according to abs(colvalue - referencevalue), so:
a = [22 0.008 6.2
2 0.008 5.2
42 0.014 4.7
23 0.016 6.7
43 0.016 7.2
3 0.016 2.2];
theta_ran = 7;
b = a;
b(:, 3) = abs(a(:, 3) - theta_ran);
[~, order] = sortrows(b, [2 3]); %sort first by 2nd column, then by distance from theta_ran in 3 rd column
a = a(order, :) %reorder a

  2 Comments

Raj Raj
Raj Raj on 20 Feb 2016
I will try it out tomorrow and update you. It seems to give my expected result. Hope the values in the first column will also be changed due to the use of sortrows(). But not the values in the second column.
Raj Raj
Raj Raj on 21 Feb 2016
Thanks Guillaume. It served the purpose with simple logic.

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More Answers (1)

Azzi Abdelmalek
Azzi Abdelmalek on 20 Feb 2016
a=[22 0.008 6.2
2 0.008 5.2
42 0.014 4.7
23 0.016 6.7
43 0.016 7.2
3 0.016 2.2]
b=a;
b(:,4)=abs(b(:,3)-7)
[ii,jj,kk]=unique(a(:,2))
out=cell2mat(accumarray(kk,(1:numel(kk))',[],@(x) {sortrows(b(x,:),4)}))
out=out(:,1:3)

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Show 4 older comments
Raj Raj
Raj Raj on 20 Feb 2016
Yes Guillaume, You are correct. It should be sorted out as 6.2 and 5.2. Since, I was using histc previously, I was blinded.
Azzi Abdelmalek
Azzi Abdelmalek on 20 Feb 2016
a=[22 0.008 6.2
2 0.008 5.2
42 0.014 4.7
23 0.016 6.7
43 0.016 7.2
3 0.016 2.2]
b=a;
[ii,jj,kk]=unique(a(:,2));
q=accumarray(kk,(1:numel(kk))',[],@(x) {sortrows(a(x,:),3)});
s=cell2mat(q);
[~,idx]=min(abs(s(:,3)-7));
n=cumsum(cellfun(@(x) size(x,1) ,q));
jdx=find(idx<=n,1);
r=sortrows([q{jdx} abs(q{jdx}(:,3)-7)],4);
q{jdx}=r(:,1:3);
out=cell2mat(q)
Raj Raj
Raj Raj on 21 Feb 2016
Thanks Azzi for your revised code. It serves the purpose. However, this code is costly compared to accepted answer. However, I voted you for your efforts.

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