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How can I write this with a for loop?

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SAHIL SAHOO
SAHIL SAHOO on 14 Feb 2023
Edited: Dyuman Joshi on 14 Feb 2023
r = (1/70).*( exp(i.*Y(:,3)) + exp(i.*Y(:,6)) + exp(i.*Y(:,9)) + exp(i.*Y(:,12)) + exp(i.*Y(:,15)) ...
+exp(i.*Y(:,18)) +exp(i.*Y(:,21)) +exp(i.*Y(:,24)) + exp(i.*Y(:,27)) + exp(i.*Y(:,30)) + exp(i.*Y(:,33)) ...
+ exp(i.*Y(:,36)) + exp(i.*Y(:,39)) +exp(i.*Y(:,42)) + exp(i.*Y(:,45)) + exp(i.*Y(:,48)) + exp(i.*Y(:,51)) ...
+ exp(i.*Y(:,54))+ exp(i.*Y(:,57)) + exp(i.*Y(:,60)) + exp(i.*Y(:,63)) + exp(i.*Y(:,66)) + exp(i.*Y(:,69)) ...
+ exp(i.*Y(:,72)) + exp(i.*Y(:,75)) + exp(i.*Y(:,78)) + exp(i.*Y(:,81)) + exp(i.*Y(:,84)) + exp(i.*Y(:,87)) ...
+ exp(i.*Y(:,90)) + exp(i.*Y(:,93)) + exp(i.*Y(:,96)) + exp(i.*Y(:,99)) + exp(i.*Y(:,102)) + exp(i.*Y(:,105))...
+ exp(i.*Y(:,108)) + exp(i.*Y(:,111)) + exp(i.*Y(:,114))+ exp(i.*Y(:,117)) + exp(i.*Y(:,120)) + exp(i.*Y(:,123))...
+ exp(i.*Y(:,126)) + exp(i.*Y(:,129)) + exp(i.*Y(:,132)) + exp(i.*Y(:,135)) + exp(i.*Y(:,138)) + exp(i.*Y(:,141))...
+ exp(i.*Y(:,144)) + exp(i.*Y(:,147)) + exp(i.*Y(:,150)) + exp(i.*Y(:,153)) + exp(i.*Y(:,156)) + exp(i.*Y(:,159)) ...
+ exp(i.*Y(:,162)) + exp(i.*Y(:,165)) + exp(i.*Y(:,168)) + exp(i.*Y(:,171)) + exp(i.*Y(:,174)) + exp(i.*Y(:,177)) ...
+ exp(i.*Y(:,180)) + exp(i.*Y(:,183))+ exp(i.*Y(:,186)) + exp(i.*Y(:,189)) + exp(i.*Y(:,192)) + exp(i.*Y(:,195)) ...
+ exp(i.*Y(:,198)) + exp(i.*Y(:,201))+ exp(i.*Y(:,204)) + exp(i.*Y(:,207)) + exp(i.*Y(:,210)));

Answers (2)

Dyuman Joshi
Dyuman Joshi on 14 Feb 2023
Edited: Dyuman Joshi on 14 Feb 2023
Vectorization would be the best approach here -
r = (1./70).*sum(exp(i*Y(:,3:3:210)),2)

Dr. W. Kurt Dobson
Dr. W. Kurt Dobson on 14 Feb 2023
Looks like the index you are using starts with 3, then increments by 3 up to 210.
Y looks like a matrix...
So, let's say the incrementing number is k, therefore you could do a for loop or just a vector computation.
For Loop Version (slower due to loop)
%
r = (1/70).*( exp(i.*Y(:,3)); % setup the first term
for k = 6:3:210 % increment by three
r = r + exp(i.*Y(:,k)); % for each term add, the prior to current
end
%
Vector Version (much faster)
k = 6:3:210; % gen k vector start at 3, increment by 6 to 210
r = (1/70).*( exp(i.*Y(:,3)); % setup the first term
r = r + exp(i.*Y(:,k)); % vector calculation on remaining terms
  1 Comment
Dr. W. Kurt Dobson
Dr. W. Kurt Dobson on 14 Feb 2023
Oops, forgot a semicolon,
k = 6:3:210; % gen k vector start at 3, increment by 6 to 210
r = (1/70).*( exp(i.*Y(:,3))); % setup the first term
r = r + exp(i.*Y(:,k)); %

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