Filter löschen
Filter löschen

Help on for loop

2 Ansichten (letzte 30 Tage)
Mark
Mark am 12 Mär. 2013
I have an R vector and a U vector. The R vector contains mostly zeros but has a couple of values. The U vector is a shorter length than the R vector and contains no zeros. I want to make a U_final vector which has the values of the U vector at the same locations where the R vector values are non-zero. My attempted code is below. It iterates through fine, but the end result is the last value in U in all locations where R is non-zero rather than all values of U in the appropriate non-zero locations of R. I'm using U = [0.0137;0.0081] and R = [0;0;0;72;0;90]. I'd like the output to be U_final = [0;0;0;0.0137;0;0.0081]
for i = 1:length(R)
for j = 1:length(U)
if R(i,1) ~= 0
U_final(i,1) = U(j,1)
else U_final(i,1) == 0
end
end
end
  1 Kommentar
Walter Roberson
Walter Roberson am 12 Mär. 2013
Note that U_final(i,1) == 0 is a comparison statement, not an assignment.

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Teja Muppirala
Teja Muppirala am 12 Mär. 2013
No need to iterate. MATLAB makes this simple with logical indexing.
U = [0.0137;0.0081];
R = [0;0;0;72;0;90];
U_final = R;
U_final(R~=0) = U
But the number of nonzeros in R needs to be equal to the length of U.
  1 Kommentar
Mark
Mark am 12 Mär. 2013
Thanks Teja. I now see it as a quick replacement of the nonzero elements. No need to overcomplicate things is apparently the lesson for the day. ;) cheers!

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

Youssef  Khmou
Youssef Khmou am 12 Mär. 2013
Bearbeitet: Youssef Khmou am 12 Mär. 2013
hi try this , BUT the number of non zeros in R must be the same as the length of U:
R=zeros(100,1);
R(2:4:55)=rand; % R contains some values ( 14 elements !)
U=rand(14,1); % U is Shorter than R and contains non zeros vals
index=find(R~=0); %locations where R has non zeros .
U_final=zeros(size(R)); % this is optional
U_final(index)=U;

Produkte

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by