Error in a for loop

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Nikolaos Zafirakis
Nikolaos Zafirakis il 14 Lug 2019
Risposto: SaiDileep Kola il 17 Lug 2019
I’m trying to run a loop through some measurement but I keep getting this error "Unable to perform assignment because the size of the left side is 3-by-1 and the size of the right side is 2-by-1." Does anyone know a way around this?
o = B(1:50,2);
for ind = 1:length(o)
a1(:,ind) = diff(o);
[c1(:,ind),d1(:,ind)] = find(a1(:,ind)>2^15);
[e1(:,ind),f1(:,ind)] = find(a1(:,ind)<-2^15);
o(c1(:,ind)+1:e1(:,ind)) = o(c1(:,ind)+1:e1(:,ind))-2^16; % Error Happens here
end
  4 Commenti
dpb
dpb il 14 Lug 2019
No idea what you're trying to explain, sorry.
Show us some data that illustrates what you're after with inputs and expected outputs and how you know those are the right answers given the input.
What is the end starting format and then the "the format I need" for the data? Bound to be a more effective way to code this if we just knew what the problem was/is...
Nikolaos Zafirakis
Nikolaos Zafirakis il 14 Lug 2019
Original data
plot(a).jpg
The result after 2 iterations of the code i showed you (I want to run what I showed you in a loop).
plotb.jpg

Accedi per commentare.

Risposte (1)

SaiDileep Kola
SaiDileep Kola il 17 Lug 2019
Hi,
I see that you get the error in 3rd line in the for loop not in the of 4rth line as you mentioned, I think your use case can be realized with the following code.
o = B(1:50,2);
for ind = 1:length(o)
a1 = diff(o);
c1 = find(a1>2^15);
e1 = find(a1<-2^15); %Error happens here
o(c1+1:e1) = o(c1+1:e1)-2^16; % Error doesn't occur here
End

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