Vectorization of For-If combination

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MAB2020
MAB2020 on 25 Jul 2020
Commented: MAB2020 on 27 Jul 2020
Requesting guidance on the proper vectorization methods for combinations of if statements within for loops. An example of the format referenced is included below. Code is utilizing a painful combination of if statements nested within for loops and appropriate vectorization is the only way in which this will run efficiency
X = zeros(1000,1)
C = 1:10
for A = 1:10
for B = 1:10
if C = 1
X(A*B*C) = 1;
elseif C = 2
X(A*B*C) = 2;
else
X(A*B*C) = 3;
end
end
end
  2 Comments
David Hill
David Hill on 25 Jul 2020
Your code does not make sense. C is an array that does not change, your code will only execute the else statement.
X = zeros(1000,1)
C = 1:10
for A = 1:10
for B = 1:10
if C == 1%need == for comparison, this is always false
X(A*B*C) = 1;
elseif C == 2%comparison C= [1 2 3 4 5 6 7 8 9 10], this will be false
X(A*B*C) = 2;
else
X(A*B*C) = 3;%this will execute but X(1:10)=3 for for loop
end
end
end
MAB2020
MAB2020 on 25 Jul 2020
% Correction, use the following code. Trying to generalize a much more complex segment.
% The overall objective is to gain a proper understanding of vectorization algorithms for
% matrix operations consisting of if statements nested within for loops
for A = 1:10
for B = 1:10
for C = 1:10
if C == 1
X(A*B*C) = 1;
elseif C == 2
X(A*B*C) = 2;
else
X(A*B*C) = 3;
end
end
end
end

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Accepted Answer

David Hill
David Hill on 25 Jul 2020
Can't think of any better way to do your problem. Only 1 loop now.
x=zeros(1000,1);
for i=1:10
x(i:i:10*i)=1;
x(2*i:2*i:20*i)=2;
x(3*i:3*i:30*i)=3;
x(4*i:4*i:40*i)=3;
x(5*i:5*i:50*i)=3;
x(6*i:6*i:60*i)=3;
x(7*i:7*i:70*i)=3;
x(8*i:8*i:80*i)=3;
x(9*i:9*i:90*i)=3;
x(10*i:10*i:100*i)=3;
end
  1 Comment
MAB2020
MAB2020 on 27 Jul 2020
Excellent, thank you very much for your input! I will implement this into my code and evaluate for relative performance improvement

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