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How to output a vector/array from a created function

Asked by Omar Sinno on 19 Jun 2019
Latest activity Edited by Stephen Cobeldick on 21 Jun 2019
In short, I have an array that has numbers from 0 to 100 using a step of 0.01, I created a function that is supposed to output another array containing 1's and -2's. While looping on the indices of my first array if the number is smaller than 40, I insert 1 in my output array, otherwise I insert a -2. I don't know what's the problem in my code:
function [outputVector] = myOutput(inputVector)
outputVector = [];
for i=1 : size(inputVector)
if inputVector(i) < 40
outputVector = [outputVector,1] ;
else
outputVector = [outputVector,-2];
end
end
end

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2 Answers

Answer by Shwetank Shrey on 20 Jun 2019
Edited by Shwetank Shrey on 21 Jun 2019
 Accepted Answer

The vector that you would be creating using 0 : 0.01 : 100 would have a size of 1x10001.
>> size(0 : 0.01 : 100)
ans =
1 10001
Your for loop on the size returns the first dimension whereas you require the second dimension.
Changing
for i=1 : size(inputVector)
to
for i=1 : size(inputVector, 2)
should work for you.

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Thanks. I wasn't aware of this and will edit the answer accordingly.

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Answer by James Tursa
on 19 Jun 2019
Edited by James Tursa
on 19 Jun 2019

Don't use size(inputVector) for your loop indexing limits since this is a vector. Use numel(inputVector) instead.
Also, you shouldn't be increasing the size of outputVector iteratively inside your loop. Each time you do that, the memory for the variable has to get copied to new memory. This can hurt the performance in a massive way. Instead, pre-allocate and use indexing. E.g., instead of this:
outputVector = [];
:
outputVector = [outputVector,1] ;
do this:
outputVector = zeros(size(inputVector));
:
outputVector(i) = 1;
There are also ways to calculate your result without using a loop.

  3 Comments

My problem is solved but I'd love to know how to do it without a loop.
"I'd love to know how to do it without a loop."
Not only are these simpler than your code, they will also be much more efficient.
Method one: basic arithmetic:
>> V = randi([1,100],1,17)
V =
93 45 83 48 1 41 14 93 43 8 41 64 46 10 21 48 65
>> 3*(V<40)-2
ans =
-2 -2 -2 -2 1 -2 1 -2 -2 1 -2 -2 -2 1 1 -2 -2
Method two: basic indexing:
>> X = [-2,1];
>> X(1+(V<40))
ans =
-2 -2 -2 -2 1 -2 1 -2 -2 1 -2 -2 -2 1 1 -2 -2
A different way that doesn't require logical arithmetic but just logical indexing:
V = randi([1, 100], 1, 17);
result = ones(size(V));
result(~(V < 40)) = -2;
% or
result = repmat(-2, size(V));
result(V < 40) = 1;

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