CODEWITHSUNDEEP
matlaboctave
LingxB
LingxB

Reputation: 497

Octave Replace elements in a vector under certain circumstances

I have two vectors as below:

p = zeros(5,1);
hx = [0.1; 0.3; 0.7; 0.9; 0.2];

the task is to replace elements in p from 0 to 1 if elements in hx >=0.5. Expeted output:

p =

   0
   0
   1
   1
   0

It can be achieved by below code, what I don't understand is: as pos = find(hx >= 0.5); gives a 2D vector, how to understand p(pos,1)=1;? How could this final line of code knows which index of p corresponding to the right element in pos? There seems no obvious connection between those two. On the other hand, how could this be done through for loop and if statement?

pos = find(hx >= 0.5);  
p(pos,1)=1;

Upvotes: 3

Views: 6363

Answers (4)

vendrick17
vendrick17

Reputation: 23

The easiest solution I could find without loops is this:

p = hx >= 0.5

It will output a vector of the same size as hx but with ones where the condition was met.

Upvotes: 1

Charles
Charles

Reputation: 255

This can be solved in a single line of code. But make sure p and hx have equal number of elements.

p(hx>=0.5)=1;

Upvotes: 1

Akmal Mukhsimov
Akmal Mukhsimov

Reputation: 11

I had the same problem and solved using following code snippet:

VectorWithZeros = VectorWithZeros + ReplaceWithScalar * (VectorWithZeros == 0)

Upvotes: 0

Jonas
Jonas

Reputation: 74930

find returns the a list of (linear) indices where the condition in the parentheses is true. In your case, this would be [3;4], since the condition is satisfied in element 3 and 4.

The second line sets the elements with rows indicated by pos and column 1 to 1.

You could do a loop

for idx = 1:length(hx);
    if hx(idx) >=0.5
       p(idx,1) = 1;
    end
end

but this would be very un-Matlab/Octave. Much nicer would be

p(hx>=0.5) = 1;

which avoids the detour via find

Upvotes: 7

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