filling a vector up with a second one

Hello,

My Problem: 2 vectors a and b. a=[0;0;0;0;0;5;5;5;4;4] b=[1;1;1;4;4;5;5;5;6;6;9;9;10;10;10;11]

My goal is to replace the 0's in a with the last entries of b and I have no idea how to realise this.

the resulting vector a should look like this: a=[9;10;10;10;11;5;5;5;4;4]

vector b is sorted thanks in advance for your participation!

[Azzi Abdelmalek]

Can you have a as [0;0;5;5;0;0;4;4;0;0]

Answer 1

[Azzi Abdelmalek]

 a=[0;0;0;0;0;5;5;5;4;4]
 b=[1;1;1;4;4;5;5;5;6;6;9;9;10;10;10;11]
 idx=find(~a);
 a(idx)=b(end-length(idx)+1:end)

Answer 2

[Image Analyst]

Try this:

lastZero = find(a==0, 1, 'last') % Find index of the last zero in a
a(1:lastZero) = b(end-lastZero+1:end) % Replace with elements from b

This will work with situations where a starts with a stretch of zeros and that is where all the zeros are. If you have a different situation, then you should say so, and give an alternate example.

Answer 3

[Matt]

Hi thanks a lot for your answers, they work both!

what could I do if a was

 a=[0;0;0;0;0;5;5;5;4;4;0;0]

and i like to receive:

a=[9;10;10;10;11;5;5;5;4;4;0;]

?

thx!

[Image Analyst]

You need to clarify the rule. The rule as I see it is the same as before except that if there is a stretch of zeros at the end you either lop off all but one, or shorten the ending stretch of zeros by one element.

[Azzi Abdelmalek]

 idx=find(~a)
 d=find(diff(idx)~=1,1);
 a(1:d)=b(end-d+1:end);
 test=0;
 k=0;
 while test==0
    if a(end-k)==0
        a(end)=[]
        k=1;
    else
        test=1;
    end
 end