How do I ignore the first few 0s in a vector but not ignore the 0s in the middle?

If i have the following vector: p = [0;0;0;0;1;2;5;4;6;0;0;0;6]

how would I code in such a way that I can ignore the first few zeros till a valid value (>0, in this case '1'). I don't want to remove/ignore the 0s in middle because I am going to be interpolating between the 2 valid values beside the 0s. so far, i have the following code:

p = [0;0;0;0;1;2;5;4;6;0;0;0;6]

for i = 1:13

    if p(i) == 0
        display('ignore')
    else
        currentValue = p(i)
        if i ~= 1
            prevValue = p(i-1)

            if(i<8)
                nextValue= p(i+1)
            end
            if (currentValue == 0 && prevValue ~= 0)
                previousindex = i-1
                speedUsed = prevValue
            end
            if currentValue == 0 %was nextValue == 0
                counter = counter+1
                i = i+1
            end

            if  prevValue == 0  && currentValue == 0 && nextValue ~= 0
                p([i+1-counter:i]) = interp(p(previousindex),nextValue,counter);
                % i = i+1

            end
        end
    end
end

but, obviously, every time p(i) = 0, it will ignore it....I'm sure I'm just missing something obvious but I really can't see it. I would appreciate any help!

Thank you

Answer 1

[Kelly Kearney]

To just drop the leading zeros, how about this?

p = p(find(p,1):end);

And is this what you're trying to achieve via the interpolation?

idx = 1:length(p);
newp = interp1(idx(p~=0), p(p~=0), idx);

[Mary]

Hi, thank you for the solution! however, is there anyway to do this without removing the 0s from a?

so if a = [0;0;0;0;1;2;5;4;6;7;0;0;0;2;5;0;2]

for i = 1:13 % go through a from a(1) till valid point. %start interpolation algorithm from valid point end

at the end of the for loop, a would still be a = [0;0;0;0;1;2;5;4;6;7;0;0;0;2;5;0;2]

thank you!

[Image Analyst]

Yes:

saved_a = a;
% Now your loop, then
a = saved_a; % Restore a.

Answer 2

[per isakson]

Hint:

   p1  = [0;0;0;0;1;2;5;4;6;0;0;0;6];
   ixf = find( p1>0, 1, 'first' );
   p2  = p1( ixf : end );

   >> p2'
   ans =
        1     2     5     4     6     0     0     0     6

[Mary]

Hi, thank you for the solution! however, is there anyway to do this without removing the 0s from a?

so if a = [0;0;0;0;1;2;5;4;6;7;0;0;0;2;5;0;2]

for i = 1:13 % go through a from a(1) till valid point. %start interpolation algorithm from valid point end

at the end of the for loop, a would still be a = [0;0;0;0;1;2;5;4;6;7;0;0;0;2;5;0;2]

thank you!