addition loops in matlab

assume i have a vector array:

a=[1 1 1 0 0 1 1 1]

how can i put loop such that if sum of any three successive elements is equal to 3, it prints 'ali'.

can it be done without loops?

[Austin Decker]

I'd have to think about a loop-less answer, but one way to get what you want is:

a = [1,1,1,0,0,1,1,1];
for i = 1:length(a)-2
    tmp = sum(a(i:i+2));
    if tmp == 3
        disp("ali");
    end
end

[ali hassan]

thanks a lot. i had the same in my mind. but i cant figure out for loop less answer.

[Austin Decker]

Well, MATLAB is likely looping behind the scenes, but you could do this:

a = [1,1,1,0,0 1,1,1];
ind1 = 1:length(a) -2;
ind2 = ind1 + 2;
result = arrayfun(@(x,y) sum(a(x:y)),ind1,ind2);
qty = sum(result == 3);
disp(join(repmat("ali",qty,1),newline));

Answer 1

[DGM]

This works easily enough. After R2016a, you can do the same with movsum().

a = [1 1 1 0 0 1 1 1];

if any(conv(a,[1 1 1],'valid')==3)
    fprintf('ali\n')
end

ali

[ali hassan]

@DGM what can be other possible ways?

[DGM]

Any sort of convolution or sliding window filter tool that implements some linear combination of elements. You could use conv(), imfilter() nlfilter(), movmean(), movsum(). I'm sure there are others.

[ali hassan]

thanks a lot @DGM. adding a bit twist here.

 a = [1 nan 1 1 0 0 nan 1 1 nan 1];

suppose it contains nan as well. how can i do the same and ignore nan?

if i use movsum, nan will be removed but the entry will still be considered. how can i also make it to ignore nan entry as well?

[DGM]

To omit nans:

a = [2 nan 1 1 0 0 nan 1 1 nan 1];

% using movsum
s = movsum(a,3,'omitnan');
any(s(2:end-1)==3)

ans = logical
   1

% using conv
a(isnan(a)) = 0;
s = conv(a,[1 1 1],'valid');
any(s==3)

ans = logical
   1

If the value range of a never extends beyond [0 1], then you could also simply treat any propagated NaNs as 0. This is simply because it wouldn't be possible to have a sum equal to 3 if the window size were also 3 and any element were anything other than 1.

a = [1 nan 1 1 0 0 nan 1 1 nan 1];

s = conv(a,[1 1 1],'valid') % this will pass NaN

s = 1×9

NaN NaN 2 1 NaN NaN NaN NaN NaN

any(s==3) % NaNs aren't equal to 3

ans = logical
   0