How do I extract the first non-zero value from array after specified number of zeros?

I have an array with 3 sections of positive values with zeros spread throughout and large sections of zeros in between the three sections (eg. [1,2,0,0,3,4,0,5,6,7,8,9,0,0,0,0,0,0,0,0,0,0,0,0,0,11,12,13,0,14,0,0,15,16,17,0,0,0,18,19,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,21,22,23,24,25,26,0,0,27,0,28,29]). Note: This isn't my actual array.

I want to extract only the first non-zero values from these three sections (eg 1,11,21 in array above) but can't figure out how to. I've tried messing around with if/or statements but can't get anything useful.

[Image Analyst]

I see you want to ignore short runs of 0s, like 1 or 2. What is the min size of the large zero runs? How do I know when the run of zeros is a "large" run? Should we just take the 3 longest runs? Does the vector ever start with a long run of zeros?

[Matt J]

How many consecutive zeros are needed before we can consider it the end of a section?

[Joshua Lucas Ewert]

For the example array here, I would say minimum 4, but for my actual array probably 20.

Answer 1

[Image Analyst]

Try this if you have the Image Processing Toolbox:

v = [1,2,0,0,3,4,0,5,6,7,8,9,0,0,0,0,0,0,0,0,0,0,0,0,0,11,12,13,0,14,0,0,15,16,17,0,0,0,18,19,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,21,22,23,24,25,26,0,0,27,0,28,29];
% Get a logical comparison to find non-zero elements.
vb = v ~= 0;
% Get rid of runs of 3 elements or less.
minRunLength = 4;
vb = ~bwareaopen(~vb, minRunLength);
% Find starting indexes of non-zeros after long runs of zeros.
indexes = [1, 1 + strfind(vb, [0, 1])]

indexes = 1×3

1 26 60

% OK, they're found.  Now let's print them out to the command window.
v(indexes)

ans = 1×3

1 11 21

[Joshua Lucas Ewert]

Thank you so much! It actually took a minRunLength of about 100 but I got my three sections' starts

Answer 2

[Matt J]

Using this FEX download,

https://www.mathworks.com/matlabcentral/fileexchange/78008-tools-for-processing-consecutive-repetitions-in-vectors

x=[1,2,0,0,3,4,0,5,6,7,8,9,0,0,0,0,0,0,0,0,0,0,0,0,0,11,12,13,0,14,0,0,15,16,17,0,0,0,18,19,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,21,22,23,24,25,26,0,0,27,0,28,29];

numzeros=5; %number of consecutive zeros constituting a section divider

[starts,stops,runlengths]=groupLims(groupTrue(~x),1);

idx=[1,stops(runlengths>numzeros)+1];


out=x(idx)

out =

1 11 21

[Joshua Lucas Ewert]

Thanks Matt! This worked great but I accepted Image Analyst's answer so I don't have to have extra function files

[Matt J]

Sure, but be aware that you do require additional toolboxes (Image Processing) for that answer to work.

[Joshua Lucas Ewert]

True true, I appreciate it

Answer 3

[Bruno Luong]

Basic for-loop programing

x=[1,2,0,0,3,4,0,5,6,7,8,9,0,0,0,0,0,0,0,0,0,0,0,0,0,11,12,13,0,14,0,0,15,16,17,0,0,0,18,19,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,21,22,23,24,25,26,0,0,27,0,28,29]

x = 1×71

1 2 0 0 3 4 0 5 6 7 8 9 0 0 0 0 0 0 0 0 0 0 0 0 0 11 12 13 0 14

k = 4; % number of 0s from which the sequence is considered as true sequence

j = [];
c = Inf;
for i=1:length(x)
    if x(i) ~= 0
        if c >= k
            j(end+1) = i;
        end
        c = 0;
    else
        c = c + 1;
    end
end
j

j = 1×3

1 26 60

x(j)

ans = 1×3

1 11 21