how to count array of 1 and 0 repeated times

I am tryig to make a counter that will look at a set of data that consists of ones and zeros and count how many times the number 1 is repeated 10 times consecutively and count that occurence as a 1 and any other occurence of a zero or less than 10 times as a zero.

Answer 1

[David Hill]

a=ones(1,10);
b=[];
for k=1:10
  b=[b,randi(2,1,20)-1,a];
end
% above just generates binary series
s=num2str(b);
s=s(s~=' ');
count=length(regexp(s,'[1]{10,}'));

Answer 2

[Matt J]

Using,

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

[~,~,runlengths]=groupLims(groupTrue(data==1),1);

count= nnz(runlengths>=10)

[Keaton Looper]

some of the functions don't work

[Matt J]

Possibly because you didn't download them? Here's an example,

e=ones(1,10);
data=[e,0,0,e 1 0 0 0 1 1];

[~,~,runlengths]=groupLims(groupTrue(data==1),1);

count= nnz(runlengths>=10)

count = 2

[Keaton Looper]

i already have a data set titled diff that contains set of ones and zeros

[Matt J]

Yes, but I don't. That's why I needed to make up my own data for the example.

[Keaton Looper]

so then would i just put

data = diff

[~,~,runlengths]=groupLims(groupTrue(data==1),1);

count= nnz(runlengths>=10)

so that it assigns my data set to the variable data?

[Matt J]

Yes, or you could just do as below, although I would advise you not to name your variable 'diff' since it conflicts with an existing Matlab function.

[~,~,runlengths]=groupLims(groupTrue(diff==1),1);
count= nnz(runlengths>=10)

[Keaton Looper]

what would i do if i wanted to add another condition to the count. Lets say if the the number one is repeated 10 times consecutively and if the respective visibility value is less than 2 miles?

[Matt J]

and if the respective visibility value is less than 2 miles

I don't think you've told us what this is.

[Keaton Looper]

So I have two conditions I need the counter to meet. When 1 is repeated consecutively 10 times and if visibility value from another data set is less than two miles. So I need it to take into account two conditions it needs to meet to count the event as a 1.

[Matt J]

Is the "visibility value" data set a vector the same size of "diff"? So <2 miles means the visibility vector elements are <2 ten times consecutively as well?

[Keaton Looper]

Yes the visibility vector is a different data set.

[Matt J]

Is the "visibility value" data set a vector the same size as the first data set? So <2 miles means the visibility vector elements are <2 ten times consecutively as well?

[Keaton Looper]

Yes

[Keaton Looper]

Both the vectors at the same length. The logically vector is the same length until the zeros are not included then it shortens

[Matt J]

[~,~,runlengths]=groupLims(groupTrue(data==1 & visibility<2),1);
count= nnz(runlengths>=10)

[Image Analyst]

"some of the functions don't work " <== if you don't want to use a File Exchange program and want to use a built-in function that will do it in a single line of code, see my answer below.

[Keaton Looper]

its giving me a count of zero when i know it should not be

[Image Analyst]

I gave an example below where I prove it worked. If it doesn't work for your vector, attach your vector in a text or mat file.

Answer 3

[Image Analyst]

If you have the Image Processing Toolbox, you can use the built-in bwlabel and do it in one line of code. If m is your matrix:

[~, count] = bwlabel(bwareafilt(logical(m), [10, inf]))

[Image Analyst]

Here is an example

m = zeros(1, 1000);
% Make 3 stretches of 10 or more:
m(100:110) = 1;
m(330:380) = 1;
m(560:590) = 1;
[~, count] = bwlabel(bwareafilt(logical(m), [10, inf]))

count = 3