counting how many times a certain value is repeated in a fixed interval

Hi,

I need a help for a relatively simple calculation, but I keep failing to solve it.

Say I have the following data, where keeping the order is necessary:

X = [0,0,1,1,0,1,1,1,1,1,0,0]

and I want to know how many times '1' is repeated for 2x, and that two '1's is considered as 1 score. So score for X should be 3.

I have a relatively long data (1x74240) with 0, 1, and -1.

With the logic above, I need to know the score for when the data shows repeated 1 and repeated -1.

Any sort of input is appriciated.

Thanks

[Tommy]

Is the score 3 because of the following three 1s?

X = [0,0,1,1,0,1,1,1,1,1,0,0]
%          ^     ^   ^
%          1     +1  +1 = 3

Would this have a score of 4?

X = [0,0,1,1,0,1,1,1,1,1,1,0]
%          ^     ^   ^   ^
%          1     +1  +1  +1 = 4

And this a score of 2?

X = [0,0,1,1,0,1,1,1,0,1,0,0]
%          ^     ^
%          1     +1 = 2

[Finda Putri]

no it does not work as pair.. I just happened to randomly assigned two-times 1's as my criteria for an example.

X = [0,0,1,1,0,1,1,1,1,1,0,0]
%         ^1    ^1  ^1  ^this 1 no longer counts as 0 comes after that..  

Answer 1

[Matt J]

If you have the Image Processing Toolbox,

reg=regionprops(X,'Area');

score = sum(structfun(@(a) floor(a/2), reg))

Answer 2

[KALYAN ACHARJYA]

Multiple Ways: One way

X=[0,0,1,1,0,1,1,1,1,1,0,0];
idx=find(X==1)
for i=1:2:length(idx)-1
  data(i)=idx(i+1)-idx(i)
end 
score=sum(data)

[Finda Putri]

I find this only work if my criteria is two 1's?

Answer 3

[madhan ravi]

According to your comment

https://www.mathworks.com/matlabcentral/answers/555889-counting-how-many-times-a-certain-value-is-repeated-in-a-fixed-interval#comment_915925

No toolboxes/loops needed:

z = (X(1:2:end)==1) + (X(2:2:end)==1) % works for even number of elements only!
score = nnz(z==2) % works only for 2 repetitions!

[Finda Putri]

This works, but yes only when the criteria is two 1's as you said, for 2 repetitions.

Answer 4

[Finda Putri]

I found a way to answer this already, but in rather long code. Thank you very much for helping me.