# How do I find no. of instances of each value till it changes?

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Advay Pawar on 18 Jul 2018
Commented: Rik on 18 Jul 2018
I have the following input: [4 4 4 4 4 4 0 0 0 4 4 4 4 4 4 0 0 0 0 0 0 0 0 4 4 4 4 4];
I want MATLAB to give me the number of 4s and 0s in each set till the digit changes. So I want my output to be: [6 3 6 8 5].
Thank you!

Rik on 18 Jul 2018
Edited: Rik on 18 Jul 2018
You can use the runlength function that Jan made.
Another option is using the code below.
input = [4 4 4 4 4 4 0 0 0 4 4 4 4 4 4 0 0 0 0 0 0 0 0 4 4 4 4 4];
diff([0 find(diff(input)) numel(input)])

Paolo on 18 Jul 2018
Edited: Paolo on 18 Jul 2018
This can be achieved nicely with regular expressions. Capture a digit and backreference back to the same digits to find the patterns of consecutive numbers.
input = [4 4 4 4 4 4 0 0 0 4 4 4 4 4 4 0 0 0 0 0 0 0 0 4 4 4 4 4];
num = regexp(char(input+'0'),'(\d)\1+','match');
n = cellfun(@numel, num)
>>n
6 3 6 8 5

Rik on 18 Jul 2018
You can run the code yourself as well, but here it is (on another pc now). Your second solution is indeed a bit faster (~25%).
input = [4 4 4 4 4 4 0 0 0 4 4 4 4 4 4 0 0 0 0 0 0 0 0 4 4 4 4 4];
t1=0;t2=0;t3=0;
for k=1:10000
tic
n=diff([0 find(diff(input)) numel(input)]);
t1=t1+toc;
tic
num = regexp(char(input+'0'),'(\d)\1+','match');
n=cellfun(@numel, num);
t2=t2+toc;
tic
num = regexp(char(input+'0'),'(\d)\1+');
n=diff([num numel(input)+1]);
t3=t3+toc;
end
disp(t1)%0.0298 sec
disp(t2)%0.2929 sec
disp(t3)%0.2220 sec
Paolo on 18 Jul 2018
Cheers! Your solution is indeed very efficient. Reason as to why I asked you to do it is because I am on mobile and do not have pc access.
Rik on 18 Jul 2018
It's a not-uncommon question, so I've seen some solutions before. My first instinct with the first I saw was a for-loop, so yours is miles better already.
And yeah, it's a bit much to copy this code to Matlab online or the app, just to have a benchmark, so I fully understand ;)