# How to find the index location of repeated consecutive numbers over a tolerance within a vector

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Nicholas Hero on 25 Feb 2021
Commented: Nicholas Hero on 26 Feb 2021
For vector A, how would you find the index value for the start of a repeated consecutive value repeated more than x times
Example 1:
A = [4 2 7 4 4 7 9 9 3 8 8 8 8 8 8 5 6 6 7 ]
if x is 5 in this case I would like the answer to give 10.
Example 2:
For vector A attached using x = 7, would like the answer outputed to be 4249
Any sugesstions are much appreciated
Nicholas Hero on 26 Feb 2021
Thanks very much for your help David, works just how intended!

David Hill on 25 Feb 2021
This should work better. Did not previously think about multi-digit numbers.
n=7;%minimum number of repeats
a=num2str(~(diff(A)==0));
a=a(a~=' ');
idx=strfind(a,repmat('0',1,n-1));

David Hill on 25 Feb 2021
Likely lots of other ways.
n=5;%minimum number of repeats
idx=strfind(cell2mat(regexp(num2str(diff(A)),'[^- ]','match')),repmat('0',1,n-1));%idx(1) would be the first occurrance