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

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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

Accepted Answer

David Hill
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));

More Answers (1)

David Hill
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
  1 Comment
Nicholas Hero
Nicholas Hero on 25 Feb 2021
Thanks David, this method seems to work on a small vector but is somehow tripping up when used with a larger vector, have attached the vector A into the question, using n = 7 it is outputing index values which are not in the vector.

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