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

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]

Thanks very much for your help David, works just how intended!

Answer 1

[David Hill]

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));

[Lauren Crew]

I hate to be that ignorant person, but I'm getting an error in strfind. Says "First argument must be text". But isn't text? I'm just trying to replicate this with my own repeated data. Thanks

[Voss]

@Lauren Crew: Can you save your A variable (or whatever variable you're trying to operate on) to a mat file and upload it here using the paperclip icon?

Also, please post the code you're running that gives the error.

Answer 2

[David Hill]

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

[Nicholas Hero]

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.