Find position of the first place a value is reapeted.

I have a large array larger than this, but I use it as an example. a=[0.9; 1; 0.6; 1.1; 0.2; 0.6; 1.1; 1; 1; 1; 1; 0.5; 0.2].

I want to find the index of the place where a number is reapeted severeal times and after each other.

So I want to find position 8.

0.6 is repeted but not after each other. 1 is the right value but I don't need it to find position 2.

Hope you understand my question. The numbers actually represent a change in number of points from a point cloud

Answer 1

[Tommy]

How about

i = find(diff(a)==0,1);

[FV]

what if it needs to be repeted three times?

[Tommy]

Try this:

i=find([diff(a,2)==0;0]&diff(a)==0,1)

On your first example:

>> a=[0.9; 1; 0.6; 1.1; 0.2; 0.6; 1.1; 1; 1; 1; 1; 0.5; 0.2];
>> i=find([diff(a,2)==0;0]&diff(a)==0,1)

i =

     8

And your second example:

>> a=[1; 2; 3; 3; 2; 4; 2; 1; 4; 6; 7; 8; 1; 1; 1; 1; 9];
>> i=find([diff(a,2)==0;0]&diff(a)==0,1)

i =

    13

Answer 2

[David Hill]

find(diff(a)==0,1);

[David Hill]

i=find(diff(a,2)==0,1);%repeated three times

[FV]

It dosn't work for me.

[FV]

It works for the array I posted but not if I try another one, like this

a=[1; 2; 3; 3; 2; 4; 2; 1; 4; 6; 7; 8; 1; 1; 1; 1; 9];

Answer 3

[Adam Danz]

This problem is not trivial, especially when working with floating decimals that potentially run into round-off error.

Here's a method below that returns the starting index of consecutve numbers with at least nConsec ocurrences in a row.

See inline comments for details.

% Create input vector (row or column)
a = [.9; 0.9; 1; 0.6; 1.1; 0.2; 1.1; 1.1; 1.1; 0.6; 1.1; 1; 1; 1; 1; 0.5; 0.2];

% Minimum number of consecutve values
nConsec = 3; 

% Find consecutive numbers.  This considers two numbers equal if their difference
% is less than 0.0000001.
% If you want to use exact matches: iseq = abs(a(1:end-1) - a(2:end)) == 0; 
iseq = abs(a(1:end-1) - a(2:end)) < 0.0000001;  

% Compute the index of the first of each consecutive string. 
cLab = cumsum(diff([0;iseq(:)]) == 1) .* iseq(:);  %concesutive labels
[counts,bin] = histcounts(cLab,1:max(cLab)+1);
startIdx = arrayfun(@(i)find(cLab==i,1,'first'), bin([counts+1>=nConsec,false]));

result

startIdx =
     7    12