How to get the starting and ending index of repeated numbers in an array?

Hi MY array is =[2 2 1 2 3 2 2 2 2 2 3 5 6 7 7 7 7 6 5]

I am trying to find the starting and index index of repeated numbers in my array

for example,

for number 2, i expect startingindex=1 and endingindex=2 as it is repeated and also startingindex=6th and endingindex=10.

for number 7, i expect startingindex=14 and endingindex=17

I have been trying using find so far but in vain

Answer 1

[Jan]

An another question from the forum concerning a run-length-encoding. With FEX: RunLength:

X = [2 2 1 2 3 2 2 2 2 2 3 5 6 7 7 7 7 6 5];
[B, N, Ind] = RunLength(X);
Ind         = [Ind, length(X)+1];
Multiple    = find(N > 1);
Start       = Ind(Multiple);
Stop        = Ind(Multiple + 1) - 1;

[yashvin]

Is runlength a separate script? I am using MATLAB R2014b..i am getting an error..

[Guillaume]

Yes, it's a separate function (not a script). Jan did provide the link to it in his answer.

[Azzi Abdelmalek]

As indicated in the above answer, you can get RunLength function in this link http://www.mathworks.com/matlabcentral/fileexchange/41813-runlength

[yashvin]

Thanks!! works perfectly

Answer 2

[Ingrid]

try using the diff function before calling find

[yashvin]

diff(MY array)

0 -1 1 1 -1 0 0 0 0 1 2 1 1 0 0 0 -1 -1

it gives me little information about the starting and ending indices of repeated arrays!

[Guillaume]

diff is going to be the starting point of any answer related to finding repeated sequences. It transforms the repeated sequences into 0 and thus transform the problem from finding transitions between arbitrary numbers into finding transitions from non-zero to zero and back. The latter is much easier to solve.

[Stephen23]

@yashvin: notice that Ingrid mentioned two functions: diff and find. You only tried the first function, but you missed the find.

>> V = [2 2 1 2 3 2 2 2 2 2 3 5 6 7 7 7 7 6 5];
>> X = diff(V)~=0;
>> B = find([true,X]) % begin of each group
B =
     1     3     4     5     6    11    12    13    14    18    19
>> E = find([X,true]) % end of each group
E =
     2     3     4     5    10    11    12    13    17    18    19
>> D = 1+E-B % the length of each group
D =
     2     1     1     1     5     1     1     1     4     1     1

Use logical indexing if you only want the groups with more than one element:

>> Y = D>1;
>> B(Y)
ans =
     1     6    14
>> E(Y)
ans =
     2    10    17

[yashvin]

Let me check it and get back to you. Thanks for your input

Answer 3

[Andrei Bobrov]

A=[2 2 1 2 3 2 2 2 2 2 3 5 6 7 7 7 7 6 5];
ii = [0, diff(A(:)')==0,0];
i1 = strfind(ii,[0 1]);
i2 = strfind(ii,[1 0]);
out = [A(i1)',i1(:),i2(:)];

[yashvin]

I tried your solution. It is working perfectly so far.

[yashvin]

@ Andrei Bobrov

But i am new to strfind. According to

http://www.mathworks.com/help/matlab/ref/strfind.html

It looks for pattern and the example in the page treats pattern in string mostly. Can you please a little description of the code you had written. Thanks!

Answer 4

[Azzi Abdelmalek]

A=[2 2 1 2 3 2 2 2 2 2 3 5 6 7 7 7 7 6 5];
d1=~diff(A).*(1:numel(A)-1);
idx1=d1.*[1  ~d1(1:end-1)];
idx2=circshift(d1.*[~d1(2:end) 1],[0 1]);
ii1=~~idx1;
out=[A(ii1);idx1(ii1) ;idx2(~~idx2)+1];