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

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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

Accepted Answer

Jan
Jan on 22 Jun 2015
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;
  4 Comments

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More Answers (3)

Ingrid
Ingrid on 22 Jun 2015
try using the diff function before calling find
  4 Comments
Stephen23
Stephen23 on 22 Jun 2015
Edited: Stephen23 on 22 Jun 2015
@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

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Andrei Bobrov
Andrei Bobrov on 22 Jun 2015
Edited: Andrei Bobrov on 22 Jun 2015
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(:)];
  2 Comments
yashvin
yashvin on 26 Jun 2015
@ Andrei Bobrov
But i am new to strfind. According to
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!

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Azzi Abdelmalek
Azzi Abdelmalek on 22 Jun 2015
Edited: Azzi Abdelmalek on 22 Jun 2015
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];

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