Find indices of repetitive values

In array a=[1,1,1,2,3,4,3]. I want to find indices of repeating (duplicate) values. Finally want to get the repeating value. Short code with commands such as "unique" and "find" is expected.

[Jan]

What is the wanted output? Is this your homework? What have you tried so far?

Answer 1

[Jan]

With FEX: RunLength:

[B, N, Index] = RunLength(sort(a));
repeated      = B(N > 1)

This is the list of values, which occur more than once.

Or:

au        = unique(a);
[N, Edge] = histcounts(a, [au, au(end) + 1]);
repeated  = au(N > 1)

[Mukund]

Nice coding, however only repeat values are obtained and not their indices.(Run of 2nd code) Original vector a=[1, 1, 1, 2, 3, 4, 3]. The output expected is: the first repeat value is 1 at places 1, 2, 3 and the second value is 3 at places 5, 7. Can you please update the response accordingly.

[Image Analyst]

What if the "a" is array a=[1,1,1,2,3,4,3, 1,4,4,2,2,3] -- what is the output in this case?

Also you never answered his question about whether this is homework or not. And WHY do you want this?

[Mukund]

Let me be clear: this not homework which I have to show. I am writinga code in which I want to analyse the result. As a novice user I could not figure it out. Also coding observed in help file was not directly applicable. So this query is put here. Whatever is the input array, the output expected is to obtain the repeating value and its indices (original location) in input array. Hope this clarifies...

[Jan]

@mukund: It would be really helpful, if you post explicitly what you want as result for the given example. Not as text, because there are different interpretations of "and its indices", but in valid Matlab syntax, because this is unique. "the first repeat value is 1 at places 1, 2, 3 and the second value is 3 at places 5, 7." So do you want this:

value   = [1, 3]
indices = {[1,2,3], [5,7]}

?

Why is "short code expected" and what exactly is "short" enough?

Of course the readers can try to guess the details, but if these guesses are wrong, posting an answer is a waste of time.

The next guess:

[N, E] = histcounts(a, unique(a));
value  = E(N > 1);
index  = cell(1, numel(value));
for k = 1:numel(value)
  index{k} = find(a == value(k));
end

[Image Analyst]

mukund, no that does not clarify, it just says what you originally said. So in my example a=[1,1,1,2,3,4,3, 1,4,4,2,2,3] let's consider the 4th 1. Is that a repeat? It's a repeat of the 1's in the first group, but it's in its own group. And the 1 in the second group of 1 is not repeated in its group - there is just one 1, not multiple 1's in the second group. I asked for the output of my example but you did not give it so I don't know how to help you in the general purpose situation.

[Mukund]

@Jan- Your guess of my expectation is right and this is the required code. Many thanks. I myself tried it in many ways, but no success. Thanks again.

@Image Analyst- Sorry for communication gap from my side. However, Thanks to you for your efforts.

Answer 2

[Image Analyst]

Use findgroups():

a=[1,1,1,2,3,4,3]
groups = findgroups(a)

Then histogram it to find groups with more than 1 count.