increment occurrences of repeated elements by one - matlab

Question

v = [1,2,1,1,1,2,2]

1 and 2 are my two groups. I'm trying to increment each group by 1 every time there is a repetition

The desired result:

v2 = 
    1  1  2  3  4  2  3

I tried the functions cumsum and movsum with splitapply, but nothing has worked so far

Does this answer your question? [Count repeating integers in an array](https://stackoverflow.com/questions/54079558/count-repeating-integers-in-an-array) — obchardon, May 23 '23 at 15:28
@obchardon I knew I had seen this (or something very similar) somewhere! However, in your question all equal values are adjacent, which simplifies the problem a little and allows for more approaches. [This](https://codegolf.stackexchange.com/q/178500/36398) PPCG challenge, linked in your question, may be a more exact "duplicate", but of course it won't contain the most practical approaches... — Luis Mendo, May 23 '23 at 15:39

Luis Mendo · Accepted Answer · 2023-05-23T00:15:10.983

2

I am using different values of v for more clarity:

v = [10, 20, 10, 10, 10, 20, 20];

Memory-efficient solution, with a loop over the unique values of v:

[a, ~, c] = unique(v);
v2 = zeros(size(v));
for k = 1:numel(a)
    ind = c==k;
    v2(ind) = 1:sum(ind); 
end

Memory-inefficient (it builds an intermediate N×N matrix, N = numel(v)), but simpler:
```
v2 = sum(triu(v==v.'));
```

Another approach, probably memory-efficient too:

s = sparse(c, 1:numel(c), 1);
v2 = nonzeros(s.*cumsum(s, 2)).';

edited May 23 '23 at 00:15

answered May 23 '23 at 00:05

Luis Mendo

Your first solution is the same as mine, but somehow more complicated. :) – Cris Luengo May 23 '23 at 00:08
@CrisLuengo You are right, I could have avoided the third output of `unique` – Luis Mendo May 23 '23 at 00:10

score 2 · Answer 2 · answered May 23 '23 at 00:07

2

If you have few groups, you could iterate over the groups. As the number of groups increases this might become expensive though.

v = [1,2,1,1,1,2,2];

v2 = zeros(size(v));
for p = unique(v)
  i = v == p;
  v2(i) = 1:sum(i);
end

answered May 23 '23 at 00:07

Cris Luengo

score 1 · Answer 3 · answered May 23 '23 at 14:45

You say you only have 2 groups, so you can just add the cumulative sums of the Boolean when equal to one or the other

idx = (v==1);
v2 = cumsum(idx).*idx + cumsum(~idx).*(~idx);

For your example,

v = [1,2,1,1,1,2,2];
v2 = [1,1,2,3,4,2,3]

In the more generic case where you have more than two values, you can generalise this using implicit expansion at the cost of memory

idx = v == (unique(v).');
v2 = sum( cumsum(idx,2) .* idx );

3 Answers3