Sort a n array for maximum number of recurring values and storing the indexes corresponding to erased components

1 vue (au cours des 30 derniers jours)
Hi,
i'm working with an array of thousands of elements and i've to limit the repeated values to 10.
Let say:
a=[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,2,0,0,0,0,0,5,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,9];
has to become:
[0,0,0,0,0,0,0,0,0,0,2,0,0,0,0,0,5,0,0,0,0,0,0,0,0,0,0,0,9].
And i want to store the indexes corresponding to the erased 10+-times recurrent values.
Any suggestion would be appreciated.
Thanks in advance,
Alessandro
  2 commentaires
Matt J
Matt J le 20 Jan 2021
And i want to store the indexes corresponding to the erased 10+-times recurrent values.
If I have a repitiion of a number, say [2 2 2 2 2] and I reduce it to [2,2,2,2] clearly I have erased one of the 2's. But which one? The first? The last? One of them in the middle?
Alessandro Togni
Alessandro Togni le 20 Jan 2021
Modifié(e) : Alessandro Togni le 20 Jan 2021
The 10th, 11th, etc of each list of 10+ occurrences!
Could be useful to say that i have to edit other vector of the same lenght in the same way on the one we're talking about.

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Réponse acceptée

Matt J
Matt J le 20 Jan 2021
Modifié(e) : Matt J le 20 Jan 2021
a=[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,2,0,0,0,0,0,5,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,9];
s=diff([inf,a])~=0;
b=ones(size(a));
b(s)=b(s)-[0,diff(find(s))];
discard=(cumsum(b)>10); %indices to discard
a_short=a(~discard) %truncated version of a
a_short = 1×28
0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 5 0 0 0 0 0 0 0 0 0 0 9

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