Azzera filtri
Azzera filtri

Maximum number of repeated values over an array

6 visualizzazioni (ultimi 30 giorni)
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].
Any suggestion would be appreciated.
Thanks in advance,
Alessandro

Risposta accettata

Stephen23
Stephen23 il 20 Gen 2021
Modificato: Stephen23 il 20 Gen 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];
x = cumsum([true;diff(a(:))~=0]);
f = @(v){v(1:min(end,10))};
c = accumarray(x,a(:),[],f);
b = vertcat(c{:}).'
b = 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
  2 Commenti
Alessandro Togni
Alessandro Togni il 20 Gen 2021
Thank you very much.
What if one would want to store the indexes of removed values?
Stephen23
Stephen23 il 20 Gen 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];
d = find([true;diff(a(:));true]);
f = @(b,e)(b+10):(e-1);
c = arrayfun(f,d(1:end-1),d(2:end),'uni',0);
x = horzcat(c{:})
x = 1×21
11 12 13 14 15 16 17 18 19 20 21 22 40 41 42 43 44 45 46 47 48

Accedi per commentare.

Più risposte (1)

Jan
Jan il 20 Gen 2021
Modificato: Jan il 20 Gen 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];
[v, n] = RunLength(a);
b = RunLength(v, min(n, 10));
If you do not have a C compiler installed, use the function RunLength_M of this submission.
Alternatively:
function out = LimitRunLength(in, nMax)
x = in(:);
d = [true; diff(x) ~= 0]; % TRUE if values change
b = x(d); % Elements without repetitions
k = find([d', true]); % Indices of changes
n = diff(k); % Number of repetitions
n = min(n, nMax); % Limit the run lengths
d = cumsum(n); % Cummulated run lengths
index = zeros(1, d(end)); % Pre-allocate
index(d(1:end-1)+1) = 1; % Get the indices where the value changes
index(1) = 1; % First element is treated as "changed" also
out = b(cumsum(index)); % Cummulated indices
% Let the output be a row vector, if the input is a row:
if size(in, 2) > 1
out = out.';
end
end
[EDITED] You ask for the indices of the removed elements:
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];
nMax = 10;
[v, n, idx] = RunLength_M(a);
b = RunLength_M(v, min(n, nMax));
crop = find(n > nMax);
idx = [idx, numel(a) + 1];
q = zeros(1, numel(a));
q(idx(crop) + nMax) = 1;
q(idx(crop + 1)) = -1;
removed = find(cumsum(q));
And as next alternative a straight forward loop:
del = false(size(a));
cur = NaN;
for k = 1:numel(a)
if a(k) == cur
len = len + 1;
del(k) = (len > nMax);
else
cur = a(k);
len = 1;
end
end
b = a(~del);
removed = find(del);

Categorie

Scopri di più su Programming in Help Center e File Exchange

Tag

Prodotti

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by