Maximum number of repeated values over an array

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 个评论
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Alessandro Togni 2021-1-20

Thank you very much.

What if one would want to store the indexes of removed values?

Stephen23 2021-1-20

在 MATLAB Online 中打开

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

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Answer 2

Jan 2021-1-20

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/721999-maximum-number-of-repeated-values-over-an-array#answer_602069

编辑：Jan 2021-1-20

在 MATLAB Online 中打开

With FileExchange: RunLength :

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);