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Matlab - 数组中的大多数重复值(不仅仅是模式)

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我有一个数字范围从1到4的数组 . 我需要知道哪些是重复更多的值 . 如果有平局我需要知道哪些是值,所以我可以进行一些操作 .

Example:

a = [1 1 1 2 2 2 3 4]
Output = [1 2]

a = [1 1 1 2 3 4]
Output = 1

a = [1 2 2 3 3 4 4]
Output = [2 3 4]

有任何想法吗?

5 回答

  • 4

    您还可以使用accumarrayunique结合使用:

    [vals,~,id] = unique(a);
    b = accumarray(id, 1);
    result = vals(b == max(b));
    
  • 3

    你可以这样做:

    a = [1 1 1 2 2 2 3 4];
    values = [1 2 3 4];
    
    counts = zeros(1,numel(values));
    
    for i=1:numel(values)
        counts(i) = sum(a == values(i));
    end
    
    output = values(counts == max(counts));
    
  • 5

    使用 histunique 的备用矢量化方法

    uVal = unique(a);
    counts = hist(a,uVal);
    out = uVal(counts == max(counts));
    

    结果:

    a = [1 1 1 2 2 2 3 4];
    
    >> out
    
    out =
    
     1     2
    
  • 4

    mode的第三个输出就是这样 . 输入向量不需要排序 .

    [~, ~, v] = mode(a);
    result = v{1};
    
  • 2

    假设输入行向量已经排序(否则您可以通过调用sort()对其进行排序),您可以使用基于 finddiffmax 的方法 -

    %// Find starting indices of each island of identical numbers being
    %// appended by the numel()+1 with the intention of getting island lengths
    %// later on by differentiating along the indices
    start_ind = [0 find(diff(a)) numel(a)]+1
    lengths = diff(start_ind)
    
    %// Look for the islands with the max island lengths. 
    %// Use those to get unique numbers associated with them for final output
    out = a(start_ind([lengths == max(lengths) false]))
    

    基准测试

    这是一个基准测试,用于比较目前列出的四种解决方案的运行时 -

    a = randi(10000,1,1000000);
    
    disp('---------------- With for-loop')
    tic
    values = unique(a);
    counts = zeros(1,numel(values));
    for i=1:numel(values)
        counts(i) = sum(a == values(i));
    end
    output = values(counts == max(counts));
    toc
    clear output counts values
    
    disp('---------------- With find+diff+max')
    tic
    sa = sort(a);
    start_ind = [0 find(diff(sa)) numel(sa)]+1;
    lengths = diff(start_ind);
    out = sa(start_ind([lengths == max(lengths) false]));
    toc
    clear out lengths start_ind sa
    
    disp('---------------- With mod')
    tic
    [~, ~, v] = mode(a);
    result = v{1};
    toc
    clear v result
    
    disp('---------------- With unique+hist+max')
    tic
    uVal = unique(a);
    counts = hist(a,uVal);
    out = uVal(counts == max(counts));
    toc
    

    运行时 -

    ---------------- With for-loop
    Elapsed time is 32.879074 seconds.
    ---------------- With find+diff+max
    Elapsed time is 0.077948 seconds.
    ---------------- With mod
    Elapsed time is 0.136005 seconds.
    ---------------- With unique+hist+max
    Elapsed time is 0.250994 seconds.
    

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