Question

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。

示例：

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]

有什么想法吗？

Answer 1

使用hist和unique

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

Answer 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find，diff＆amp;的方法。 max -

%// 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

这是一个基准测试，用于比较目前为止列出的四种解决方案的运行时间 -

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.

Answer 3

mode的第三项结果就是这样。输入向量不需要排序。

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

Answer 4

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));

Answer 5

您还可以将accumarray与unique结合使用：

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

Matlab - 数组中的大多数重复值（不仅仅是模式）

5 个答案:

基准