将矢量中的唯一值分组并将它们放在矩阵中

Question

我有一个包含重复数字的向量,如下所示:

[1 1 1 1 5 5 5 5 93 93 93 6 6 6 6 6 6]等等.我想要做的是将相似的值(1,5等)分组.我希望将每个唯一值放在一个大矩阵的行中,例如:

[ 1  1  1  1  0  0

  5  5  5  5  0  0

 93 93 93  0  0  0

  6  6  6  6  6  6]

我不知道唯一值的最大出现次数,因此可以创建一个具有大量列的初始零矩阵(我确信它大于唯一值的最大出现次数).任何帮助都非常感谢.

Answer 1

这个怎么样？

A = [1 1 1 1 5 5 5 5 93 93 93 6 6 6 6 6 6];
[a,b] = hist(A,unique(A))

f = @(x) [ones(1,a(x)) zeros(1,max(a)-a(x))]

X = cell2mat( arrayfun(@(x) {f(x)*b(x)}, 1:numel(b) )' )

收益:

X =

     1     1     1     1     0     0
     5     5     5     5     0     0
     6     6     6     6     6     6
    93    93    93     0     0     0

我知道顺序不同,那重要吗？除此以外:

n = hist(A,1:max(A))           % counts how often every number apperas
[a b] = unique(A,'stable')     % gets all unique numbers
n = n(a)                       % correlates count and numbers

f = @(x) [ones(1,n(x)) zeros(1,max(n)-n(x))]     % creates the logical index 
                                                 % vector for every single row

X = cell2mat( arrayfun(@(x) {f(x)*b(x)}, 1:numel(b) )' )  %fills the rows

或者受路易斯·门多的回答更短一点:

n = hist(A,1:max(A));
a = unique(A,'stable')
n = n(a)
Y = repmat(a',1,max(n)).*bsxfun(@le, cumsum(ones(max(n),numel(n))), n)'

收益:

X =

     1     1     1     1     0     0
     5     5     5     5     0     0
    93    93    93     0     0     0
     6     6     6     6     6     6

对于那些无聊的人来说,有一个单线解决方案:

X = getfield(cell2mat(arrayfun(@(x,y) padarray( padarray(x,[0 y],'replicate','pre'),[0 max(hist(A,1:max(A)))-y],'post'),1:max(A),hist(A,1:max(A)),'uni',0)'),{unique(A,'stable'),2:1+max(hist(A,1:max(A)))})

或者几乎可爱的双线:

n = hist(A,1:max(A))
X = getfield(cell2mat(arrayfun(@(x,y) padarray( padarray(x,[0 y],'replicate',...
    'pre'),[0 max(n)-y],'post'),1:max(A),n,'uni',0)'),...
    {unique(A,'stable'),2:1+max(n)})

纯娱乐 ;)

Answer 2

矢量化解决方案(无循环):

x = [1 1 1 1 5 5 5 5 93 93 93 6 6 6 6 6 6]; %// data
ind = [find(diff(x)) numel(x)]; %// end of each run of equal values
values = x(ind); %// unique values (maintaining order)
count = diff([0 ind]); %// count of each value
result = bsxfun(@le, meshgrid(1:max(count),1:numel(values)), count.'); %'// mask
result = bsxfun(@times, result, values.'); %'// fill with the values

编辑:

避免第二种方法的替代程序bsxfun:

x = [1 1 1 1 5 5 5 5 93 93 93 6 6 6 6 6 6]; %// data
ind = [find(diff(x)) numel(x)];
values = x(ind); %// unique values (maintaining order)
count = diff([0 ind]); %// count of each value
mask = bsxfun(@le, ndgrid(1:max(count),1:numel(values)), count);
result = zeros(size(mask)); %// pre-allocate and pre-shape (transposed) result
result(mask) = x; %// fill in values
result = result.';