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是否有一种简单（理想情况下没有多个 for 循环）的方法来根据 Matlab 中的一组类别对值向量进行分组？

我有表格中的数据矩阵

CATEG_A    CATEG_B   CATEG_C  ...   VALUE

   1          1        1      ...   0.64
   1          2        1      ...   0.86
   1          1        1      ...   0.74
   1          1        2      ...   0.56
  ...

等等

我想要的是一个N维数组

 all_VALUE( CATEG_A, CATEG_B, CATEG_C, ..., index ) = VALUE_i

当然，可能有任意数量的值具有相同的类别组合，size(end)最大类别中的值的数量也是如此——其余项目将用nan.

或者我会很高兴

 all_VALUE { CATEG_A, CATEG_B, CATEG_C, ... } ( index )

即向量的单元格数组。我想这有点像创建一个数据透视表，但使用 n 维，而不是计算mean.

我在帮助中找到了这个功能

A = accumarray(subs,val,[],@(x) {x})

但我无法理解如何让它做我想做的事！

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2 回答 2

2

这是一团糟，但这是一个解决方案

[U,~,subs] = unique(X(:,1:end-1),'rows');

sz = max(U);
Uc = mat2cell(U, size(U,1), ones(1,size(U,2)));
%// Uc is converted to cell matrices so that we can take advantage of the {:} notation which returns a comma-separated-list which allows us to pass a dynamic number of arguments to functions like sub2ind

I = sub2ind(sz, Uc{:});

G = accumarray(subs, X(:,end),[],@(x){x});

A{prod(max(U))} = [];  %// Pre-assign the correct number of cells to A so we can reshape later
A(I) = G;
reshape(A, sz)

在您的示例数据（忽略...s）上，这将返回：

A(:,:,1) = 

    [2x1 double]    [0.8600]


A(:,:,2) = 

    [0.5600]    []

A(1,1,1)在哪里[0.74; 0.64]

于 2015-02-11T15:36:14.270 回答
2

这也是一团糟，但有效。它采用 ND 阵列方式。

X = [1        1        1        0.64
     1        2        1        0.86
     1        1        1        0.74
     1        1        2        0.56]; %// data
N = size(X,1); %// number of values
[~, ~, label] = unique(X(:,1:end-1),'rows'); %// unique labels for indices
cumLabel = cumsum(sparse(1:N, label, 1),1); %// used for generating a cumulative count
    %// for each label. The trick here is to separate each label in a different column
lastInd = full(cumLabel((1:N).'+(label-1)*N)); %'// pick appropriate values from 
    %// cumLabel to generate the cumulative count, which will be used as last index
    %// for the result array
sizeY = [max(X(:,1:end-1),[],1) max(lastInd)]; %// size of result
Y = NaN(sizeY); %// initiallize result with NaNs
ind = mat2cell([X(:,1:end-1) lastInd], ones(1,N)); %// needed for comma-separated list
Y(sub2ind(sizeY, ind{:})) = X(:,end); %// linear indexing of values into Y

您的示例中的结果是以下 4D 数组：

>> Y
Y(:,:,1,1) =
    0.6400    0.8600
Y(:,:,2,1) =
    0.5600       NaN
Y(:,:,1,2) =
    0.7400       NaN
Y(:,:,2,2) =
   NaN   NaN
于 2015-02-11T15:10:52.033 回答