将类别值快速扩展为矩阵中的归一化特征向量行

Question

我正在寻找一种有效的矢量化方法，将类别编号的列向量扩展为标准化特征矩阵。

例如，我可能有：

octave:16> false_vals
false_vals =

  -0.10000  -0.20000  -0.50000

octave:17> true_vals
true_vals =

   0.90000   0.80000   0.50000

octave:18> cats
cats =

   1
   2
   3

我想将它们组合起来创建一个矩阵，其中每一行与false_vals 相同，但由cats 标识的索引位置设置为true_vals 中的相应值。例如，它看起来像这样：

octave:19> X
X =

   0.90000  -0.20000  -0.50000
  -0.10000   0.80000  -0.50000
  -0.10000  -0.20000   0.50000

我发现this question and answer 可以扩展给我一个随机分配：

octave:28> X = repmat(false_vals,3,1)
X =

  -0.10000  -0.20000  -0.50000
  -0.10000  -0.20000  -0.50000
  -0.10000  -0.20000  -0.50000

octave:29> Y = repmat(true_vals,3,1)
Y =

   0.90000   0.80000   0.50000
   0.90000   0.80000   0.50000
   0.90000   0.80000   0.50000

octave:30> L = ( rand(3,3) > 0.5 ) % This stands in for the actual logical matrix I want
L =

   1   1   1
   1   1   1
   0   0   1

octave:31> X(L) = Y(L)
X =

   0.90000   0.80000   0.50000
   0.90000   0.80000   0.50000
  -0.10000  -0.20000   0.50000

但我随后被困在如何用一些可以将我的cats 向量转换为的函数替换上面的rand

L = 
 1  0  0
 0  1  0
 0  0  1

即我知道我可以用逻辑矩阵做我想做的事，但不知道如何从我的向量cats 到正确的逻辑矩阵（对于L 的每一行，列位置由该行的值标识cats 向量应该是真的，所以我可以在最后一个语句中使用它。

Answer 1

您可以使用repmat 重复该行，然后使用linear indexing 替换所需的值：

n = numel(cats);
result = repmat(false_vals, n, 1); %// repeat row n times
result((cats.'-1)*n+(1:n)) = true_vals; %'// replace desired values

Answer 2

代码

N = 3; %// size of false_vals and true_vals

%// Expand false_vals to create X
X = repmat(false_vals,N,1)

%// Find linear indices for expanded matrix, where true_vals are to be put
%// For this, you would need two things - row and column indices.
%// 1. Row indices would be just [1:N], as we are looking to get one replacement
%// per row.
%// 2. Column indices would be either of 1 or 2 or .. N
ind = round((N-1)*rand(1,N)) %// ind would be column indices minus 1
true_vals_idx = N*ind+[1:N] %// [1:N] represent the row indices

%// Replace the elements at those linearly indexed positions with true_vals 
X(true_vals_idx) = true_vals

如果除了之前每行一个替换的标准，您还希望每列有一个替换，您可以修改ind 计算部分，有点像这样 -

%// Create unique numbers for the interval [1,N], which could also be calculated with 
%// randperm, but that being slow has been replaced by a sort-based implementation
[~,ind] = sort(rand(N,1)) 
true_vals_idx = N*(ind-1)+[1:N]

%// .. Rest of the code stays the same

注意：“重新创建”randperm 的基于排序的实现是基于this solution。