Question

我为异常值检测构建了一个函数，它运行得很好，但是考虑到我正在处理的大量数据，我需要删除“for循环”，所以这里我们有矢量化版本（或者至少是什么我认为是我的代码的矢量化版本）。调用该函数以下参数由用户初始化，我正在使用以下内容：

alpha=3
gamma=0.5
k=5

系列“价格”存在于工作空间中，在调用函数时链接。我想我几乎做到了，但我遇到了问题这是一段代码：

[n] = size(price,1);
x = price;
[j1]=find(x); %output is a column vector with size (n,1) of the following form j1=[1:1:n]
matrix_left=zeros(n, k,'double');
matrix_right=zeros(n, k,'double');
toc
matrix_left(j1(k+1:end),:)=x(j1-k:j1-1);

％这里返回以下错误：下标索引必须是实数正整数或逻辑。

matrix_right(j1(1:end-k),:)=x(j1+1:j1+k);

％此处，它说明以下内容：订阅的分配维度不匹配。

matrix_group=[matrix_left matrix_right];
trimmed_mean=trimmean(matrix_group,10,'round',2);
score=bsxfun(@minus,x,trimmed_mean);
sd=std(matrix_group,2);
temp = abs(score) > (alpha .* sd + gamma);
outmat = temp*1;

我想拥有的是：如果k = 5

left_matrix (3443,5):  
[100.25  103.5   102.25    102.75   103]  <---5 left neighbouring observations of the 15th row of **x**
[103.5   102.25  102.75    103    103.5]  <---5 left neighbouring observations of the 16th row of **x**

right_matrix(3443,5):  
[103.75  104.25  104   104.75  104.25]  <---5 right neighbouring observations of the 15th row of **x**
[104.25  104   104.75  104.25   104.5]  <---5 right neighbouring observations of the 16th row of **x**

以下是一小部分数据：

x = Price; price size = (3443, 1)
[...]  
100.25       %// '*suppose here we are at the 10th row*' 
103.5
102.25
102.75  
103  
103.5        %// '*here we are at the 15th row*'
103.75   
104.25  
104  
104.75  
104.25  
104.5  
[...]

Time (3443,1) %// the same as price, it reports the time of the transaction (HH:MM:SS).  
j1 (3443,1)
1  
2  
[...]  
3442  
3443

提前感谢大家，
乔治

Answer 1

这是答案，要走的路是（再一次）bsxfun：

[n] = size(price,1);
x = price;

idxArray_left=bsxfun(@plus,(k+1:n)',-k:-1);
idxArray_fill_left=bsxfun(@plus,(1:k)',1:k);
matrix_left=[idxArray_fill_left; idxArray_left];
idxArray_right=bsxfun(@plus,(1:n-k)',1:k);
idxArray_fill_right=bsxfun(@plus,(n-k+1:n)',-k:-1);
matrix_right=[idxArray_right; idxArray_fill_right];      
idx_matrix=[matrix_left matrix_right];
neigh_matrix=x(idx_matrix);
trimmed_mean=trimmean(neigh_matrix,10,'round',2);
score=bsxfun(@minus,x,trimmed_mean);
sd=std(neigh_matrix,0,2);
temp = abs(score) > (alpha .* sd + gamma);
outmat = temp*1;

感谢Jonas给了我很好的解决问题的提示

Matlab：使用来自列向量的移动间隔填充矩阵行，而不使用for循环

1 个答案: