Question

我正在尝试计算所有样本的变量的所有对组合之间的Pearson系数。

假设我有一个m * n矩阵，其中m是变量，n是样本

我想计算我的数据的每个变量与每个其他变量的相关性。

所以，我设法用嵌套循环来做到这一点：

X = rand[1000 100];
for i = 1:1000
base = X(i, :);
    for j = 1:1000
    target = X(j, :);
    correlation = corrcoef(base, target);
    correlation = correlation(2, 1);
    corData(1, j) = correlation
    end
totalCor(i, :) = corData
end

它有效，但需要花费太多时间才能运行

我试图找到一种在行的基础上运行corrcoef函数的方法，这意味着可能创建一个带有基值的repmat的附加矩阵，并使用一些FUN函数与X数据相关联。

无法弄清楚如何使用数组输入的乐趣，在个人行/列之间运行

帮助将不胜感激

Answer 1

这篇文章涉及一些 hacking ，所以请耐心等待！

阶段＃0 要开始，我们有 -

for i = 1:N
    base = X(i, :);
    for j = 1:N
        target = X(j, :);
        correlation = corrcoef(base, target);
        correlation = correlation(2, 1)
        corData(1, j) = correlation;
    end
end

第1阶段在其源代码中的corrcoef文档中：

如果C是协方差矩阵，C = COV（X），则CORRCOEF（X）是矩阵的第（i，j）个元素是：C(i,j)/SQRT(C(i,i)*C(j,j))。

在侵入covariance的代码后，我们看到对于一个输入的默认情况，协方差公式只是 -

[m,n] = size(x);
xc = bsxfun(@minus,x,sum(x,1)/m);
xy = (xc' * xc) / (m-1);

因此，混合这两个定义并将它们放入手头的问题中，我们有 -

m = size(X,2);
for i = 1:N
    base = X(i, :);
    for j = 1:N
        target = X(j, :);
        BT = [base(:) target(:)];
        xc = bsxfun(@minus,BT,sum(BT,1)/m);
        C = (xc' * xc) / (m-1); %//'
        corData = C(2,1)/sqrt(C(2,2)*C(1,1))
    end
end

阶段＃2 这是我们使用真正有趣又名bsxfun来杀死所有循环的最后阶段，例如所以 -

%// Broadcasted subtract of each row by the average of it.
%// This corresponds to "xc = bsxfun(@minus,BT,sum(BT,1)/m)"
p1 = bsxfun(@minus,X,mean(X,2));

%// Get pairs of rows from X and get the dot product. 
%// Thus, a total of "N x N" such products would be obtained.
p2 = sum(bsxfun(@times,permute(p1,[1 3 2]),permute(p1,[3 1 2])),3);

%// Scale them down by "size(X,2)-1". 
%// This was for the part : "C = (xc' * xc) / (m-1)".
p3 = p2/(size(X,2)-1);

%// "C(2,2)" and "C(1,1)" are diagonal elements from "p3", so store them.
dp3 = diag(p3);

%// Get "sqrt(C(2,2)*C(1,1))" by broadcasting elementwise multiplication 
%// of "dp3". Finally do elementwise division of "p3" by it.
totalCor_out = p3./sqrt(bsxfun(@times,dp3,dp3.'));

基准

本节将原始方法与建议方法进行比较，并验证输出。这是基准测试代码 -

disp('---------- With original approach')
tic
X = rand(1000,100);
corData = zeros(1,1000);
totalCor = zeros(1000,1000);
for i = 1:1000
    base = X(i, :);
    for j = 1:1000
        target = X(j, :);
        correlation = corrcoef(base, target);
        correlation = correlation(2, 1);
        corData(1, j) = correlation;
    end
    totalCor(i, :) = corData;
end
toc

disp('---------- With the real fun aka BSXFUN')
tic
p1 = bsxfun(@minus,X,mean(X,2));
p2 = sum(bsxfun(@times,permute(p1,[1 3 2]),permute(p1,[3 1 2])),3);
p3 = p2/(size(X,2)-1);
dp3 = diag(p3);
totalCor_out = p3./sqrt(bsxfun(@times,dp3,dp3.')); %//'
toc

error_val = max(abs(totalCor(:)-totalCor_out(:)))

输出 -

---------- With original approach
Elapsed time is 186.501746 seconds.
---------- With the real fun aka BSXFUN
Elapsed time is 1.423448 seconds.
error_val =
    4.996e-16

两个矩阵之间的相关系数找到相互关系

1 个答案:

基准