在R中对具有连续索引的矩阵和向量进行向量叉积的向量化处理

Question

一些背景：我有一个n×1的向量e和一个 n×n 的相关矩阵 R。我正在尝试高效地计算在 R 中以下数量的值，而不使用循环：。到目前为止，我已经能够使用单个循环完成，使用以下代码：

nom <- 0; denom <- 0
for(i in 1:n){
  nom <- nom + e[i]*(R[i,-(1:i)]%*%e[-(1:i)])
  denom <- denom + e[i]*sum(e[-(1:i)])
}
beta <- nom/denom

这样做是可以的，实际上我的计算不会太长，因为对于手头的任务来说，n 的大小最多只有11或12（所以改进可能不会在测量时间性能方面产生很大差异）。然而，我很好奇如何更有效地完成这项工作，因为：

a）R 是对称的，计算只需要使用 R 的主对角线以上（或以下）的部分；

b）我计划将其用于一些大规模的 MC 模拟中，因此任何节省的计算时间都可能对整个计划产生影响。

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为了复制/计算目的，这里提供一个可能的数值示例：

e <- c(0.4972,0.0902,0.02822,0.1688,0.0149,0.0028,0.01411,0.02733,0.0151,0.0391,0.01301,0.0894)
R <- matrix(data = c(1,0.9,0.4,0.75,0.5,0.3,0.4,0.4,0.25,0.25,0.5,0.4,0.9,1,0.5,0.9,0.5,0.3,0.4,0.35,0.2,0.2,0.5,0.3,0.4,0.5,1,0.3,0.5,0.4,0.25,0.2,0.2,0.2,0.3,0.3,0.75,0.9,0.3,1,0.3,0.3,0.4,0.25,0.25,0.2,0.3,0.75,0.5,0.5,0.5,0.3,1,0.5,0.35,0.8,0.8,0.3,0.7,0.45,0.3,0.3,0.4,0.3,0.5,1,0.3,0.4,0.3,0.2,0.45,0.35,0.4,0.4,0.25,0.4,0.35,0.3,1,0.3,0.3,0.2,0.5,0.5,0.4,0.35,0.2,0.25,0.8,0.4,0.3,1,0.8,0.2,0.6,0.8,0.25,0.2,0.2,0.25,0.8,0.3,0.3,0.8,1,0.3,0.7,0.8,0.25,0.2,0.2,0.2,0.3,0.2,0.2,0.2,0.3,1,0.25,0.3,0.5,0.5,0.3,0.3,0.7,0.45,0.5,0.6,0.7,0.25,1,0.7,0.4,0.3,0.3,0.75,0.45,0.35,0.5,0.8,0.8,0.3,0.7,1), nrow = 12, ncol = 12)

- Emil

Answer 1

你可以做以下事情：

ee <- tcrossprod(e)
ee0 <- ee; diag(ee0) <- 0

sum(R*ee0)/sum(ee0)

针对您的示例数据：

> sum(R*ee0)/sum(ee0)
[1] 0.5647038

一点小的计算优化： sum(ee0) 相当于

sum(e)^2 - sum(e^2)

这是另一种变体：

d <- diag(R)
diag(R) <- 0
sum(e*crossprod(e, R))/(sum(e)^2 - sum(e^2))

- jogo

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这是一个非常好的答案，代码也非常简洁。非常感谢！ - Emil

@Emil，我添加了另一个变量。 - jogo