an alternative to nested for loops in r OR possible Rcpp needed?

Question

I have a nested loop as

X <- matrix(c(0.5,0,0,0.75), nrow = 2)
k = nrow(X)

ans1 <- 0
ans2 <- 0
for (aa in 1:k) {
  for (bb in 1:k) {
    for (cc in 1:k) {
      for (dd in 1:k) {
        ans1 = ans1 + (0.45 * X[aa,bb] * X[cc,dd])
        for (xx in 1:k) {
          for (yy in 1:k){ 
            ans2 = ans2 + (1.7*X[aa,bb]*X[xx,yy]*X[cc,dd] + 0.2*X[aa,xx]*X[bb,yy]*X[cc,dd])  
          }
        }
      }
    }
  }
}

But the matrix X which must be a square matrix can be of very high dimensions. Which would therefore slow the loop considerably. e.g. X = matrix(rnorm(10000),nrow = 100,byrow = T)

I am wondering if there is a shorter way to compress this. That would be readable and most importantly a slightly faster. I have tried expand.grid but this doesn't help much.

For instance for ans1

library(tidyverse)
an1 <- expand.grid(rep(list(seq(length(X))),2)) %>% arrange_all()
an11 <- t(apply(an1, 1, function(x) as.vector(t(X))[x]))

But as I mentioned, this doesn't improve the speed. Any suggestions? I am also thinking Rcpp might help but I am not sure and I have not tried that (not very good with the c++ syntax).

Answer 1

您根本不需要使用循环。由于您的ans1和代码ans2只是术语的总和，并且这些术语根本不相互作用，因此表达式简化为

ans1simple <- 0.45*sum(X)^2
ans2simple <- 1.9*sum(X)^3

您可以对随机数据进行测试。X如果您不相信，请更改 的种子或大小：

ans1simple <- 0.45*sum(X)^2
ans2simple <- 1.9*sum(X)^3

^{由reprex 包(v1.0.0)于 2021 年 4 月 19 日创建}

差异只是舍入误差。