benchamrk

Question

我必须遵循以下代码：

   WBP2=lapply(1:2500,function(j){as.data.frame(lapply(1:10000,function(i)
   {rowSums(data.frame((mapply(`*`,Theta[[j]],t(W[i,]),SIMPLIFY=FALSE))))}))})

W是矩阵80列和10000行，Theta是2500个矩阵，包含1000行和80列。我们的目标是为每2500次模拟创建10000个数据帧。每个data.frame将是一行W（W [i，]）乘以theta [[j]]的所有行的结果（因此是80列和1000行的组合）。我们只需要这个组合的每一行的总和，因此需要一列和1000行的data.frame

这是一个可重现的代码

W=matrix((1:80), nrow =10, ncol=4)

theta= lapply(1:5,function(j){matrix((41:60),nrow=5,ncol=4)})

WBP=lapply(1:5,function(j){as.data.frame(lapply(1:10,function(i)  
{rowSums(data.frame(as.matrix(theta[[j]])%*%as.numeric(t(W[i,]))))}))})

View(WBP[[1]])

正如您所看到的，它非常沉重，我的电脑无法实现这样的功能。

有没有办法可以让它更高效，更省时？

谢谢!!!

Answer 1

你只能使用一个lapply

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wbp1=function()lapply(1:5,function(j){as.data.frame(as.matrix(theta[[j]])%*%as.matrix(t(W)))})

benchamrk

你的变种

tcrossprod

结果相同（仅名称不同）wbp2=function()lapply(1:5,function(j){as.data.frame(tcrossprod(as.matrix(theta[[1]]),as.matrix(W)))})

转换R中的代码以提高效率

1 个答案:

benchamrk