r - 在 R 中并行化滚动窗口回归

Question

我正在运行与以下代码非常相似的滚动回归:

library(PerformanceAnalytics)
library(quantmod)
data(managers)

FL <- as.formula(Next(HAM1)~HAM1+HAM2+HAM3+HAM4)
MyRegression <- function(df,FL) {
  df <- as.data.frame(df)
  model <- lm(FL,data=df[1:30,])
  predict(model,newdata=df[31,])
}

system.time(Result <- rollapply(managers, 31, FUN="MyRegression",FL,
    by.column = FALSE, align = "right", na.pad = TRUE))

我有一些额外的处理器，所以我试图找到一种方法来并行化滚动窗口。如果这是一个非滚动回归，我可以使用 apply 系列函数轻松地将它并行化......

Answer 1

显而易见的是使用 lm.fit()而不是 lm()所以你不会在处理公式等过程中产生所有的开销。

更新:所以当我说明显的时候，我想说的是 显而易见但看似难以实现 !

经过一番摆弄，我想出了这个

library(PerformanceAnalytics)
library(quantmod)
data(managers)

第一阶段是实现模型矩阵可以预先构建，因此我们这样做并将其转换回 Zoo 对象以供 rollapply() 使用。 :

mmat2 <- model.frame(Next(HAM1) ~ HAM1 + HAM2 + HAM3 + HAM4, data = managers, 
                     na.action = na.pass)
mmat2 <- cbind.data.frame(mmat2[,1], Intercept = 1, mmat2[,-1])
mmatZ <- as.zoo(mmat2)

现在我们需要一个函数来使用 lm.fit()无需在每次迭代时创 build 计矩阵即可完成繁重的工作:

MyRegression2 <- function(Z) {
    ## store value we want to predict for
    pred <- Z[31, -1, drop = FALSE]
    ## get rid of any rows with NA in training data
    Z <- Z[1:30, ][!rowSums(is.na(Z[1:30,])) > 0, ]
    ## Next() would lag and leave NA in row 30 for response
    ## but we precomputed model matrix, so drop last row still in Z
    Z <- Z[-nrow(Z),]
    ## fit the model
    fit <- lm.fit(Z[, -1, drop = FALSE], Z[,1])
    ## get things we need to predict, in case pivoting turned on in lm.fit
    p <- fit$rank
    p1 <- seq_len(p)
    piv <- fit$qr$pivot[p1]
    ## model coefficients
    beta <- fit$coefficients
    ## this gives the predicted value for row 31 of data passed in
    drop(pred[, piv, drop = FALSE] %*% beta[piv])
}

时序对比:

> system.time(Result <- rollapply(managers, 31, FUN="MyRegression",FL,
+                                 by.column = FALSE, align = "right", 
+                                 na.pad = TRUE))
   user  system elapsed 
  0.925   0.002   1.020 
> 
> system.time(Result2 <- rollapply(mmatZ, 31, FUN = MyRegression2,
+                                  by.column = FALSE,  align = "right",
+                                  na.pad = TRUE))
   user  system elapsed 
  0.048   0.000   0.05

与原始版本相比，这提供了相当合理的改进。现在检查结果对象是否相同:

> all.equal(Result, Result2)
[1] TRUE

享受!