在R中实现滚动窗口回归的并行化

Question

在R中实现滚动窗口回归的并行化

12

我正在运行一个滚动回归，与以下代码非常相似：

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，所以我正在尝试找到一种并行化滚动窗口的方法。如果这是非滚动回归，我可以轻松地使用apply函数族来并行化它...

- Zach

2个回答

2

新答案

我编写了一个名为rollRegres的软件包，可以更快地完成此操作。它比Gavin Simpson的答案快约58倍。以下是一个例子：

# simulate data
set.seed(101)
n <- 10000
wdth <- 50
X <- matrix(rnorm(10 * n), n, 10)
y <- drop(X %*% runif(10)) + rnorm(n)
Z <- cbind(y, X)

# assign other function
lm_version <- function(Z, width = wdth) {
  pred <- Z[width + 1, -1, drop = FALSE]
  ## fit the model
  Z <- Z[-nrow(Z), ]
  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 next obs
  drop(pred[, piv, drop = FALSE] %*% beta[piv])
}

# show that they yield the same
library(rollRegres) # the new package
library(zoo)
all.equal(
  rollapply(Z, wdth + 1, FUN = lm_version,
            by.column = FALSE,  align = "right", fill = NA_real_),
  roll_regres.fit(
    x = X, y = y, width = wdth, do_compute = "1_step_forecasts"
    )$one_step_forecasts)
#R [1] TRUE

# benchmark
library(compiler)
lm_version <- cmpfun(lm_version)
microbenchmark::microbenchmark(
  newnew = roll_regres.fit(
    x = X, y = y, width = wdth, do_compute = "1_step_forecasts"),
  prev   = rollapply(Z, wdth + 1, FUN = lm_version,
                     by.column = FALSE,  align = "right", fill = NA_real_),
  times = 10)
#R Unit: milliseconds
#R   expr       min        lq      mean    median        uq       max neval
#R newnew  10.27279  10.48929  10.91631  11.04139  11.13877  11.87121    10
#R   prev 555.45898 565.02067 582.60309 582.22285 602.73091 605.39481    10

旧答案

您可以通过更新分解来减少运行时间。这样，每次迭代的成本为 $\mathcal{O}(p^2)$ ，而不是 $\mathcal{O}(np^2 + p^3)$ ，其中n是您窗口宽度。以下是一个比较两者的代码。用C++实现可能会更快，但是LINPACK的dchud和dchdd并没有包含在R中，所以您需要编写一个软件包来执行操作。此外，我记得读过您可以通过其他实现比LINPACK的dchud和dchdd更快地更新R。

library(SamplerCompare) # for LINPACK `chdd` and `chud`
roll_forcast <- function(X, y, width){
  n <- nrow(X)
  p <- ncol(X)
  out <- rep(NA_real_, n)

  is_first <- TRUE
  i <- width 
  while(i < n){
    if(is_first){
      is_first <- FALSE
      qr. <- qr(X[1:width, ])
      R <- qr.R(qr.)

      # Use X^T for the rest
      X <- t(X)

      XtY <- drop(tcrossprod(y[1:width], X[, 1:width]))
    } else {
      x_new <- X[, i]
      x_old <- X[, i - width]

      # update R 
      R <- .Fortran(
        "dchud", R, p, p, x_new, 0., 0L, 0L, 
        0., 0., numeric(p), numeric(p), 
        PACKAGE = "SamplerCompare")[[1]]

      # downdate R
      R <- .Fortran(
        "dchdd", R, p, p, x_old, 0., 0L, 0L, 
        0., 0., numeric(p), numeric(p), integer(1),
        PACKAGE = "SamplerCompare")[[1]]

      # update XtY
      XtY <- XtY + y[i] * x_new - y[i - width] * x_old
    }

    coef. <- .Internal(backsolve(R, XtY, p, TRUE, TRUE))
    coef. <- .Internal(backsolve(R, coef., p, TRUE, FALSE))

    i <- i + 1
    out[i] <- X[, i] %*% coef.
  }

  out
}

# simulate data
set.seed(101)
n <- 10000
wdth = 50
X <- matrix(rnorm(10 * n), n, 10)
y <- drop(X %*% runif(10)) + rnorm(n)
Z <- cbind(y, X)

# assign other function
lm_version <- function(Z, width = wdth) {
  pred <- Z[width + 1, -1, drop = FALSE]
  ## fit the model
  Z <- Z[-nrow(Z), ]
  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])
}

# show that they yield the same
library(zoo)
all.equal(
  rollapply(Z, wdth + 1, FUN = lm_version,  
            by.column = FALSE,  align = "right", fill = NA_real_),
  roll_forcast(X, y, wdth))
#R> [1] TRUE

# benchmark
library(compiler)
roll_forcast <- cmpfun(roll_forcast)
lm_version <- cmpfun(lm_version)
microbenchmark::microbenchmark(
  new =  roll_forcast(X, y, wdth),
  prev = rollapply(Z, wdth + 1, FUN = lm_version,  
                   by.column = FALSE,  align = "right", fill = NA_real_), 
  times = 10)
#R> Unit: milliseconds
#R> expr      min       lq     mean   median       uq      max neval cld
#R>  new 113.7637 115.4498 129.6562 118.6540 122.4930 230.3414    10  a 
#R> prev 639.6499 674.1677 682.1996 678.6195 686.8816 763.8034    10   b

- Benjamin Christoffersen

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- Gavin Simpson · Accepted Answer

显而易见的方法是使用lm.fit()而不是lm()，这样您就不需要承担处理公式等所有开销。

更新：所以当我说“显然”时，我想说的是非常明显但实现起来却具有欺骗性难度！

经过一番摆弄后，我想出了这个方法

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

首先需要认识到模型矩阵是可以预先构建的，因此我们进行了预构建并将其转换回Zoo对象以便在rollapply()中使用：

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() 来完成繁重的工作，而不必在每次迭代时创建设计矩阵：

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])
}

一个时间比较：

> 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

这比原来的要好得多。现在检查一下生成的对象是否相同：

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

享受吧！