使R中的滚动窗口回归平行

问题描述

| 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。如果这是非滚动回归，那么我可以使用apply系列函数轻松并行化它。

解决方法

很明显的一个方法是使用lm.fit()而不是lm()，这样就不会在处理公式等方面产生所有开销。更新：所以当我说的很明显时，我的意思是说的很明显，但是看似难以实现！经过一番摆弄之后，我想到了这个

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

第一步是认识到可以预先构建模型矩阵，因此我们将其转换回Zoo对象以用于rollapply()：

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，我们需要一个函数使用employ1进行繁重的工作，而不必在每次迭代时都创建设计矩阵：

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[,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])
}

计时比较：

> 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,FUN = MyRegression2,+                                  by.column = FALSE,+                                  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

请享用！ ,新答案我写了一个10英镑的包裹，这样做的速度要快得多。比盖文·辛普森的答案快58倍。这是一个例子

# 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,drop = FALSE]
  ## fit the model
  Z <- Z[-nrow(Z),]
  fit <- lm.fit(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[,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,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,do_compute = \"1_step_forecasts\"),prev   = rollapply(Z,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。这将在每次迭代中产生一个成本，而不是n是您的窗口宽度。下面是比较两者的代码。在C ++中这样做可能会快得多，但是R中不包含LINPACKdchud和dchdd，因此您必须编写一个程序包才能这样做。此外，我记得阅读过的文章，对于R更新，您可能会比LINPACKdchud和dchdd更快地使用其他实现

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,x_new,0.,0L,numeric(p),PACKAGE = \"SamplerCompare\")[[1]]

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

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

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

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

  out
}

# simulate data
set.seed(101)
n <- 10000
wdth = 50
X <- matrix(rnorm(10 * n),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[,drop = FALSE] %*% beta[piv])
}

# show that they yield the same
library(zoo)
all.equal(
  rollapply(Z,roll_forcast(X,wdth))
#R> [1] TRUE

# benchmark
library(compiler)
roll_forcast <- cmpfun(roll_forcast)
lm_version <- cmpfun(lm_version)
microbenchmark::microbenchmark(
  new =  roll_forcast(X,wdth),prev = rollapply(Z,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

回归回归回归平行滚动滚动滚动窗口窗口窗口

使R中的滚动窗口回归平行

问题描述

解决方法

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