在多列上滚动回归

Question

我在寻找最有效的方法来计算对具有多列的 xts 对象的滚动线性回归时遇到问题。我已经在 stackoverflow 上搜索并阅读了几个以前的问题。

这个question and answer 接近但在我看来还不够，因为我想计算多个回归，而因变量在所有回归中都保持不变。我试图用随机数据重现一个例子：

require(xts)
require(RcppArmadillo)  # Load libraries

data <- matrix(sample(1:10000, 1500), 1500, 5, byrow = TRUE)  # Random data
data[1000:1500, 2] <- NA  # insert NAs to make it more similar to true data
data <- xts(data, order.by = as.Date(1:1500, origin = "2000-01-01"))

NR <- nrow(data)  # number of observations
NC <- ncol(data)  # number of factors
obs <- 30  # required number of observations for rolling regression analysis
info.names <- c("res", "coef")

info <- array(NA, dim = c(NR, length(info.names), NC))
colnames(info) <- info.names

创建数组是为了存储多个变量（残差、系数等）随时间和每个因子的变化。

loop.begin.time <- Sys.time()

for (j in 2:NC) {
  cat(paste("Processing residuals for factor:", j), "\n")
  for (i in obs:NR) {
    regression.temp <- fastLm(data[i:(i-(obs-1)), j] ~ data[i:(i-(obs-1)), 1])
    residuals.temp <- regression.temp$residuals
    info[i, "res", j] <- round(residuals.temp[1] / sd(residuals.temp), 4)
    info[i, "coef", j] <- regression.temp$coefficients[2]
  } 
}

loop.end.time <- Sys.time()
print(loop.end.time - loop.begin.time)  # prints the loop runtime

正如循环所示，这个想法是每次针对其他因素之一运行 30 个观察值滚动回归，并将 data[, 1] 作为因变量（因子）。我必须将 30 个残差存储在一个临时对象中以便将它们标准化，因为 fastLm 不计算标准化残差。

如果 xts 对象中的列数（因子）增加到大约 100 到 1,000 列，那么循环会非常缓慢并且会变得很麻烦。我希望有一个更有效的代码来在大型数据集上创建滚动回归。

Answer 1

这是使用rollRegres 包更快的方法

library(xts)
library(RcppArmadillo)

#####
# simulate data
set.seed(50554709)
data <- matrix(sample(1:10000, 1500), 1500, 5, byrow = TRUE)  # Random data
# data[1000:1500, 2] <- NA # only focus on the parts that are computed
data <- xts(data, order.by = as.Date(1:1500, origin = "2000-01-01"))

#####
# setup for solution in OP
NR <- nrow(data)
NC <- ncol(data)
obs <- 30L
info.names <- c("res", "coef")

info <- array(NA, dim = c(NR, length(info.names), NC))
colnames(info) <- info.names

#####
# solve with rollRegres
library(rollRegres)

loop.begin.time <- Sys.time()

X <- cbind(1, drop(data[, 1]))
out <- lapply(2:NC, function(j){
  fit <- roll_regres.fit(
    y = data[, j], x = X, width = obs, do_compute = c("sigmas"))

  # are you sure you want the residual of the first and not the last
  # observation in each window?
  idx <- 1:(nrow(data) - obs + 1L)
  idx_tail <- idx + obs - 1L
  resids <- c(rep(NA_real_, obs - 1L),
                  data[idx, j] - rowSums(fit$coefs[idx_tail, ] * X[idx, ]))

  # the package uses the unbaised estimator so we have to time by this factor
  # to get the same
  sds <-  fit$sigmas * sqrt((obs - 2L) / (obs - 1L))

  unclass(cbind(coef = fit$coefs[, 2L], res = drop(round(resids / sds, 4))))
})

loop.end.time <- Sys.time()
print(loop.end.time - loop.begin.time)
#R Time difference of 0.03123808 secs

#####
# solve with original method
loop.begin.time <- Sys.time()

for (j in 2:NC) {
  cat(paste("Processing residuals for factor:", j), "\n")
  for (i in obs:NR) {
    regression.temp <- fastLm(data[i:(i-(obs-1)), j] ~ data[i:(i-(obs-1)), 1])
    residuals.temp <- regression.temp$residuals
    info[i, "res", j] <- round(residuals.temp[1] / sd(residuals.temp), 4)
    info[i, "coef", j] <- regression.temp$coefficients[2]
  }
}
#R Processing residuals for factor: 2
#R Processing residuals for factor: 3
#R Processing residuals for factor: 4
#R Processing residuals for factor: 5

loop.end.time <- Sys.time()
print(loop.end.time - loop.begin.time)  # prints the loop runtime
#R Time difference of 7.554767 secs

#####
# check that results are the same
all.equal(info[, "coef", 2L], out[[1]][, "coef"])
#R [1] TRUE
all.equal(info[, "res" , 2L], out[[1]][, "res"])
#R [1] TRUE

all.equal(info[, "coef", 3L], out[[2]][, "coef"])
#R [1] TRUE
all.equal(info[, "res" , 3L], out[[2]][, "res"])
#R [1] TRUE

all.equal(info[, "coef", 4L], out[[3]][, "coef"])
#R [1] TRUE
all.equal(info[, "res" , 4L], out[[3]][, "res"])
#R [1] TRUE

all.equal(info[, "coef", 5L], out[[4]][, "coef"])
#R [1] TRUE
all.equal(info[, "res" , 5L], out[[4]][, "res"])
#R [1] TRUE

请注意上述解决方案中的此注释

# are you sure you want the residual of the first and not the last
# observation in each window?

---

这是与Sameer's answer的比较

library(rollRegres)
require(xts)

data <- matrix(sample(1:10000, 1500000, replace=T), 1500, 1000, byrow = TRUE)  # Random data
data <- xts(data, order.by = as.Date(1:1500, origin = "2000-01-01"))

NR <- nrow(data)  # number of observations
NC <- ncol(data)  # number of factors
obs <- 30  # required number of observations for rolling regression analysis

loop.begin.time <- Sys.time()

X <- cbind(1, drop(data[, 1]))
out <- lapply(2:NC, function(j){
  fit <- roll_regres.fit(
    y = data[, j], x = X, width = obs, do_compute = c("sigmas"))

  # are you sure you want the residual of the first and not the last
  # observation in each window?
  idx <- 1:(nrow(data) - obs + 1L)
  idx_tail <- idx + obs - 1L
  resids <- c(rep(NA_real_, obs - 1L),
              data[idx, j] - rowSums(fit$coefs[idx_tail, ] * X[idx, ]))

  # the package uses the unbaised estimator so we have to time by this factor
  # to get the same
  sds <-  fit$sigmas * sqrt((obs - 2L) / (obs - 1L))

  unclass(cbind(coef = fit$coefs[, 2L], res = drop(round(resids / sds, 4))))
})

loop.end.time <- Sys.time()
print(loop.end.time - loop.begin.time)
#R Time difference of 0.9019711 secs

时间包括用于计算标准化残差的时间。

Answer 2

如果您深入到线性回归的数学级别，应该会很快。如果 X 是自变量，Y 是因变量。系数由

给出

Beta = inv(t(X) %*% X) %*% (t(X) %*% Y)

对于您希望哪个变量成为依赖变量以及哪个变量是独立变量，我有点困惑，但希望解决下面的类似问题也能对您有所帮助。

在下面的示例中，我采用 1000 个变量而不是原来的 5 个变量，并且不引入任何 NA。

require(xts)

data <- matrix(sample(1:10000, 1500000, replace=T), 1500, 1000, byrow = TRUE)  # Random data
data <- xts(data, order.by = as.Date(1:1500, origin = "2000-01-01"))

NR <- nrow(data)  # number of observations
NC <- ncol(data)  # number of factors
obs <- 30  # required number of observations for rolling regression analysis

现在我们可以使用 Joshua 的 TTR 包计算系数。

library(TTR)

loop.begin.time <- Sys.time()

in.dep.var <- data[,1]
xx <- TTR::runSum(in.dep.var*in.dep.var, obs)
coeffs <- do.call(cbind, lapply(data, function(z) {
    xy <- TTR::runSum(z * in.dep.var, obs)
    xy/xx
}))

loop.end.time <- Sys.time()

print(loop.end.time - loop.begin.time)  # prints the loop runtime

时差 3.934461 秒

res.array = array(NA, dim=c(NC, NR, obs))
for(z in seq(obs)) {
  res.array[,,z] = coredata(data - lag.xts(coeffs, z-1) * as.numeric(in.dep.var))
}
res.sd <- apply(res.array, c(1,2), function(z) z / sd(z))

如果我在索引res.sd 中没有犯任何错误，应该会给你标准化的残差。请随时修复此解决方案以纠正任何错误。