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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。

这个问题和答案很接近，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

创建数组是为了随时间和每个因子存储多个变量（残差、系数等）。

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

data[, 1]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，并将其作为因变量（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，则循环非常缓慢并且会变得很麻烦，这需要很长时间。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。

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2 回答 2

10

如果你深入到线性回归的数学水平，它应该会很快。如果 X 是自变量，Y 是因变量。系数由下式给出

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

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

在下面的示例中，我采用 1000 个变量而不是原来的 5 个，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))

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

于 2012-08-09T14:12:29.547 回答
0

rollRegres这是使用包的一种更快的方法

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评论

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

这是与Sameer的答案的比较

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。

于 2018-07-02T17:39:08.717 回答