假设您有一个VAR（）回归操作返回的类'varrest'的模型对象。 我想将模型保存到文件中，但不是所有用于估计系数的数据。
如何在没有训练数据的情况下保存模型规范？ 因为当我保存模型时，它的文件大小超过1GB，因此加载确实需要时间。 可以保存没有某些属性的对象吗？
答案 0 :(得分：5)
predict.varest函数从这段代码开始：
K <- object$K
p <- object$p
obs <- object$obs
type <- object$type
data.all <- object$datamat
ynames <- colnames(object$y)
然后，您可以调查可能会实现多少修剪：
data(Canada)
tcan <-
VAR(Canada, p = 2, type = "trend")
names(tcan)
# [1] "varresult" "datamat" "y" "type" "p"
# [6] "K" "obs" "totobs" "restrictions" "call"
object.size(tcan[c("K","p", "obs", "type", "datamat", "y")] )
#15080 bytes
object.size(tcan)
#252032 bytes
所以区别很大，但只保存这些项目是不够的，因为predict.varest中的下一行是：
B <- Bcoef(object)
您需要将该对象添加到上面的列表中，然后构造一个新的预测函数，该函数接受的内容少于模型对象的大“varresult”节点。还发现有一个下游调用需要存储的内部函数。 （您需要提前决定预测所需的时间间隔。）
tsmall <- c( tcan[c("K","p", "obs", "type", "datamat", "y", "call")] )
tsmall[["Bco"]] <- Bcoef(tcan)
tsmall$sig.y <- vars:::.fecov(x = tcan, n.ahead = 10)
修改后的predict函数将是：
sm.predict <- function (object, ..., n.ahead = 10, ci = 0.95, dumvar = NULL)
{
K <- object$K
p <- object$p
obs <- object$obs
type <- object$type
data.all <- object$datamat
ynames <- colnames(object$y)
n.ahead <- as.integer(n.ahead)
Z <- object$datamat[, -c(1:K)]
# This used to be a call to Bcoef(object)
B <- object$Bco
if (type == "const") {
Zdet <- matrix(rep(1, n.ahead), nrow = n.ahead, ncol = 1)
colnames(Zdet) <- "const"
}
else if (type == "trend") {
trdstart <- nrow(Z) + 1 + p
Zdet <- matrix(seq(trdstart, length = n.ahead), nrow = n.ahead,
ncol = 1)
colnames(Zdet) <- "trend"
}
else if (type == "both") {
trdstart <- nrow(Z) + 1 + p
Zdet <- matrix(c(rep(1, n.ahead), seq(trdstart, length = n.ahead)),
nrow = n.ahead, ncol = 2)
colnames(Zdet) <- c("const", "trend")
}
else if (type == "none") {
Zdet <- NULL
}
if (!is.null(eval(object$call$season))) {
season <- eval(object$call$season)
seas.names <- paste("sd", 1:(season - 1), sep = "")
cycle <- tail(data.all[, seas.names], season)
seasonal <- as.matrix(cycle, nrow = season, ncol = season -
1)
if (nrow(seasonal) >= n.ahead) {
seasonal <- as.matrix(cycle[1:n.ahead, ], nrow = n.ahead,
ncol = season - 1)
}
else {
while (nrow(seasonal) < n.ahead) {
seasonal <- rbind(seasonal, cycle)
}
seasonal <- seasonal[1:n.ahead, ]
}
rownames(seasonal) <- seq(nrow(data.all) + 1, length = n.ahead)
if (!is.null(Zdet)) {
Zdet <- as.matrix(cbind(Zdet, seasonal))
}
else {
Zdet <- as.matrix(seasonal)
}
}
if (!is.null(eval(object$call$exogen))) {
if (is.null(dumvar)) {
stop("\nNo matrix for dumvar supplied, but object varest contains exogenous variables.\n")
}
if (!all(colnames(dumvar) %in% colnames(data.all))) {
stop("\nColumn names of dumvar do not coincide with exogen.\n")
}
if (!identical(nrow(dumvar), n.ahead)) {
stop("\nRow number of dumvar is unequal to n.ahead.\n")
}
if (!is.null(Zdet)) {
Zdet <- as.matrix(cbind(Zdet, dumvar))
}
else {
Zdet <- as.matrix(dumvar)
}
}
Zy <- as.matrix(object$datamat[, 1:(K * (p + 1))])
yse <- matrix(NA, nrow = n.ahead, ncol = K)
# This used to be a call to vars:::.fecov
sig.y <- object$sig.y
for (i in 1:n.ahead) {
yse[i, ] <- sqrt(diag(sig.y[, , i]))
}
yse <- -1 * qnorm((1 - ci)/2) * yse
colnames(yse) <- paste(ci, "of", ynames)
forecast <- matrix(NA, ncol = K, nrow = n.ahead)
lasty <- c(Zy[nrow(Zy), ])
for (i in 1:n.ahead) {
lasty <- lasty[1:(K * p)]; print(lasty); print(B)
Z <- c(lasty, Zdet[i, ]) ;print(Z)
forecast[i, ] <- B %*% Z
temp <- forecast[i, ]
lasty <- c(temp, lasty)
}
colnames(forecast) <- paste(ynames, ".fcst", sep = "")
lower <- forecast - yse
colnames(lower) <- paste(ynames, ".lower", sep = "")
upper <- forecast + yse
colnames(upper) <- paste(ynames, ".upper", sep = "")
forecasts <- list()
for (i in 1:K) {
forecasts[[i]] <- cbind(forecast[, i], lower[, i], upper[,
i], yse[, i])
colnames(forecasts[[i]]) <- c("fcst", "lower", "upper",
"CI")
}
names(forecasts) <- ynames
result <- list(fcst = forecasts, endog = object$y, model = object,
exo.fcst = dumvar)
class(result) <- "varprd"
return(result)
}
答案 1 :(得分：4)
要么
save()函数。