我正在使用forecast::auto.arima
运行时间序列预测,我正在尝试查看是否有提取分配给p
,d
,{{1}的值的方法从拟合的时间序列对象中(和季节性的,如果适用的话)。例如:
q
说fit <- auto.arima(mydata)
选择auto.arima()
模型。有没有办法从适合中提取ARIMA(1,1,0)(0,1,1)[12]
,p
,d
(以及q
,P
,D
)的值?最后,我想自动分配六个变量,如下所示:
Q
答案 0 :(得分:4)
如果查看?auto.arima
,您将知道它返回与stats::arima
相同的对象。如果您进一步查看?arima
,您会看到所需信息可以从返回值的$model
中找到。 $model
的详细信息可以从?KalmanLike
:
phi, theta: numeric vectors of length >= 0 giving AR and MA parameters.
Delta: vector of differencing coefficients, so an ARMA model is
fitted to ‘y[t] - Delta[1]*y[t-1] - ...’.
所以,你应该这样做:
p <- length(fit$model$phi)
q <- length(fit$model$theta)
d <- fit$model$Delta
来自?auto.arima
的示例:
library(forecast)
fit <- auto.arima(WWWusage)
length(fit$model$phi) ## 1
length(fit$model$theta) ## 1
fit$model$Delta ## 1
fit$coef
# ar1 ma1
# 0.6503760 0.5255959
或者(实际上更好),您可以参考$arma
值:
arma: A compact form of the specification, as a vector giving the
number of AR, MA, seasonal AR and seasonal MA coefficients,
plus the period and the number of non-seasonal and seasonal
differences.
但是你需要正确和仔细地匹配它们。对于上面的例子,有:
fit$arma
# [1] 1 1 0 0 1 1 0
使用符号ARIMA(p,d,q)(P,D,Q)[m]
,我们可以为清晰呈现添加名称属性:
setNames(fit$arma, c("p", "q", "P", "Q", "m", "d", "D"))
# p q P Q m d D
# 1 1 0 0 1 1 0