这是我的data.frame:
data <- matrix(rnorm(10*5),nrow=25)
GDP <- data.frame(data )
GDP
X1 X2
1 -0.37000725 2.53311407
2 1.54825124 0.15811930
3 2.32926402 0.75203918
4 1.39942457 -0.42772401
5 -0.94124582 -0.73874833
6 0.83330085 0.14364736
7 0.73488659 -0.71502188
8 0.12321817 1.31648567
9 1.55536358 -1.57426731
10 1.42325808 0.04616108
11 -0.35875716 -0.02854382
12 -0.49774322 1.41312880
13 -1.88498804 0.82919301
14 -1.13962628 0.18335208
15 -0.45672902 1.33955701
16 1.17333357 1.20232913
17 -0.32018730 0.87183555
18 0.04167326 -0.11642683
19 -0.17698318 0.34282848
20 2.28473762 -0.98547134
21 -0.80361048 1.12771148
22 1.23063390 0.22982985
23 -0.03444458 0.91857055
24 -0.66244086 -0.21407559
25 -0.24960018 -2.72181616
有任何包,可以帮助我做一个简单的自回归线性回归,而不必在我的data.frame中创建另一列?这就是我想要的:
X1 = X1(t-1) + X2(t-2)
谢谢。
以下是一些方法:
1)达因
library(dyn)
Lag <- function(x, k = 1) lag(x, -k)
dyn$lm(X1 ~ Lag(X1) + Lag(X2) - 1, as.zoo(GDP))
给:Call:
lm(formula = dyn(X1 ~ Lag(X1) + Lag(X2) - 1), data = as.zoo(GDP))
Coefficients:
Lag(X1) Lag(X2)
-0.1876 0.0772
请注意,这也可以工作,但定义Lag
我们上面所做的使它看起来更漂亮。
dyn$lm(X1 ~ lag(X1, -1) + lag(X2, -1) - 1, as.zoo(GDP))
2) var
library(vars)
VAR(GDP, type = "none")
给:VAR Estimation Results:
=======================
Estimated coefficients for equation X1:
=======================================
Call:
X1 = X1.l1 + X2.l1
X1.l1 X2.l1
-0.18755204 0.07719922
Estimated coefficients for equation X2:
=======================================
Call:
X2 = X1.l1 + X2.l1
X1.l1 X2.l1
0.4433822 0.2558610
或者如果我们只想看第一个方程:
VAR(GDP, type = "none")[[1]]$X1
给:Call:
lm(formula = y ~ -1 + ., data = datamat)
Coefficients:
X1.l1 X2.l1
-0.1876 0.0772
3) No packages
n <- nrow(GDP)
lm(X1[-1] ~ X1[-n] + X2[-n] - 1, GDP)
给:Call:
lm(formula = X1[-1] ~ X1[-n] + X2[-n] - 1, data = GDP)
Coefficients:
X1[-n] X2[-n]
-0.1876 0.0772
注在上面的例子中,我们使用了以下的GDP:
GDP <-
structure(list(X1 = c(-0.480007101227991, -0.710506834821923,
-1.4008090378277, 0.234161619712456, 0.0798157911638669, -0.835197270889505,
0.598254213927639, -1.14352681562672, 1.03688327045929, 0.660297071029499,
-0.351328818974587, 0.790545641075689, -0.792678099784052, -0.357614703160382,
0.314291502993829, -0.431642261560374, 0.316918597548564, 0.5209261331865,
1.0013650951443, 1.05596920913398, -0.753506630185664, -1.4890660967781,
1.43183749932514, -0.423639570640277, 0.637317561276307), X2 = c(0.474962739361749,
-2.39846608215569, -0.98006715899912, -0.0271182048898923, 0.0296705736957689,
-1.24925308595335, -0.893230759394588, 0.241972221010069, -0.431946104440377,
-0.638101222832251, 0.844712933353179, 0.883298568281938, 0.996083349802754,
1.89504374477663, -0.148165464503539, 1.15286878557205, -0.425104835813157,
-1.38572745123415, 1.52226162248381, 1.55272897266444, 1.35700497284096,
0.389532599186254, 0.256357476163037, 1.29116051537444, -0.440232029770923
)), .Names = c("X1", "X2"), row.names = c(NA, -25L), class = "data.frame")