自动回归线性回归 data.frame [关闭]答案

【问题标题】：Auto regressive Linear Regression data.frame [closed]自动回归线性回归 data.frame [关闭]
【发布时间】：2016-01-17 01:03:34
【问题描述】：

这是我的数据框：

    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

是否有任何软件包可以帮助我进行简单的自回归线性回归，而无需在我的 data.frame 中创建另一列？这就是我想要的：

X1 = X1(t-1) + X2(t-2)

谢谢。

【问题讨论】：

看起来 dynlm 根据 stackoverflow.com/questions/13096787/… 具有滞后变量

标签： r dataframe linear-regression

【解决方案1】：

这里有几种方法：

1) 动态

library(dyn)
Lag <- function(x, k = 1) lag(x, -k)
dyn$lm(X1 ~ Lag(X1) + Lag(X2) - 1, as.zoo(GDP))

给予：

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，但是像我们上面所做的那样定义 Lag 使它看起来更漂亮一些。

dyn$lm(X1 ~ lag(X1, -1) + lag(X2, -1) - 1, as.zoo(GDP))

2) 变量

library(vars)
VAR(GDP, type = "none")

给予：

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

给予：

Call:
lm(formula = y ~ -1 + ., data = datamat)

Coefficients:
  X1.l1    X2.l1  
-0.1876   0.0772

3) 没有包

n <- nrow(GDP)
lm(X1[-1] ~ X1[-n] + X2[-n] - 1, GDP)

给予：

Call:
lm(formula = X1[-1] ~ X1[-n] + X2[-n] - 1, data = GDP)

Coefficients:
 X1[-n]   X2[-n]  
-0.1876   0.0772

注意我们在上面的例子中使用了以下 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")

【讨论】：