Question

我正在尝试基于示例here构建滚动回归函数，但除了返回预测值之外，我还想返回一些滚动模型诊断（即系数，t值和mabye R ^ 2）。我希望根据结果类型在离散对象中返回结果。上面链接中提供的示例成功地创建了滚动预测，但我需要一些帮助打包并写出滚动模型诊断：

最后，我希望函数返回三（3）个对象：

预测
系数
T值
R ^ 2

以下是代码：

require(zoo)
require(dynlm)

## Create Some Dummy Data
set.seed(12345)
x <- rnorm(mean=3,sd=2,100)
y <- rep(NA,100)
y[1] <- x[1]
for(i in 2:100) y[i]=1+x[i-1]+0.5*y[i-1]+rnorm(1,0,0.5)
int <- 1:100
dummydata <- data.frame(int=int,x=x,y=y)
zoodata <- as.zoo(dummydata)


rolling.regression <- function(series) {
  mod <- dynlm(formula = y ~ L(y) + L(x), data = as.zoo(series)) # get model

  nextOb <- max(series[,'int'])+1 # To get the first row that follows the window
  if (nextOb<=nrow(zoodata)) {   # You won't predict the last one

    # 1) Make Predictions
    predicted <- predict(mod,newdata=data.frame(x=zoodata[nextOb,'x'],y=zoodata[nextOb,'y']))
    attributes(predicted) <- NULL
    c(predicted=predicted,square.res <-(predicted-zoodata[nextOb,'y'])^2)

    # 2) Extract coefficients
    #coefficients <- coef(mod)

    # 3) Extract rolling coefficient t values
    #tvalues <- ????(mod)

    # 4) Extract rolling R^2
    #rsq <-


  }
}    

rolling.window <- 20
results.z <-  rollapply(zoodata, width=rolling.window, FUN=rolling.regression, by.column=F, align='right')

因此，在弄清楚如何从模型中提取t值（即mod）之后，我需要做些什么来使函数返回三（3）个单独的对象（即预测，系数和T值）？

我对R来说相当新，对功能很陌生，对动物园来说是新手，而且我被卡住了。

非常感谢任何协助。

Answer 1

我希望我能帮到你，但这是对你功能的一个小编辑：

rolling.regression <- function(series) {
  mod <- dynlm(formula = y ~ L(y) + L(x), data = as.zoo(series)) # get model

  nextOb <- max(series[,'int'])+1 # To get the first row that follows the window
  if (nextOb<=nrow(zoodata)) {   # You won't predict the last one
    # 1) Make Predictions
    predicted=predict(mod,newdata=data.frame(x=zoodata[nextOb,'x'],y=zoodata[nextOb,'y']))
    attributes(predicted)<-NULL
    #Solution 1; Quicker to write
    #     c(predicted=predicted, 
    #       square.res=(predicted-zoodata[nextOb,'y'])^2,
    #       summary(mod)$coef[, 1],
    #       summary(mod)$coef[, 3],
    #       AdjR = summary(mod)$adj.r.squared)

    #Solution 2; Get column names right
    c(predicted=predicted, 
      square.res=(predicted-zoodata[nextOb,'y'])^2,
      coef_intercept = summary(mod)$coef[1, 1],
      coef_Ly = summary(mod)$coef[2, 1],
      coef_Lx = summary(mod)$coef[3, 1],
      tValue_intercept = summary(mod)$coef[1, 3],
      tValue_Ly = summary(mod)$coef[2, 3],
      tValue_Lx = summary(mod)$coef[3, 3],
      AdjR = summary(mod)$adj.r.squared)
  }
}



rolling.window <- 20
results.z <-  rollapply(zoodata, width=rolling.window, FUN=rolling.regression, by.column=F, align='right')

    head(results.z)
   predicted square.res coef_intercept   coef_Ly  coef_Lx tValue_intercept tValue_Ly tValue_Lx      AdjR
20 10.849344   0.721452     0.26596465 0.5798046 1.049594       0.38309211  7.977627  13.59831 0.9140886
21 12.978791   2.713053     0.26262820 0.5796883 1.039882       0.37741499  7.993014  13.80632 0.9190757
22  9.814676  11.719999     0.08050796 0.5964808 1.073941       0.12523824  8.888657  15.01353 0.9340732
23  5.616781  15.013297     0.05084124 0.5984748 1.077133       0.08964998  9.881614  16.48967 0.9509550
24  3.763645   6.976454     0.26466039 0.5788949 1.068493       0.51810115 11.558724  17.22875 0.9542983
25  9.433157  31.772658     0.38577698 0.5812665 1.034862       0.70969330 10.728395  16.88175 0.9511061

要了解它是如何工作的，请使用回归做一个小例子：

x <- rnorm(1000); y <- 2*x + rnorm(1000)
reg <- lm(y ~ x)
summary(reg)$coef
              Estimate Std. Error    t value Pr(>|t|)
(Intercept) 0.02694322 0.03035502  0.8876033 0.374968
x           1.97572544 0.03177346 62.1816310 0.000000

正如您所看到的，首先调用summary然后获取它的系数（coef(summary(reg))也可以正常工作）会为您提供一个包含估计值，标准误差和t值的表格。所以估计值保存在该表的第1列，第3列中的t值。这就是我在更新的rolling.regression函数中获取它们的方式。

修改

我更新了我的解决方案;现在它还包含调整后的R2。如果您只想要正常的R2，请删除.adj。

编辑2

快速而肮脏的黑客如何命名列：

rolling.regression <- function(series) { mod <- dynlm(formula = y ~ L(y) + L(x), data = as.zoo(series)) # get model nextOb <- max(series[,'int'])+1 # To get the first row that follows the window if (nextOb<=nrow(zoodata)) { # You won't predict the last one # 1) Make Predictions predicted=predict(mod,newdata=data.frame(x=zoodata[nextOb,'x'],y=zoodata[nextOb,'y'])) attributes(predicted)<-NULL #Get variable names strVar <- c("Intercept", paste0("L", 1:(nrow(summary(mod)$coef)-1))) vec <- c(predicted=predicted, square.res=(predicted-zoodata[nextOb,'y'])^2, AdjR = summary(mod)$adj.r.squared, summary(mod)$coef[, 1], summary(mod)$coef[, 3]) names(vec)[4:length(vec)] <- c(paste0("Coef_", strVar), paste0("tValue_", strVar)) vec } }

滚动回归返回多个对象

1 个答案: