滚动回归返回多个对象 [英] Rolling regression return multiple objects
问题描述
最后，我希望函数返回三(3)个对象:
- 预测
- 系数
- T值
- R ^ 2
下面是代码:
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(即mod)中提取t值之后，我该怎么做才能使函数返回三(3)个独立的对象(即预测，系数和T值)?
我对R相当陌生，对于功能真的很陌生，而对于Zoo则是陌生的，所以我陷入了困境.
任何帮助将不胜感激.
我希望我能正确理解您的名字，但这是对您的函数的一个小修改:
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
要查看其工作原理，请做一个带有回归的小例子:
x <- rnorm(1000); y <- 2*x + rnorm(1000)
reg <- lm(y ~ x)
summary(reg)$coef
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.02694322 0.03035502 0.8876033 0.374968
x 1.97572544 0.03177346 62.1816310 0.000000
如您所见，
首先调用summary，然后获取其系数(coef(summary(reg))也可以使用)，这将为您提供一个包含估计值，标准误差和t值的表.因此，估算值保存在该表的第1列中，t值保存在第3列中.这就是我在更新的rolling.regression函数中获取它们的方式.
编辑
我更新了解决方案；现在它还包含调整后的R2.如果只需要普通的R2，请摆脱.adj.
编辑2
快速又肮脏的黑客如何命名列:
rolling.regression <- function(series) {
mod <- dynlm(formula = y ~ L(y) + L(x), data = as.zoo(series)) # get model
nextOb <- max(series[,'int'])+1 # To get the first row that follows the window
if (nextOb<=nrow(zoodata)) { # You won't predict the last one
# 1) Make Predictions
predicted=predict(mod,newdata=data.frame(x=zoodata[nextOb,'x'],y=zoodata[nextOb,'y']))
attributes(predicted)<-NULL
#Get variable names
strVar <- c("Intercept", paste0("L", 1:(nrow(summary(mod)$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
}
}
I am trying to build a rolling regression function based on the example here, but in addition to returning the predicted values, I would like to return the some rolling model diagnostics (i.e. coefficients, t-values, and mabye R^2). I would like the results to be returned in discrete objects based on the type of results. The example provided in the link above sucessfully creates thr rolling predictions, but I need some assistance packaging and writing out the rolling model diagnostics:
In the end, I would like the function to return three (3) objects:
- Predictions
- Coefficients
- T values
- R^2
Below is the code:
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')
So after figuring out how to extract t values from model (i.e. mod) , what do I need to do to make the function return three (3) seperate objects (i.e. Predictions, Coefficients, and T-values)?
I am fairly new to R, really new to functions, and extreemly new to zoo, and I'm stuck.
Any assistance would be greatly appreciated.
I hope I got you correctly, but here is a small edit of your function:
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)$adj.r.squared)
}
}
rolling.window <- 20
results.z <- rollapply(zoodata, width=rolling.window, FUN=rolling.regression, by.column=F, align='right')
head(results.z)
predicted square.res coef_intercept coef_Ly coef_Lx tValue_intercept tValue_Ly tValue_Lx AdjR
20 10.849344 0.721452 0.26596465 0.5798046 1.049594 0.38309211 7.977627 13.59831 0.9140886
21 12.978791 2.713053 0.26262820 0.5796883 1.039882 0.37741499 7.993014 13.80632 0.9190757
22 9.814676 11.719999 0.08050796 0.5964808 1.073941 0.12523824 8.888657 15.01353 0.9340732
23 5.616781 15.013297 0.05084124 0.5984748 1.077133 0.08964998 9.881614 16.48967 0.9509550
24 3.763645 6.976454 0.26466039 0.5788949 1.068493 0.51810115 11.558724 17.22875 0.9542983
25 9.433157 31.772658 0.38577698 0.5812665 1.034862 0.70969330 10.728395 16.88175 0.9511061
To see how it works, make a small example with a regression:
x <- rnorm(1000); y <- 2*x + rnorm(1000)
reg <- lm(y ~ x)
summary(reg)$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
As you can see, calling summary first and then getting the coefficients of it (coef(summary(reg)) works as well) gives you a table with estimates, standard errors, and t-values. So estimates are saved in column 1 of that table, t-values in column 3. And that's how I obtain them in the updated rolling.regression function.
EDIT
I updated my solution; now it also contains the adjusted R2. If you just want the normal R2, get rid of the .adj.
EDIT 2
Quick and dirty hack how to name the columns:
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
}
}
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