R中的滚动回归xts对象 [英] Rolling regression xts object in R
问题描述
我正在尝试对 xts 对象执行滚动 100 天回归并返回所有日期的斜率系数的 t 统计量.我有一个 xts 对象，价格:
I am attempting to perform a rolling 100 day regression on an xts object and return the t statistic of the slope coefficient for all dates. I have an xts object, prices:
> tail(prices)
DBC EEM EFA GLD HYG IEF IWM IYR MDY TLT
2012-11-02 27.14 41.60 53.69 162.60 92.41 107.62 81.19 64.50 179.99 122.26
2012-11-05 27.37 41.80 53.56 163.23 92.26 107.88 81.73 64.02 181.10 122.95
2012-11-06 27.86 42.13 54.07 166.30 92.40 107.39 82.34 64.16 182.69 121.79
2012-11-07 27.34 41.44 53.26 166.49 91.85 108.29 80.34 63.84 178.90 124.00
2012-11-08 27.38 40.92 52.78 167.99 91.55 108.77 79.21 63.19 176.37 125.84
2012-11-09 27.60 41.00 52.80 167.82 91.39 108.78 79.38 62.98 176.80 125.98
还有一个函数，用于从最近 100 个价格的回归中生成 t 值，并作为与价格具有相同列名的 xts 对象返回:
And a function to generate the t value from a regression of the last 100 prices and return as xts object with the same column names as prices:
compute.regression <- function(x)
{
last100 = last(x,100)
fit <- lm (last100~time(last100))
tval <- unlist(lapply( coef(summary(fit)), "[" ,"(Intercept)" ,"t value"))
tval.mat <-as.matrix(tval)
tval.mat2 <- matrix(c(tval.mat), nrow=1, ncol=10)
new2 <- xts(tval.mat2, Sys.Date())
colnames(new2) <- tickers
return(new2)
}
> compute.regression(prices)
IWM EFA HYG EEM IYR IEF TLT DBC GLD MDY
2012-11-10 -7.642781 -14.33474 -16.28911 -14.52982 -15.85337 5.732489 -8.026495 -1.960392 -11.82474 8.686045
但是，此函数仅返回当前日期的 t 值.我需要一个 xts 对象，其中包含价格 xts 对象中包含的所有日期的 t 值.我正在尝试使用 rollapply 但没有得到任何地方:
However, this function only returns the t values for the current date. I need a xts object that includes the t values for all dates included in the prices xts object. I am attempting to use rollapply but not getting anywhere:
fit.r <- function (x)
{
hard <- lm (x~time(x))
return(hard)
}
compute.r <- function(x)
{
l100.ra <- rollapply(x, 100, fit.r, fill=NA, align='right')
tval <- unlist(lapply( coef(summary(l100.ra)), "[" ,"(Intercept)" ,"t value"))
tval.mat <-as.matrix(tval)
tval.mat2 <- matrix(c(tval.mat), nrow=1, ncol=10)
new2 <- xts(tval.mat2, Sys.Date())
colnames(new2) <- tickers
return(new2)
}
但这会返回错误:
compute.r(价格)zoo(rval, index(x)[i]) 中的错误:x":尝试定义无效的动物园对象
compute.r(prices) Error in zoo(rval, index(x)[i]) : "x" : attempt to define invalid zoo object
我知道问题出在 l100.ra 行中，但无法修复.有什么想法吗?
I know the problem is in the l100.ra line but cannot fix it. Any thoughts?
编辑 1
提供的功能如下:
x <- prices
f <- function (x) {
res <- coef(summary(lm(x ~ time(x)))
sapply(res, "[" ,"time(x)" ,"t value")
}
r <- rollapplyr(x, 100, f, by.column=FALSE)
运行良好，但将日期/时间格式更改为以下格式:
works well but changes the date/time format to that below:
> last(r,5)
Response MDY Response TLT
1352160000 15.38380 -7.913764
1352246400 14.81888 -7.957261
1352332800 13.96203 -7.666699
1352419200 13.28624 -7.299532
1352678400 12.75266 -7.100558
我一直在阅读时间戳，但不明白为什么会发生转换.另外，当我尝试:
I've been reading up on timestamps but don't understand why the conversion took place. Also, when I try:
last(index(r),5)
[1] "2012-11-06 GMT" "2012-11-07 GMT" "2012-11-08 GMT" "2012-11-09 GMT" "2012-11-12 GMT"
它以 POSIXct 格式显示时间戳.这里有什么线索吗?我需要转换回上面价格中的格式.
It shows the timestamps in POSIXct format. Any clues here? I need to convert back to the format in prices above.
推荐答案
您的 rollapply 调用应该调用一个返回 t 值的函数.现在它返回一个 lm 对象，这不是动物园对象的 coredata 的有效值.
Your rollapply call should call a function that returns the t-values. Right now it returns a lm object, which is not a valid value for the coredata of a zoo object.
让你的函数只返回 t 值，它就会起作用.
Make your function only return the t-values and it will work.
library(quantmod)
getSymbols("SPY;IEF")
x <- merge(Cl(SPY),Cl(IEF))
f <- function (x) {
res <- coef(summary(lm(x ~ time(x))))
sapply(res, "[" ,"(Intercept)" ,"t value")
}
r <- rollapplyr(x, 100, f, by.column=FALSE)
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