问题描述
对于 5 个货币对,我有 1033 个每日回报点的 xts,我想在这些货币对上运行滚动窗口回归,但是 rollapply 不适用于我定义的使用 lm() 的函数.这是我的数据:
I have an xts of 1033 daily returns points for 5 currency pairs on which I want to run a rolling window regression, but rollapply is not working for my defined function which uses lm(). Here is my data:
> head(fxr)
USDZAR USDEUR USDGBP USDCHF USDCAD
2007-10-18 -0.005028709 -0.0064079963 -0.003878743 -0.0099537170 -0.0006153215
2007-10-19 -0.001544470 0.0014275520 -0.001842564 0.0023058211 -0.0111410271
2007-10-22 0.010878027 0.0086642116 0.010599365 0.0051899551 0.0173792230
2007-10-23 -0.022783987 -0.0075236355 -0.010804304 -0.0041668499 -0.0144788687
2007-10-24 -0.006561223 0.0008545792 0.001024275 -0.0004261666 0.0049525483
2007-10-25 -0.014788901 -0.0048523001 -0.001434280 -0.0050425302 -0.0046422944
> tail(fxr)
USDZAR USDEUR USDGBP USDCHF USDCAD
2012-02-10 0.018619309 0.007548205 0.005526184 0.006348533 0.0067151342
2012-02-13 -0.006449463 -0.001055966 -0.002206810 -0.001638002 -0.0016995755
2012-02-14 0.006320364 0.006843933 0.006605875 0.005992935 0.0007001751
2012-02-15 -0.001666872 0.004319096 -0.001568874 0.003686840 -0.0015009759
2012-02-16 0.006419616 -0.003401364 -0.005194817 -0.002709588 -0.0019044761
2012-02-17 -0.004339687 -0.003675992 -0.003319899 -0.003043481 0.0000000000
我可以轻松地对整个数据集运行 lm 以针对其他货币对对 USDZAR 进行建模:
I can easily run an lm on it for the whole data set to model USDZAR against the other pairs:
> lm(USDZAR ~ ., data = fxr)$coefficients
(Intercept) USDEUR USDGBP USDCHF USDCAD
-1.309268e-05 5.575627e-01 1.664283e-01 -1.657206e-01 6.350490e-01
但是我想运行一个滚动的 62 天窗口来获得这些系数随时间的演变,因此我创建了一个函数 dolm 来执行此操作:
However I want to run a rolling 62 day window to get the evolution of these coefficients over time, so I create a function dolm which does this:
> dolm
function(x) {
return(lm(USDZAR ~ ., data = x)$coefficients)
}
但是,当我对此运行 rollapply 时,我得到以下信息:
However when I run rollapply on this I get the following:
> rollapply(fxr, 62, FUN = dolm)
Error in terms.formula(formula, data = data) :
'.' in formula and no 'data' argument
即使 dolm(fxr) 本身工作正常:
that is even though dolm(fxr) on its own works fine:
> dolm(fxr)
(Intercept) USDEUR USDGBP USDCHF USDCAD
-1.309268e-05 5.575627e-01 1.664283e-01 -1.657206e-01 6.350490e-01
这是怎么回事?如果 dolm 是一个更简单的函数,它似乎工作正常,例如意思:
What's going on here? It seems to work fine if dolm is a simpler function for example mean:
> dolm <- edit(dolm)
> dolm
function(x) {
return(mean(x))
}
> rollapply(fxr, 62, FUN = dolm)
USDZAR USDEUR USDGBP USDCHF USDCAD
2007-11-29 -1.766901e-04 -6.899297e-04 6.252596e-04 -1.155952e-03 7.021468e-04
2007-11-30 -1.266130e-04 -6.512204e-04 7.067767e-04 -1.098413e-03 7.247315e-04
2007-12-03 8.949942e-05 -6.406932e-04 6.637066e-04 -1.154806e-03 8.727564e-04
2007-12-04 2.042046e-04 -5.758493e-04 5.497422e-04 -1.116308e-03 7.124593e-04
2007-12-05 7.343586e-04 -4.899982e-04 6.161819e-04 -1.057904e-03 9.915495e-04
非常感谢任何帮助.基本上我想要的是在滚动的 62 天窗口中获得 USDZAR ~ USDEUR + USDGBP + USDCHF + USDCAD 回归的权重.
Any help much appreciated. Essentially what I want is to get the weightings for the regression of USDZAR ~ USDEUR + USDGBP + USDCHF + USDCAD over a rolling 62-day window.
推荐答案
这里有几个问题:
rollapply
传递一个矩阵,但lm
需要一个data.frame
.rollapply
将函数分别应用于每一列,除非我们指定by.column=FALSE
.- 您可能希望也可能不希望结果是与日期右对齐,但如果您使用
rollapplyr
:
rollapply
passes a matrix butlm
requires adata.frame
.rollapply
applies the function to each column separately unless wespecifyby.column=FALSE
.- you may or may not want the result to beright aligned with the dates but if you do use
rollapplyr
:
1) 结合以上我们有:
dolm <- function(x) coef(lm(USDZAR ~ ., data = as.data.frame(x))))
rollapplyr(fxr, 62, dolm, by.column = FALSE)
2) 上面 dolm
中 lm
的替代方法是使用 lm.fit
适用于矩阵并且速度更快:
2) An alternative to the lm
in the dolm
above is to use lm.fit
which directly works with matrices and is also faster:
dolm <- function(x) coef(lm.fit(cbind(Intercept = 1, x[,-1]), x[,1]))
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