逐行对列表中的 xts 元素求和

Question

我有一个名为 data 的 xts 对象，其中包含从 2015-01-01 17:00:00 到 2015-12-31 17:00:00 期间的 5 分钟回报。每个交易日从17:00:00开始，到次日同一时间结束，共计288个每日收益[（24小时*60分钟）/5分钟=288个日内收益]。回报表示为

head(data, 5)
                          DPRICE
2015-01-01 17:00:00 0.000000e+00
2015-01-01 17:05:00 9.797714e-05
2015-01-01 17:10:00 2.027022e-04
2015-01-01 17:15:00 2.735798e-04
2015-01-01 17:20:00 7.768653e-05

tail(data, 5)
                          DPRICE
2015-12-31 16:40:00 0.0001239429
2015-12-31 16:45:00 0.0001272704
2015-12-31 16:50:00 0.0010186764
2015-12-31 16:55:00 0.0006841370
2015-12-31 17:00:00 0.0002481227

根据 McMillan 和 Speight Daily FX Volatility Forecasts (2012)，我正在尝试通过每 5 分钟日内间隔的平均绝对值对数据进行标准化。

数学公式是：

我的*代码是

library(xts)
std_data = abs(data) #create absolute returns
D <- split(std_data, "days") #splits data to days
mts.days <- lapply(seq_along(D) - 1, function(i) {
  if (i > 0) rbind(D[[i]]["T17:00:00/T23:55:00"], D[[i + 1]]["T00:00:00/T16:55:00"])
}) #creates a list with 365 elements each containing 288 unique returns
dummy = mapply(sum, mts.days) #add the first,second... observations from each element

使用这段代码，我创建了一个包含 365 个 xts 元素的列表，每个元素都有维度

> dim(mts.days[[2]])
[1] 288   1 

我想从每个元素中添加相同的观察结果来创建上述函数的分母。

Answer 1

我不明白你的要求，但还是会试一试。

## generate bogus data
library(quantmod)
set.seed(123)
ndays <- 3
ndatperday <- 288
data <- cumsum(do.call("rbind", lapply(13:15, function(dd){
    xts(rnorm(ndatperday)/1e4,
        seq(as.POSIXct(paste0("2016-08-",dd," 17:00:00")),
            length = ndatperday, by = 300))

})))
colnames(data) <- "DPRICE"

## calculate percentage returns
ret <- ROC(data, type="discrete")

## this is probably not what you need: returns divided by the overall mean
ret/mean(abs(ret), na.rm=T)

## I suspect indeed that you need returns divided by the daily mean return
library(dplyr)
ret.df <- data.frame(ret)
## create a factor identifying the 3 days of bogus data
ret.df$day <- rep(paste0("2016-08-",13:15),each=ndatperday)
## compute daily mean return
dail <- ret.df %>%
    group_by(day) %>%
    summarise(mean=mean(abs(DPRICE), na.rm=TRUE))
## attach daily mean returns to the days they actually are associated to
ret.df <- ret.df %>% left_join(dail)
## normalize
ret.df$DPRICE <- ret.df$DPRICE/ret.df$mean

%%%%%%%%%

第二个镜头：读完论文 (http://onlinelibrary.wiley.com/doi/10.1002/for.1222/full) 我可能已经明白你的目的了：

library(quantmod)
library(dplyr)
set.seed(123)

## generate bogus 5-min series
ndays <- 365
ndatperday <- 288
data <- as.xts(zoo(0.1+cumsum(rt(ndays*ndatperday, df=3))/1e4,
                  seq(as.POSIXct("2015-01-01 17:00"),
                      as.POSIXct("2015-12-31 17:00"), by=300)))
colnames(data) <- "DPRICE"

## calculate 5-min percentage returns
ret <- ROC(data, type="discrete")

## create a factor identifying the 5-minute intra-day interval
ret.df <- as.data.frame(ret)
ret.df$intra5 <- strftime(index(ret), format="%H:%M")

## compute mean returns (over the year) for each of the 288 5-minute intra-day intervals
dail <- ret.df %>%
    group_by(intra5) %>%
    summarise(mean=mean(abs(DPRICE), na.rm=TRUE))

## attach mean returns to each datapoint
ret.df <- ret.df %>% left_join(dail)

## normalize
ret.df$DPRICE <- ret.df$DPRICE/ret.df$mean