Question

我有以下zoo个对象Z，代表每日股票回报的时间序列：

     structure(c(NA, 0.00559252442304969, 0.00654699049630403, -0.00398657154846838, 
-0.00624956112632535, -0.00777275296777835, -0.017518871955562, 
0.0108002319512903, -0.00709931875224101, -0.0104723385815398, 
0.0148894241296256, 0.00287666307083789, 0.00107565435640011, 
-0.0126790830945559, -0.000145106290357688, 0.00928815035193398, 
-0.00582356747429735, 0.00665316748625977, 0.00933908045977017, 
0.0151601423487544, -0.00371590829418778, 0.00570021111893038, 
0.00412847246518799, -0.00689895470383284, 0.00456108343274164, 
-0.00523889354568319, 0.0019661540622149, 0.012684841264279, 
0.0114186851211073, -0.00520013684570642), .Names = c("1986-01-01", 
"1986-01-02", "1986-01-03", "1986-01-06", "1986-01-07", "1986-01-08", 
"1986-01-09", "1986-01-10", "1986-01-13", "1986-01-14", "1986-01-15", 
"1986-01-16", "1986-01-17", "1986-01-20", "1986-01-21", "1986-01-22", 
"1986-01-23", "1986-01-24", "1986-01-27", "1986-01-28", "1986-01-29", 
"1986-01-30", "1986-01-31", "1986-02-03", "1986-02-04", "1986-02-05", 
"1986-02-06", "1986-02-07", "1986-02-10", "1986-02-11"), index = structure(c(5844, 
5845, 5846, 5849, 5850, 5851, 5852, 5853, 5856, 5857, 5858, 5859, 
5860, 5863, 5864, 5865, 5866, 5867, 5870, 5871, 5872, 5873, 5874, 
5877, 5878, 5879, 5880, 5881, 5884, 5885), class = "Date"), class = "zoo")

我需要计算每个月前10天的平均回报率。我使用了以下代码（其中apply.rolling是PerformanceAnalytics包中的函数）：

apply.rolling(Z, 10)

但是，这样可以在滚动的基础上返回10天而不会忽略当月的休息日。

任何想法如何执行？

Answer 1

以下是tapply的方法：

# Some example data:
library(zoo)
set.seed(1)
dates <- as.Date("2014-01-01") + seq(0, 58)
Z <- zoo(rnorm(59), dates)

# Calculate mean of days 1-10 for each month:
tapply(Z, format(time(Z), "%Y-%m"), 
       function(x) mean(x[as.integer(format(time(x), "%d")) <= 10]))

#     2014-01     2014-02 
#  0.13220278 -0.03159011

Answer 2

使用lubridate而不是rollapply的{{1}}和zoo略有不同的方法：

apply.rolling

library(zoo) library(lubridate) ## set.seed(123) Df <- data.frame( Date=seq.Date( from=as.Date("1986-01-01"), by="day", length.out=365), Value=c(NA,rnorm(364))) ## Z <- zoo(Df$Value,Df$Date) ## ## ath thru bth days of month fDays <- function(x,a=1,b=10) { x[(mday(x) %in% a:b)] } ## > rollapply(fDays(Z),FUN=mean,width=10,by=10,na.rm=TRUE,align="left") 1986-01-01 1986-02-01 1986-03-01 1986-04-01 1986-05-01 1986-06-01 1986-07-01 1986-08-01 1.324354e-01 3.220446e-01 -1.401720e-01 6.546170e-01 -2.057204e-01 1.273702e-01 6.596781e-05 -2.615902e-01 1986-09-01 1986-10-01 1986-11-01 1986-12-01 -4.955370e-02 -5.366626e-02 6.447457e-02 2.297708e-01不会影响价值的地方，只显示当月的第1天而不是第5天（align="left"，默认值）或第10天（{{ 1}}）。

Answer 3

试试这个：

aggregate(Z, as.yearmon, function(x) mean(head(x, 10), na.rm = TRUE))

计算前10天的回报

3 个答案: