计算R期间1年以上时间窗口的累计产品

我有问题要求我计算一系列1期间收益的滚动产品。滚动窗口的长度是可变的。目的是获得尽可能接近12个月窗口的1期回报滚动产品。计算R期间1年以上时间窗口的累计产品

我已经能够通过使用蛮力通过for循环和if语句生成工作解决方案，但是我想知道是否有优雅的解决方案。我花了很多时间尝试rollapply和其他类似的功能，但我一直无法获得解决方案。

以下数据说明了该问题。

date  rt_1_period rt_12_mth_window 
1 04-04-13 NA   NA 
2 10-04-13 0.729096362 NA 
3 24-05-13 1.002535647 NA 
4 30-05-13 0.993675716 NA 
5 21-07-13 1.002662843 NA 
6 03-08-13 1.009516582 NA 
7 01-09-13 0.963099395 NA 
8 20-10-13 1.012470278 NA 
9 25-10-13 1.01308502 NA 
10 03-11-13 1.005440704 NA 
11 01-01-14 1.024208021 NA 
12 11-01-14 0.996613924 NA 
13 17-02-14 1.009811368 NA 
14 24-02-14 1.008139557 NA 
15 30-03-14 1.002794709 NA 
16 30-04-14 0.998745849 1.042345473 
17 02-05-14 1.002324076 1.044767963 
18 27-06-14 0.997741026 1.046389027 
19 24-08-14 1.015767546 1.050072129 
20 05-09-14 1.014405005 1.106010894 
21 02-11-14 1.013830296 1.09319212 
22 09-11-14 1.013127219 1.101549487 
23 16-11-14 1.012614177 1.115444628 
24 18-01-15 0.986893629 1.078458006 
25 24-01-15 1.028120919 1.108785236 
26 10-04-15 0.912452762 0.991025615 
27 09-08-15 1.004676152 0.981376513 
28 07-01-16 1.004236123 0.934086003 
29 01-04-16 1.02341302 0.94215696

在12个月内返回行29的示例中被计算为从行26 1期间返回到29，因为15年2月4日（365天从16年1月4日）的产物被包含间第25行和第26行。另一方面，第15行的12个月的回报是NA，因为30-03-13（从30-03-14的365天）超出了我有可观察到的1期回报的时间窗口之外。

如果有人能提出某种解决这个问题的方法，我会很高兴。

只是为了清楚起见，如果提供的数据没有多大意义，那是因为这是为了说明目的而创建的较大数据库的减少版本。

来源

2016-04-24 Abdd

你可以使用xts和lubridate为simplyfying日期操作

数据：

require(xts) 
require(lubridate) 


DF = read.csv(text=" 
date,rt_1_period,rt_12_mth_window 
04-04-13,   , 
10-04-13,0.729096362, 
24-05-13,1.002535647, 
30-05-13,0.993675716, 
21-07-13,1.002662843, 
03-08-13,1.009516582, 
01-09-13,0.963099395, 
20-10-13,1.012470278, 
25-10-13,1.01308502 , 
03-11-13,1.005440704, 
01-01-14,1.024208021, 
11-01-14,0.996613924, 
17-02-14,1.009811368, 
24-02-14,1.008139557, 
30-03-14,1.002794709, 
30-04-14,0.998745849,1.042345473 
02-05-14,1.002324076,1.044767963 
27-06-14,0.997741026,1.046389027 
24-08-14,1.015767546,1.050072129 
05-09-14,1.014405005,1.106010894 
02-11-14,1.013830296,1.09319212 
09-11-14,1.013127219,1.101549487 
16-11-14,1.012614177,1.115444628 
18-01-15,0.986893629,1.078458006 
24-01-15,1.028120919,1.108785236 
10-04-15,0.912452762,0.991025615 
09-08-15,1.004676152,0.981376513 
07-01-16,1.004236123,0.934086003 
01-04-16,1.02341302 ,0.94215696",header=TRUE,stringsAsFactors=FALSE,na.strings="") 


#Convert to xts time series for ease in date manipulation 

DF_xts = xts(DF[,-1],order.by = as.Date(DF[,1],format="%d-%m-%y")) 
head(DF_xts) 
# 
#    rt_1_period rt_12_mth_window 
#2013-04-04   NA    NA 
#2013-04-10 0.729096362    NA 
#2013-05-24 1.002535647    NA 
#2013-05-30 0.993675716    NA 
#2013-07-21 1.002662843    NA 
#2013-08-03 1.009516582    NA 




#set lag period as 1 year 
lagPeriod = 1

累积12米产品：

对于每一天构建一个窗口[prevYearDate，日期] ，子集1m返回位于此窗口中，计算累积产品并选择最后一个产品

rt_12_mth_window_Calc = do.call(rbind,lapply(as.Date(index(DF_xts)),function(x) { 

prevYearDate = x-years(lagPeriod) 

rt_12_mth_window_Calc = last(cumprod(DF_xts[paste0(prevYearDate,"/",x),"rt_1_period"])) 
colnames(rt_12_mth_window_Calc) = "rt_12_mth_window_Calc" 

return(rt_12_mth_window_Calc) 

}))

最终数据集：

#Merge with original time series for final dataset 

new_DF = merge.xts(DF_xts,rt_12_mth_window_Calc) 

#Calculate difference in original and calculated 12 month returns 
new_DF$delta = new_DF$rt_12_mth_window_Calc - new_DF$rt_12_mth_window 
new_DF 

#   rt_1_period rt_12_mth_window rt_12_mth_window_Calc   delta 
#2013-04-04   NA    NA     NA    NA 
#2013-04-10 0.729096362    NA     NA    NA 
#2013-05-24 1.002535647    NA     NA    NA 
#2013-05-30 0.993675716    NA     NA    NA 
#2013-07-21 1.002662843    NA     NA    NA 
#2013-08-03 1.009516582    NA     NA    NA 
#2013-09-01 0.963099395    NA     NA    NA 
#2013-10-20 1.012470278    NA     NA    NA 
#2013-10-25 1.013085020    NA     NA    NA 
#2013-11-03 1.005440704    NA     NA    NA 
#2014-01-01 1.024208021    NA     NA    NA 
#2014-01-11 0.996613924    NA     NA    NA 
#2014-02-17 1.009811368    NA     NA    NA 
#2014-02-24 1.008139557    NA     NA    NA 
#2014-03-30 1.002794709    NA     NA    NA 
#2014-04-30 0.998745849  1.042345473   1.042345470 -2.64001643e-09 
#2014-05-02 1.002324076  1.044767963   1.044767960 -2.54864396e-09 
#2014-06-27 0.997741026  1.046389027   1.046389025 -1.97754613e-09 
#2014-08-24 1.015767546  1.050072129   1.050072127 -1.66086833e-09 
#2014-09-05 1.014405005  1.106010894   1.106010893 -1.34046041e-09 
#2014-11-02 1.013830296  1.093192120   1.093192120 -6.47777387e-11 
#2014-11-09 1.013127219  1.101549487   1.101549488 5.99306826e-10 
#2014-11-16 1.012614177  1.115444628   1.115444628 -1.89856353e-10 
#2015-01-18 0.986893629  1.078458006   1.078458005 -1.15637744e-09 
#2015-01-24 1.028120919  1.108785236   1.108785235 -9.57268265e-10 
#2015-04-10 0.912452762  0.991025615   0.991025613 -1.54581248e-09 
#2015-08-09 1.004676152  0.981376513   0.996850412 1.54738992e-02 
#2016-01-07 1.004236123  0.934086003   0.934086002 -9.15302278e-10 
#2016-04-01 1.023413020  0.942156960   0.942156960 -1.82048598e-10

的计算，一部开拓创新的值非常接近的，除了 2015年8月9日的所有意见，在价值观的偏差为1.55％，你能确认你这段时期的计算

来源

2016-04-24 12:48:21 OdeToMyFiddle

非常感谢您的回答！非常感激！你对2015-08-09是正确的。我使用的数据是模拟Excel电子表格的复制粘贴，并且您的版本是正确的。我已经开始使用你的代码，似乎运作良好。谢谢！ – Abdd

刚刚尝试运行真实数据集上的代码，这是我发布的已删除版本的代码，并且出现以下错误 '应用程序错误（coredata（x），2，函数（y）cumprod（y））：暗淡（X）必须有一个正面长度' 你知道什么可能是起源？在此先感谢 – Abdd

您是否可以更新您的问题，并包括您正在执行的函数调用以及dput（头（数据））'wherre'data'是您的输入数据。只有这样我们才能复制错误 – OdeToMyFiddle

这是一个解决方案，只取决于xts，它可能是更直接的一些。

library(xts) 
x <- as.xts(read.zoo(text="date,rt_1_period,rt_12_mth_window 
04-04-13,   , 
10-04-13,0.729096362, 
24-05-13,1.002535647, 
30-05-13,0.993675716, 
21-07-13,1.002662843, 
03-08-13,1.009516582, 
01-09-13,0.963099395, 
20-10-13,1.012470278, 
25-10-13,1.013085020, 
03-11-13,1.005440704, 
01-01-14,1.024208021, 
11-01-14,0.996613924, 
17-02-14,1.009811368, 
24-02-14,1.008139557, 
30-03-14,1.002794709, 
30-04-14,0.998745849,1.042345473 
02-05-14,1.002324076,1.044767963 
27-06-14,0.997741026,1.046389027 
24-08-14,1.015767546,1.050072129 
05-09-14,1.014405005,1.106010894 
02-11-14,1.013830296,1.09319212 
09-11-14,1.013127219,1.101549487 
16-11-14,1.012614177,1.115444628 
18-01-15,0.986893629,1.078458006 
24-01-15,1.028120919,1.108785236 
10-04-15,0.912452762,0.991025615 
09-08-15,1.004676152,0.981376513 
07-01-16,1.004236123,0.934086003 
01-04-16,1.023413020,0.94215696", header=TRUE, sep=",", format="%d-%m-%y")) 
ix <- index(x)  # index values 
ixlag <- ix-365  # 1-year lag index values 
x$rt_12 <- NA_real_ # initialize result column 
for(i in which(ixlag > ix[1])) { 
    # 1-year subset 
    xyear <- window(x, start=ixlag[i], end=ix[i]) 
    # calculate product and update result column 
    x[i,"rt_12"] <- prod(xyear[,"rt_1_period"]) 
}

来源

2016-04-24 21:50:44

感谢您的替代解决方案！ – Abdd

计算R期间1年以上时间窗口的累计产品

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