接下来几天的移动平均值，不包括当天

Question

我有一个数据 table，其中包含 return 一家公司的数据，如下所示：

df=structure(list(Date = structure(c(13236, 13237, 13238, 13239, 
                                      13240, 13241, 13242, 13243, 13244, 13245, 13246, 13247, 13248, 
                                      13249, 13250, 13251), class = "Date"), IBES = c("@O5G", "@O5G", 
                                                                                      "@O5G", "@O5G", "@O5G", "@O5G", "@O5G", "@O5G", "@O5G", "@O5G", 
                                                                                      "@O5G", "@O5G", "@O5G", "@O5G", "@O5G", "@O5G"), MktAdjReturn = c(-0.00381466643441897, 
                                                                                                                                                        -0.00834070809256926, -0.0193226301897589, NA, NA, -0.00885564092195712, 
                                                                                                                                                        -0.051612619547402, -0.0065292323057804, 0.042244140103735, 0.003100395243401, 
                                                                                                                                                        NA, NA, -0.00486229222347689, -0.0184708840023963, 0.00273824763632391, 
                                                                                                                                                        -0.00510010246255499)), .Names = c("Date", "IBES", "MktAdjReturn"
                                                                                                                                                        ), class = c("data.table", "data.frame"), row.names = c(NA, -16L
                                                                                                                                                        ))

我想计算除当天以外的接下来5天的移动平均线。因此，代码应该跳过 NA 并获取下一个可用的 returns。例如。对于第一个日期 2006-03-29，5 天移动平均线应包含：-0.008340708 ; -0.019322630 ; -0.008855641 ; -0.051612620 ; -0,006529232.

我试过：

rollapply(MktAdjReturn,width = 5,FUN=mean,align = "left",fill = NA,na.rm=T), by=c("IBES")

但是，它也包括当前日期，我不确定它对 fill = NA 的确切作用。

Answer 1

您需要做一些更改。假设，您需要按 IBES 对结果进行分组，然后您可以使用 dplyr。

首先，您应该删除值为 NA 的行，因为 OP 不想考虑那些宽度为 roll 的行。执行计算后，我们将 join 与 main data.frame 的结果。

要排除当前行并考虑下一行，您可以对滚动平均值使用 lead。

解决方案应如下所示：

library(tidyverse)
library(zoo)


df %>% group_by(IBES) %>%
filter(!is.na(MktAdjReturn)) %>% #Remove rows with NA
mutate(rollmean = lead(rollapply(MktAdjReturn, 5, FUN=mean, na.rm=TRUE,
                                 align = "left", fill = NA))) %>%
  right_join(df, by=c("Date", "IBES")) %>% #Join back
  mutate(LastNonNaMean = max(which(!is.na(rollmean)))) %>%
  fill(rollmean) %>%  #Fill to populate rows with NA values for mean. 
  mutate(rollmean = ifelse(row_number()>LastNonNaMean, NA, rollmean)) %>%
  select(Date, IBES, MktAdjReturn = MktAdjReturn.x, rollmean)

# # A tibble: 16 x 4
# # Groups: IBES [1]
# Date       IBES  MktAdjReturn  rollmean
# <date>     <chr>        <dbl>     <dbl>
# 1 2006-03-29 @O5G     - 0.00381 - 0.0189 
# 2 2006-03-30 @O5G     - 0.00834 - 0.00882
# 3 2006-03-31 @O5G     - 0.0193  - 0.00433
# 4 2006-04-01 @O5G      NA       - 0.00433
# 5 2006-04-02 @O5G      NA       - 0.00433
# 6 2006-04-03 @O5G     - 0.00886 - 0.00353
# 7 2006-04-04 @O5G     - 0.0516    0.00310
# 8 2006-04-05 @O5G     - 0.00653   0.00495
# 9 2006-04-06 @O5G       0.0422  - 0.00452
# 10 2006-04-07 @O5G       0.00310  NA      
# 11 2006-04-08 @O5G      NA        NA      
# 12 2006-04-09 @O5G      NA        NA      
# 13 2006-04-10 @O5G     - 0.00486  NA      
# 14 2006-04-11 @O5G     - 0.0185   NA      
# 15 2006-04-12 @O5G       0.00274  NA      
# 16 2006-04-13 @O5G     - 0.00510  NA
#

接下来几天的移动平均值，不包括当天

Moving mean for the following days not including the current day

r

moving-average