Question

我想计算R值随窗口大小变化的移动平均值。更具体地说：移动平均线应在三年内计算，但数据（时间序列）的使用频率较高，并且每个三年窗口的窗口大小可能会有所不同。

假设以下数据集：

library(data.table)
set.seed(1)   # reproduceable data
dataset <- data.table(ID=c(rep("A",2208),rep("B",2208)),
x = c(rnorm(2208*2)), time=c(seq(as.Date("1988/03/15"),
as.Date("2000/04/16"), "day"),seq(as.Date("1988/03/15"),
as.Date("2000/04/16"), "day")))

应该为两个ID（人A和B）计算变量x的三年移动平均值。可以最好使用zoo和datatable来做到这一点吗？但是任何解决方案都可以。

请注意，我知道如何使用固定的窗口大小执行此操作，这里的问题是窗口大小不断变化。

Answer 1

如果我理解正确，那么OP希望精确地跨度3年。由于这可能包括leap年，因此窗口大小可以是1095天或1096天。

这可以通过将聚集在非等额联接中以及lubridate的回滚日期算术来解决。

library(data.table)
library(lubridate)
# create 3 years windows for each ID for later non-equi join
win <- dataset[, CJ(ID = ID, start = time, unique = TRUE)][
  # make sure to pick
  , end := start %m+% years(3) - days(1)][
    # remove windows which end out of date range
    end <= max(start)]
win

      ID      start        end
   1:  A 1988-03-15 1991-03-14
   2:  A 1988-03-16 1991-03-15
   3:  A 1988-03-17 1991-03-16
   4:  A 1988-03-18 1991-03-17
   5:  A 1988-03-19 1991-03-18
  ---                         
6638:  B 1997-04-13 2000-04-12
6639:  B 1997-04-14 2000-04-13
6640:  B 1997-04-15 2000-04-14
6641:  B 1997-04-16 2000-04-15
6642:  B 1997-04-17 2000-04-16

# check window lengths
win[, .N, by = .(days = end - start + 1L)]

        days    N
1: 1095 days 2166
2: 1096 days 4476

# see what happens in leap years
win[leap_year(start) & month(start) == 2 & day(start) %in% 28:29, 
  .(start, end, days = end - start + 1L)]

        start        end      days
1: 1992-02-28 1995-02-27 1096 days
2: 1992-02-29 1995-02-27 1095 days
3: 1996-02-28 1999-02-27 1096 days
4: 1996-02-29 1999-02-27 1095 days
5: 1992-02-28 1995-02-27 1096 days
6: 1992-02-29 1995-02-27 1095 days
7: 1996-02-28 1999-02-27 1096 days
8: 1996-02-29 1999-02-27 1095 days

win[leap_year(end) & month(end) == 2 & day(end) %in% 28:29,
    .(start, end, days = end - start + 1L)]

        start        end      days
1: 1989-03-01 1992-02-29 1096 days
2: 1993-03-01 1996-02-29 1096 days
3: 1997-03-01 2000-02-29 1096 days
4: 1989-03-01 1992-02-29 1096 days
5: 1993-03-01 1996-02-29 1096 days
6: 1997-03-01 2000-02-29 1096 days

# aggregate in a non-equi-join
dataset[win, on = .(ID, time >= start, time <= end), by = .EACHI, .(avg = mean(x))]

      ID       time       time         avg
   1:  A 1988-03-15 1991-03-14 -0.01184078
   2:  A 1988-03-16 1991-03-15 -0.01317813
   3:  A 1988-03-17 1991-03-16 -0.01179571
   4:  A 1988-03-18 1991-03-17 -0.01006100
   5:  A 1988-03-19 1991-03-18 -0.01221798
  ---                                     
6638:  B 1997-04-13 2000-04-12 -0.03412214
6639:  B 1997-04-14 2000-04-13 -0.03604176
6640:  B 1997-04-15 2000-04-14 -0.03556291
6641:  B 1997-04-16 2000-04-15 -0.03392185
6642:  B 1997-04-17 2000-04-16 -0.03393674

Answer 2

正如@ A.Suliman在评论中提到的那样，示例数据的窗口宽度是固定的，但让我们假设您的真实数据没有，这就是您的文字所说的。

width中的参数rollapply不必是常数，因此我们可以先计算宽度，确保将窗口向左对齐，然后运行rollaply。

library(zoo)
library(tidyverse)
dataset %>% 
  arrange(ID,time) %>%
  group_by(ID) %>%
  mutate(avg = rollapply(x, FUN = mean, align = "left",
                         width = map_dbl(time, ~which.max(time[time < .x + 3*365.25])) - row_number()+1))
#                          
# # A tibble: 8,832 x 4
# # Groups:   ID [2]
#         ID       x       time    avg
#     <fctr>   <dbl>     <date>  <dbl>
#   1      A -0.0258 1988-03-15 0.0109
#   2      A -0.0258 1988-03-15 0.0109
#   3      A -0.1562 1988-03-16 0.0115
#   4      A -0.1562 1988-03-16 0.0115
#   5      A  0.8193 1988-03-17 0.0115
#   6      A  0.8193 1988-03-17 0.0112
#   7      A -1.1136 1988-03-18 0.0102
#   8      A -1.1136 1988-03-18 0.0107
#   9      A -0.9336 1988-03-19 0.0105
#  10      A -0.9336 1988-03-19 0.0109

窗口变化的移动平均线

2 个答案: