在 R 中展开/估算时间序列答案

【问题标题】：Expand / impute time series in R在 R 中展开/估算时间序列
【发布时间】：2021-04-26 07:50:59
【问题描述】：

我有一个每周需要的时间序列，但我目前拥有每季度的数据。例如：

 R> test     
   Quarter       week totA totB totC totD
 1       1 2015-12-28 1745 1720   11 1714
 2       2 2016-03-28 1736 1718    7 1710
 3       3 2016-06-27 1777 1768    5 1750
 4       4 2016-09-26 1833 1815   13 1795
 5       1 2016-12-26 1708 1697    6 1677
 R>

我想要的是每周获得信息，每个总数（totA 到 totD）需要除以下一个季度之前的周数（即 13，因为有 13 2016 年季度中的周数 - 但如果有一年有 53 周（例如 2015 年），则偶尔可能是 14 周），因此季度总数相同。因此，根据上面的示例，前 26 周变为：

 1            1      2015-12-28 134.9 132.3 0.8462 131.8
 2            2      2016-01-04 134.9 132.3 0.8462 131.8
 3            3      2016-01-11 134.9 132.3 0.8462 131.8
 4            4      2016-01-18 134.9 132.3 0.8462 131.8
 5            5      2016-01-25 134.9 132.3 0.8462 131.8
 6            6      2016-02-01 134.9 132.3 0.8462 131.8
 7            7      2016-02-08 134.9 132.3 0.8462 131.8
 8            8      2016-02-15 134.9 132.3 0.8462 131.8
 9            9      2016-02-22 134.9 132.3 0.8462 131.8
 10          10      2016-02-29 134.9 132.3 0.8462 131.8
 11          11      2016-03-07 134.9 132.3 0.8462 131.8
 12          12      2016-03-14 134.9 132.3 0.8462 131.8
 13          13      2016-03-21 134.9 132.3 0.8462 131.8
 14          14      2016-03-28 133.5 132.2 0.5385 131.5
 15          15      2016-04-04 133.5 132.2 0.5385 131.5
 16          16      2016-04-11 133.5 132.2 0.5385 131.5
 17          17      2016-04-18 133.5 132.2 0.5385 131.5
 18          18      2016-04-25 133.5 132.2 0.5385 131.5
 19          19      2016-05-02 133.5 132.2 0.5385 131.5
 20          20      2016-05-09 133.5 132.2 0.5385 131.5
 21          21      2016-05-16 133.5 132.2 0.5385 131.5
 22          22      2016-05-23 133.5 132.2 0.5385 131.5
 23          23      2016-05-30 133.5 132.2 0.5385 131.5
 24          24      2016-06-06 133.5 132.2 0.5385 131.5
 25          25      2016-06-13 133.5 132.2 0.5385 131.5
 26          26      2016-06-20 133.5 132.2 0.5385 131.5

 R>

这是使用以下方法获得的：

rbind(
    data.frame(Week_number=c(1:13),
               Week_commencing=seq(as.Date("2015-12-28"), by=7, len=13),
               totA=rep(1754/13,13),
               totB=rep(1720/13,13),
               totC=rep(11/13,13),
               totD=rep(1714/13,13)
               ),
    data.frame(Week_number=c(14:26),
               Week_commencing=seq(as.Date("2016-03-28"), by=7, len=13),
               totA=rep(1736/13,13),
               totB=rep(1718/13,13),
      totC=rep(7/13,13),
      totD=rep(1710/13,13)
      )
)

But there's clearly a better way of doing it rather than manually... The data set is, of course, much larger!

I've tried a few things, but other than creating a sequence of weeks and then filling it in manually as above, I'm going around in circles. I'm sure there's a way to do it in the tidyverse, but I can't figure out how (most of my R is self-taught, and from before when tidyverse was available). Any help would be appreciated!

【问题讨论】：

我无法理解您的需求。如果您可以提供有关不同列是什么的更多信息，例如，您可能会得到更好的答案。现有的week 列显示了什么。提供所需输出的示例也将大有帮助！
谢谢。我试图添加更多说明。现有的“周”只是本季度的第一个星期一，但我需要一年中每一周的数据。

标签： r time-series imputation

【解决方案1】：

由于此解决方案中使用的所有信息均来自数据，因此假设最后一个季度的天数（或周数）与其前一个季度相同。所以你可能想检查一下......

library(dplyr)
library(lubridate)
library(purrr)
library(tidyr)

test %>%
  mutate(week = ymd(week),
         weekCount = if_else(week != max(week),
                             as.double(abs(week - lead(week)))/7,
                             as.double(abs(week - lag(week)))/7),
         weekInQuarter = map(weekCount, ~ seq_len(.)),
         across(totA:totD, ~ ./weekCount)) %>%
  unnest(weekInQuarter) %>%
  mutate(week = week + weeks(weekInQuarter - 1)) %>%
  select(- weekCount)

#     Quarter week        totA  totB  totC  totD weekInQuarter
#       <dbl> <date>     <dbl> <dbl> <dbl> <dbl>         <int>
#   1       1 2015-12-28  134.  132. 0.846  132.             1
#   2       1 2016-01-04  134.  132. 0.846  132.             2
#   3       1 2016-01-11  134.  132. 0.846  132.             3
#   4       1 2016-01-18  134.  132. 0.846  132.             4
#   5       1 2016-01-25  134.  132. 0.846  132.             5
#   6       1 2016-02-01  134.  132. 0.846  132.             6
#   7       1 2016-02-08  134.  132. 0.846  132.             7
#   8       1 2016-02-15  134.  132. 0.846  132.             8
#   9       1 2016-02-22  134.  132. 0.846  132.             9
#  10       1 2016-02-29  134.  132. 0.846  132.            10

【讨论】：