在分组的时间序列中填充缺失的日期-tidyverse方式？

问题描述

给出一个包含一个时间序列和一个或一个矿石分组字段的data.frame。因此，我们有几个时间序列-每个分组组合都有一个。
但是缺少一些日期。
那么，将这些日期与正确的分组值相加的最简单的方法（就最 tidyverse而言）？

Given a data.frame that contains a time series and one or ore grouping fields. So we have several time series - one for each grouping combination.But some dates are missing.So, what's the easiest (in terms of the most "tidyverse way") of adding these dates with the right grouping values?

通常我会说我生成一个包含所有日期的data.frame并使用我的时间序列进行full_join。但是，现在我们必须对分组值的每个组合进行此操作-并填写分组值。

Normally I would say I generate a data.frame with all dates and do a full_join with my time series. But now we have to do it for each combination of grouping values -- and fill in the grouping values.

让我们看一个示例：

首先，我创建一个缺少值的data.frame：

First I create a data.frame with missing values:

library(dplyr)
library(lubridate)

set.seed(1234)
# Time series should run vom 2017-01-01 til 2017-01-10
date <- data.frame(date = seq.Date(from=ymd("2017-01-01"), to=ymd("2017-01-10"), by="days"), v = 1)
# Two grouping dimensions
d1   <- data.frame(d1 = c("A", "B", "C", "D"), v = 1)
d2   <- data.frame(d2 = c(1, 2, 3, 4, 5), v = 1)

# Generate the data.frame
df <- full_join(date, full_join(d1, d2)) %>%
  select(date, d1, d2) 
# and ad to value columns
df$v1 <- runif(200)
df$v2 <- runif(200)

# group by the dimension columns
df <- df %>% 
  group_by(d1, d2)

# create missing dates
df.missing <- df %>%
  filter(v1 <= 0.8)

# So now  2017-01-01 and 2017-01-10, A, 5 are missing now
df.missing %>%
  filter(d1 == "A" & d2 == 5)

# A tibble: 8 x 5
# Groups:   d1, d2 [1]
        date     d1    d2         v1        v2
      <date> <fctr> <dbl>      <dbl>     <dbl>
1 2017-01-02      A     5 0.21879954 0.1335497
2 2017-01-03      A     5 0.32977018 0.9802127
3 2017-01-04      A     5 0.23902573 0.1206089
4 2017-01-05      A     5 0.19617465 0.7378315
5 2017-01-06      A     5 0.13373890 0.9493668
6 2017-01-07      A     5 0.48613541 0.3392834
7 2017-01-08      A     5 0.35698708 0.3696965
8 2017-01-09      A     5 0.08498474 0.8354756

因此要添加缺少的日期，我会生成一个包含所有日期的data.frame：

So to add the missing dates I generate a data.frame with all dates:

start <- min(df.missing$date)
end   <- max(df.missing$date)

all.dates <- data.frame(date=seq.Date(start, end, by="day"))

不，我想做类似的事情（记住：df.missing是group_by（d1，d2））

No I want to do something like (remember: df.missing is group_by(d1, d2))

df.missing %>%
  do(my_join())

所以我们定义my_join（）：

So let's define my_join():

my_join <- function(data) {
  # get value of both dimensions
  d1.set <- data$d1[[1]]
  d2.set <- data$d2[[1]]

  tmp <- full_join(data, all.dates) %>%
    # First we need to ungroup.  Otherwise we can't change d1 and d2 because they are grouping variables
    ungroup() %>%
    mutate(
      d1 = d1.set,
      d2 = d2.set 
    ) %>%
    group_by(d1, d2)

  return(tmp)
}

现在我们可以为每种组合调用my_join（）并查看 A / 5

Now we can call my_join() for each combination and have a look at "A/5"

df.missing %>%
  do(my_join(.)) %>%
  filter(d1 == "A" & d2 == 5)

# A tibble: 10 x 5
# Groups:   d1, d2 [1]
         date     d1    d2         v1        v2
       <date> <fctr> <dbl>      <dbl>     <dbl>
 1 2017-01-02      A     5 0.21879954 0.1335497
 2 2017-01-03      A     5 0.32977018 0.9802127
 3 2017-01-04      A     5 0.23902573 0.1206089
 4 2017-01-05      A     5 0.19617465 0.7378315
 5 2017-01-06      A     5 0.13373890 0.9493668
 6 2017-01-07      A     5 0.48613541 0.3392834
 7 2017-01-08      A     5 0.35698708 0.3696965
 8 2017-01-09      A     5 0.08498474 0.8354756
 9 2017-01-01      A     5         NA        NA
10 2017-01-10      A     5         NA        NA

太好了！这就是我们想要的。
但是我们需要在my_join中定义d1和d2，这感觉有点笨拙。

Great! That's what we were looking for.But we need to define d1 and d2 in my_join and it feels a little bit clumsy.

那么，该解决方案是否有种种方法？

So, is there any tidyverse-way of this solution?

PS：我已将代码放入要点：

P.S.: I've put the code into a gist: https://gist.github.com/JerryWho/1bf919ef73792569eb38f6462c6d7a8e

推荐答案

有一些解决此类问题的好工具。看看 。

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

want <- df.missing %>% 
  ungroup() %>%
  complete(nesting(d1, d2), date = seq(min(date), max(date), by = "day"))

want %>% filter(d1 == "A" & d2 == 5) 

#> # A tibble: 10 x 5
#>        d1    d2       date         v1        v2
#>    <fctr> <dbl>     <date>      <dbl>     <dbl>
#>  1      A     5 2017-01-01         NA        NA
#>  2      A     5 2017-01-02 0.21879954 0.1335497
#>  3      A     5 2017-01-03 0.32977018 0.9802127
#>  4      A     5 2017-01-04 0.23902573 0.1206089
#>  5      A     5 2017-01-05 0.19617465 0.7378315
#>  6      A     5 2017-01-06 0.13373890 0.9493668
#>  7      A     5 2017-01-07 0.48613541 0.3392834
#>  8      A     5 2017-01-08 0.35698708 0.3696965
#>  9      A     5 2017-01-09 0.08498474 0.8354756
#> 10      A     5 2017-01-10         NA        NA

这篇关于在分组的时间序列中填充缺失的日期-tidyverse方式？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！