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r - dplyr / left_join 中的嵌套管道链

问题描述

在尝试获取分组滞后变量的过程中（仅使用 是不可能的lag），建议的解决方案是将数据拉出，滞后不同的行，然后重新加入它。

我更喜欢在不创建中间对象的情况下执行此操作，并且希望在链中执行此操作。然而它似乎并没有像我预期的那样工作，问题似乎是.在 left_join 中使用和嵌套链之间的一些交互。 

require(tidyverse)
#> Loading required package: tidyverse
df <- data.frame(Team = c("A", "A", "A", "A", "B", "B", "B", "C", "C", "D", "D"),
                 Date = c("2016-05-10","2016-05-10", "2016-05-10", "2016-05-10",
                          "2016-05-12", "2016-05-12", "2016-05-12",
                          "2016-05-15","2016-05-15",
                          "2016-05-30", "2016-05-30"), 
                 Points = c(1,4,3,2,1,5,6,1,2,3,9)
)


#This works:
df %>% left_join(x = ., y = df %>% 
                   distinct(Team, Date) %>% 
                   mutate(Date_Lagged = lag(Date)))
#> Joining, by = c("Team", "Date")
#>    Team       Date Points Date_Lagged
#> 1     A 2016-05-10      1        <NA>
#> 2     A 2016-05-10      4        <NA>
#> 3     A 2016-05-10      3        <NA>
#> 4     A 2016-05-10      2        <NA>
#> 5     B 2016-05-12      1  2016-05-10
#> 6     B 2016-05-12      5  2016-05-10
#> 7     B 2016-05-12      6  2016-05-10
#> 8     C 2016-05-15      1  2016-05-12
#> 9     C 2016-05-15      2  2016-05-12
#> 10    D 2016-05-30      3  2016-05-15
#> 11    D 2016-05-30      9  2016-05-15

#And this works:
df %>% left_join(x = ., y = .)
#> Joining, by = c("Team", "Date", "Points")
#>    Team       Date Points
#> 1     A 2016-05-10      1
#> 2     A 2016-05-10      4
#> 3     A 2016-05-10      3
#> 4     A 2016-05-10      2
#> 5     B 2016-05-12      1
#> 6     B 2016-05-12      5
#> 7     B 2016-05-12      6
#> 8     C 2016-05-15      1
#> 9     C 2016-05-15      2
#> 10    D 2016-05-30      3
#> 11    D 2016-05-30      9

#This doesn't work despite the fact that `.` is df.  
df %>% left_join(x = ., y = . %>% 
                   distinct(Team, Date) %>% 
                   mutate(Date_Lagged = lag(Date)))
#> Error in UseMethod("tbl_vars"): no applicable method for 'tbl_vars' applied to an object of class "c('fseq', 'function')"



#Desired output
distinct(df, Team, Date) %>%
  mutate(Date_Lagged = lag(Date)) %>%
  right_join(., df) %>%
  select(Team, Date, Points, Date_Lagged)
#> Joining, by = c("Team", "Date")
#>    Team       Date Points Date_Lagged
#> 1     A 2016-05-10      1        <NA>
#> 2     A 2016-05-10      4        <NA>
#> 3     A 2016-05-10      3        <NA>
#> 4     A 2016-05-10      2        <NA>
#> 5     B 2016-05-12      1  2016-05-10
#> 6     B 2016-05-12      5  2016-05-10
#> 7     B 2016-05-12      6  2016-05-10
#> 8     C 2016-05-15      1  2016-05-12
#> 9     C 2016-05-15      2  2016-05-12
#> 10    D 2016-05-30      3  2016-05-15
#> 11    D 2016-05-30      9  2016-05-15

由reprex 包（v0.2.0）于 2018 年 6 月 12 日创建。

标签： rdplyr

解决方案

为了使您的代码正常工作，您需要在参数周围使用花括号y，如下所示 

  df %>% left_join(x = ., y = {.} %>% 
                   distinct(Team, Date) %>% 
                   mutate(Date_Lagged = lag(Date)))

Joining, by = c("Team", "Date")
   Team       Date Points Date_Lagged
1     A 2016-05-10      1        <NA>
2     A 2016-05-10      4        <NA>
3     A 2016-05-10      3        <NA>
4     A 2016-05-10      2        <NA>
5     B 2016-05-12      1  2016-05-10
6     B 2016-05-12      5  2016-05-10
7     B 2016-05-12      6  2016-05-10
8     C 2016-05-15      1  2016-05-12
9     C 2016-05-15      2  2016-05-12
10    D 2016-05-30      3  2016-05-15
11    D 2016-05-30      9  2016-05-15

你可以做

df %>% left_join(df%>% 
                   distinct(Team, Date) %>% 
                   mutate(Date_Lagged = lag(Date)))

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