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在R中的分组日期内填充_R_Date_Time Series_Lubridate_Padr - Fatal编程技术网

在R中的分组日期内填充

在R中的分组日期内填充,r,date,time-series,lubridate,padr,R,Date,Time Series,Lubridate,Padr,这个错误对我来说没有多大意义。我正在寻找如下所示的输出: df %>% group_by(`Action Item ID`) %>% pad() #> Error: Not all grouping variables are column names of x. #>#一个tible:?x4 #>`Action Item ID``当前状态`Type日期 #> #>ABC截止日期:2019-01-01

这个错误对我来说没有多大意义。我正在寻找如下所示的输出:

df %>% group_by(`Action Item ID`) %>% pad()
#> Error: Not all grouping variables are column names of x.
#>#一个tible:?x4
#>`Action Item ID``当前状态`Type日期
#>                                      
#>ABC截止日期:2019-01-01
#>ABC NA 2019-01-02
#>ABC NA 2019-01-03
#>  ...              ...              ...          ...
#>ABC截止日期2019-01-15
#>DEF关闭日期创建于2019-01-01
#>DEF NA 2019-01-02
#>  ...              ...              ...          ...
#>DEF NA 2019-05-30
#>DEF关闭日期2019-05-31
#>GHI关闭日期:2019-06-01
#>  ...              ...              ...          ...

有人知道出了什么问题吗?

根据
?pad
,有一个
参数

group—指定分组变量的可选字符向量。填充将在不同的组中进行。未指定interval时,将确定对datetime变量整体应用get_interval,忽略组(请参见最后一个示例)

因此,最好利用该参数

#> # A tibble: ? x 4
#>  `Action Item ID` `Current Status` Type         Date      
#>  <chr>            <chr>            <chr>        <date>    
#>  ABC              Closed           Date Created 2019-01-01
#>  ABC              NA               NA           2019-01-02
#>  ABC              NA               NA           2019-01-03
#>  ...              ...              ...          ...
#>  ABC              Closed           Date Closed  2019-01-15
#>  DEF              Closed           Date Created 2019-01-01
#>  DEF              NA               NA           2019-01-02
#>  ...              ...              ...          ...
#>  DEF              NA               NA           2019-05-30
#>  DEF              Closed           Date Closed  2019-05-31
#>  GHI              Closed           Date Created 2019-06-01
#>  ...              ...              ...          ...
库(dplyr)
图书馆(padr)
df%>%
pad(组=“操作项ID”)
#A tibble:233x4
#`Action Item ID``当前状态`Type日期
#                                       
#1 ABC截止日期创建于2019-01-01
#2 ABC 2019-01-02
#3 ABC 2019-01-03
#4 ABC 2019-01-04
#5 ABC 2019-01-05
#6 ABC 2019-01-06
#7 ABC 2019-01-07
#8 ABC 2019-01-08
#9 ABC 2019-01-09
#10 ABC 2019-01-10
#…还有223行

我不熟悉
padr
,但如果您将列名更改为语法上有效的列名,即“action\u item\u id”,则它会起作用。那可能是引擎盖下的小问题
df%>%gatitor::clean_names()%%>%group_by(action_item_id)%%>%pad()有关于NAs的警告,但没有errors@camille这就解开了谜团。很好的侦探工作。你发布的错误消息告诉我,某个地方的列名丢失了。包含错误消息使帮助更容易的完美示例!
#> # A tibble: ? x 4
#>  `Action Item ID` `Current Status` Type         Date      
#>  <chr>            <chr>            <chr>        <date>    
#>  ABC              Closed           Date Created 2019-01-01
#>  ABC              NA               NA           2019-01-02
#>  ABC              NA               NA           2019-01-03
#>  ...              ...              ...          ...
#>  ABC              Closed           Date Closed  2019-01-15
#>  DEF              Closed           Date Created 2019-01-01
#>  DEF              NA               NA           2019-01-02
#>  ...              ...              ...          ...
#>  DEF              NA               NA           2019-05-30
#>  DEF              Closed           Date Closed  2019-05-31
#>  GHI              Closed           Date Created 2019-06-01
#>  ...              ...              ...          ...
library(dplyr)
library(padr)
df %>% 
   pad(group = "Action Item ID")
# A tibble: 233 x 4
#  `Action Item ID` `Current Status` Type         Date      
#   <chr>            <chr>            <chr>        <date>    
# 1 ABC              Closed           Date Created 2019-01-01
# 2 ABC              <NA>             <NA>         2019-01-02
# 3 ABC              <NA>             <NA>         2019-01-03
# 4 ABC              <NA>             <NA>         2019-01-04
# 5 ABC              <NA>             <NA>         2019-01-05
# 6 ABC              <NA>             <NA>         2019-01-06
# 7 ABC              <NA>             <NA>         2019-01-07
# 8 ABC              <NA>             <NA>         2019-01-08
# 9 ABC              <NA>             <NA>         2019-01-09
#10 ABC              <NA>             <NA>         2019-01-10
# … with 223 more rows