如何使用 dplyr R 选择名称为日期的列答案

【问题标题】：How to select columns with names as dates using dplyr R如何使用 dplyr R 选择名称为日期的列
【发布时间】：2021-07-20 09:00:40
【问题描述】：

我有数据框，其中一些标题有名称、字符和日期。我想选择所有没有日期作为标题的列以及所有日期小于当前日期或系统日期（sys.date（））的列。如何在 dplyr 中使用 select 语句进行精简。

下面是数据框

> dput(job_times[1:5,])
structure(list(Skill = c("KAC", "KAC", "KAC", "KAC", "KAC"), 
    Patch = c("A1", "A2", "A3", "A4", "A5"), `Work Code` = c("W01", 
    "W01", "W01", "W01", "W01"), Product = c("KAC Repair", "KAC Repair", 
    "KAC Repair", "KAC Repair", "KAC Repair"), `Visit Time` = c(45.68, 
    42.55, 46.45, 51.86, 43.49), Travel = c(32.5, 21.66, 26.33, 
    28.63, 27.03), Success = c(0.69, 0.66, 0.67, 0.65, 0.67), 
    `Completion Time` = c(1.9, 1.61, 1.8, 2.05, 1.74), `28-12-2020` = c(1.9, 
    1.61, 1.8, 2.05, 1.74), `04-01-2021` = c(1.9, 1.61, 1.8, 
    2.05, 1.74), `11-01-2021` = c(1.9, 1.61, 1.8, 2.05, 1.74), 
    `18-01-2021` = c(1.9, 1.61, 1.8, 2.05, 1.74), `25-01-2021` = c(1.9, 
    1.61, 1.8, 2.05, 1.74), `01-02-2021` = c(1.9, 1.61, 1.8, 
    2.05, 1.74), `08-02-2021` = c(1.9, 1.61, 1.8, 2.05, 1.74), 
    `15-02-2021` = c(1.9, 1.61, 1.8, 2.05, 1.74), `22-02-2021` = c(1.9, 
    1.61, 1.8, 2.05, 1.74), `01-03-2021` = c(1.9, 1.61, 1.8, 
    2.05, 1.74), `08-03-2021` = c(1.9, 1.61, 1.8, 2.05, 1.74), 
    `15-03-2021` = c(1.9, 1.61, 1.8, 2.05, 1.74), `22-03-2021` = c(1.9, 
    1.61, 1.8, 2.05, 1.74), `29-03-2021` = c(1.9, 1.61, 1.8, 
    2.05, 1.74), `05-04-2021` = c(1.9, 1.61, 1.8, 2.05, 1.74), 
    `12-04-2021` = c(1.9, 1.61, 1.8, 2.05, 1.74), `19-04-2021` = c(1.9, 
    1.61, 1.8, 2.05, 1.74), `26-04-2021` = c(1.9, 1.61, 1.8, 
    2.05, 1.74), `03-05-2021` = c(1.9, 1.61, 1.8, 2.05, 1.74), 
    `10-05-2021` = c(1.9, 1.61, 1.8, 2.05, 1.74), `17-05-2021` = c(1.9, 
    1.61, 1.8, 2.05, 1.74), `24-05-2021` = c(1.9, 1.61, 1.8, 
    2.05, 1.74), `31-05-2021` = c(1.9, 1.61, 1.8, 2.05, 1.74), 
    `07-06-2021` = c(1.9, 1.61, 1.8, 2.05, 1.74), `14-06-2021` = c(1.9, 
    1.61, 1.8, 2.05, 1.74), `21-06-2021` = c(1.9, 1.61, 1.8, 
    2.05, 1.74), `28-06-2021` = c(1.9, 1.61, 1.8, 2.05, 1.74), 
    `05-07-2021` = c(1.9, 1.61, 1.8, 2.05, 1.74), `12-07-2021` = c(1.9, 
    1.61, 1.8, 2.05, 1.74), `19-07-2021` = c(1.9, 1.61, 1.8, 
    2.05, 1.74), `26-07-2021` = c(1.9, 1.61, 1.8, 2.05, 1.74), 
    `02-08-2021` = c(1.9, 1.61, 1.8, 2.05, 1.74), `09-08-2021` = c(1.9, 
    1.61, 1.8, 2.05, 1.74), `16-08-2021` = c(1.9, 1.61, 1.8, 
    2.05, 1.74), `23-08-2021` = c(1.9, 1.61, 1.8, 2.05, 1.74), 
    `30-08-2021` = c(1.9, 1.61, 1.8, 2.05, 1.74), `06-09-2021` = c(1.9, 
    1.61, 1.8, 2.05, 1.74), `13-09-2021` = c(1.9, 1.61, 1.8, 
    2.05, 1.74), `20-09-2021` = c(1.9, 1.61, 1.8, 2.05, 1.74), 
    `27-09-2021` = c(1.9, 1.61, 1.8, 2.05, 1.74), `04-10-2021` = c(1.9, 
    1.61, 1.8, 2.05, 1.74), `11-10-2021` = c(1.9, 1.61, 1.8, 
    2.05, 1.74), `18-10-2021` = c(1.9, 1.61, 1.8, 2.05, 1.74), 
    `25-10-2021` = c(1.9, 1.61, 1.8, 2.05, 1.74), `01-11-2021` = c(1.9, 
    1.61, 1.8, 2.05, 1.74), `08-11-2021` = c(1.9, 1.61, 1.8, 
    2.05, 1.74), `15-11-2021` = c(1.9, 1.61, 1.8, 2.05, 1.74), 
    `22-11-2021` = c(1.9, 1.61, 1.8, 2.05, 1.74), `29-11-2021` = c(1.9, 
    1.61, 1.8, 2.05, 1.74), `06-12-2021` = c(1.9, 1.61, 1.8, 
    2.05, 1.74), `13-12-2021` = c(1.9, 1.61, 1.8, 2.05, 1.74), 
    `20-12-2021` = c(1.9, 1.61, 1.8, 2.05, 1.74), `27-12-2021` = c(1.9, 
    1.61, 1.8, 2.05, 1.74), `03-01-2022` = c(1.9, 1.61, 1.8, 
    2.05, 1.74), `10-01-2022` = c(1.9, 1.61, 1.8, 2.05, 1.74), 
    `17-01-2022` = c(1.9, 1.61, 1.8, 2.05, 1.74), `24-01-2022` = c(1.9, 
    1.61, 1.8, 2.05, 1.74), `31-01-2022` = c(1.9, 1.61, 1.8, 
    2.05, 1.74), `07-02-2022` = c(1.9, 1.61, 1.8, 2.05, 1.74), 
    `14-02-2022` = c(1.9, 1.61, 1.8, 2.05, 1.74), `21-02-2022` = c(1.9, 
    1.61, 1.8, 2.05, 1.74), `28-02-2022` = c(1.9, 1.61, 1.8, 
    2.05, 1.74), `07-03-2022` = c(1.9, 1.61, 1.8, 2.05, 1.74), 
    `14-03-2022` = c(1.9, 1.61, 1.8, 2.05, 1.74), `21-03-2022` = c(1.9, 
    1.61, 1.8, 2.05, 1.74), `28-03-2022` = c(1.9, 1.61, 1.8, 
    2.05, 1.74), `04-04-2022` = c(1.9, 1.61, 1.8, 2.05, 1.74), 
    `11-04-2022` = c(1.9, 1.61, 1.8, 2.05, 1.74), `18-04-2022` = c(1.9, 
    1.61, 1.8, 2.05, 1.74), `25-04-2022` = c(1.9, 1.61, 1.8, 
    2.05, 1.74), `02-05-2022` = c(1.9, 1.61, 1.8, 2.05, 1.74), 
    `09-05-2022` = c(1.9, 1.61, 1.8, 2.05, 1.74), `16-05-2022` = c(1.9, 
    1.61, 1.8, 2.05, 1.74), `23-05-2022` = c(1.9, 1.61, 1.8, 
    2.05, 1.74), `30-05-2022` = c(1.9, 1.61, 1.8, 2.05, 1.74), 
    `06-06-2022` = c(1.9, 1.61, 1.8, 2.05, 1.74), `13-06-2022` = c(1.9, 
    1.61, 1.8, 2.05, 1.74), `20-06-2022` = c(1.9, 1.61, 1.8, 
    2.05, 1.74), `27-06-2022` = c(1.9, 1.61, 1.8, 2.05, 1.74), 
    `04-07-2022` = c(1.9, 1.61, 1.8, 2.05, 1.74), `11-07-2022` = c(1.9, 
    1.61, 1.8, 2.05, 1.74), `18-07-2022` = c(1.9, 1.61, 1.8, 
    2.05, 1.74), `25-07-2022` = c(1.9, 1.61, 1.8, 2.05, 1.74), 
    `01-08-2022` = c(1.9, 1.61, 1.8, 2.05, 1.74), `08-08-2022` = c(1.9, 
    1.61, 1.8, 2.05, 1.74), `15-08-2022` = c(1.9, 1.61, 1.8, 
    2.05, 1.74), `22-08-2022` = c(1.9, 1.61, 1.8, 2.05, 1.74), 
    `29-08-2022` = c(1.9, 1.61, 1.8, 2.05, 1.74), `05-09-2022` = c(1.9, 
    1.61, 1.8, 2.05, 1.74), `12-09-2022` = c(1.9, 1.61, 1.8, 
    2.05, 1.74), `19-09-2022` = c(1.9, 1.61, 1.8, 2.05, 1.74), 
    `26-09-2022` = c(1.9, 1.61, 1.8, 2.05, 1.74), `03-10-2022` = c(1.9, 
    1.61, 1.8, 2.05, 1.74), `10-10-2022` = c(1.9, 1.61, 1.8, 
    2.05, 1.74), `17-10-2022` = c(1.9, 1.61, 1.8, 2.05, 1.74), 
    `24-10-2022` = c(1.9, 1.61, 1.8, 2.05, 1.74), `31-10-2022` = c(1.9, 
    1.61, 1.8, 2.05, 1.74), `07-11-2022` = c(1.9, 1.61, 1.8, 
    2.05, 1.74), `14-11-2022` = c(1.9, 1.61, 1.8, 2.05, 1.74), 
    `21-11-2022` = c(1.9, 1.61, 1.8, 2.05, 1.74), `28-11-2022` = c(1.9, 
    1.61, 1.8, 2.05, 1.74), `05-12-2022` = c(1.9, 1.61, 1.8, 
    2.05, 1.74), `12-12-2022` = c(1.9, 1.61, 1.8, 2.05, 1.74), 
    `19-12-2022` = c(1.9, 1.61, 1.8, 2.05, 1.74), `26-12-2022` = c(1.9, 
    1.61, 1.8, 2.05, 1.74)), row.names = c(NA, -5L), class = c("tbl_df", 
"tbl", "data.frame"))

我想要技能、补丁、工作代码、产品、访问时间、旅行、成功、完成时间列以及日期小于或等于 sys.Date() 的所有列。使用 dplyr 和 select 语句。

【问题讨论】：

标签： r dplyr data-analysis

【解决方案1】：

这就是我要解决的方法-

cols <- grep('\\d{2}-\\d{2}-\\d{2}', names(job_times), value = TRUE)
result <- job_times[, c(setdiff(names(job_times), cols), 
              cols[Sys.Date() > as.Date(cols, '%d-%m-%Y')])]

您可以将其集成到dplyr 管道中作为 -

library(dplyr)

job_times %>%
  select({
    cols <- grep('\\d{2}-\\d{2}-\\d{2}', names(.), value = TRUE)
    c(setdiff(names(.), cols),
      cols[Sys.Date() > as.Date(cols, '%d-%m-%Y')])
  })

【讨论】：

【解决方案2】：

我建议创建一个辅助函数，然后您可以像这样使用 select：

library(tidyverse)
library(lubridate)

is_before_today <- function(x) {
  (dmy(x, quiet = TRUE) < Sys.Date()) %>% coalesce(FALSE)
}

df %>% 
  select(
    matches("^\\D"), all_of(colnames(.) %>% keep(is_before_today))
  )
#> # A tibble: 5 x 38
#>   Skill Patch `Work Code` Product   `Visit Time` Travel Success `Completion Tim~
#>   <chr> <chr> <chr>       <chr>            <dbl>  <dbl>   <dbl>            <dbl>
#> 1 KAC   A1    W01         KAC Repa~         45.7   32.5    0.69             1.9 
#> 2 KAC   A2    W01         KAC Repa~         42.6   21.7    0.66             1.61
#> 3 KAC   A3    W01         KAC Repa~         46.4   26.3    0.67             1.8 
#> 4 KAC   A4    W01         KAC Repa~         51.9   28.6    0.65             2.05
#> 5 KAC   A5    W01         KAC Repa~         43.5   27.0    0.67             1.74
#> # ... with 30 more variables: 28-12-2020 <dbl>, 04-01-2021 <dbl>,
#> #   11-01-2021 <dbl>, 18-01-2021 <dbl>, 25-01-2021 <dbl>, 01-02-2021 <dbl>,
#> #   08-02-2021 <dbl>, 15-02-2021 <dbl>, 22-02-2021 <dbl>, 01-03-2021 <dbl>,
#> #   08-03-2021 <dbl>, 15-03-2021 <dbl>, 22-03-2021 <dbl>, 29-03-2021 <dbl>,
#> #   05-04-2021 <dbl>, 12-04-2021 <dbl>, 19-04-2021 <dbl>, 26-04-2021 <dbl>,
#> #   03-05-2021 <dbl>, 10-05-2021 <dbl>, 17-05-2021 <dbl>, 24-05-2021 <dbl>,
#> #   31-05-2021 <dbl>, 07-06-2021 <dbl>, 14-06-2021 <dbl>, 21-06-2021 <dbl>,
#> #   28-06-2021 <dbl>, 05-07-2021 <dbl>, 12-07-2021 <dbl>, 19-07-2021 <dbl>

^{由reprex package (v1.0.0) 于 2021 年 7 月 20 日创建}

【讨论】：

非常有帮助的评论杰森。需要详细说明吗？
那么你的答案仍然返回 113 列
感谢您告诉我 - 刚刚修复了错误。

【解决方案3】：

Ronak Shah 的回答非常棒。以下是我的做法。

    ## Get the List of All Column Names
    ColumnNames <- names(TestDF)

    ## Retain Only those don't have Dates
    CharacterColumnNames <- ColumnNames[grepl( "[[:alpha:]]" , names( TestDF ) )] 

    ## Get the List of all Date Column Names
    DateColumns <- setdiff(names(TestDF),CharacterColumnNames)

    ## Filter Required Date Column Names
    RequiredDateColumns <- DateColumns[ Sys.Date() > as.Date(DateColumns, '%d-%m-%Y')]

    ## Get the Modified DF
    ModifiedDF <- TestDF[, c(CharacterColumnNames, RequiredDateColumns)]

【讨论】：