R：按日期排序（按年份，按月）

2 回答

• 2

  在tidyverse（w / lubridate添加）：

  library(tidyverse)
  library(lubridate)
  
  dfYrMon <- 
      df1 %>% 
      mutate(date = parse_date_time(month, "my"),
             year = year(date),
             month = month(date)
             ) %>% 
      arrange(year, month) %>% 
      select(date, year, month, result)
  

  有了数据：

  df1 <- tibble(month = c("01/2018", "02/2018", "03/2018", "04/2018", "05/2018", "11/2017", "12/2017"), 
                result = c(96.13636, 96.4, 94, 97.92857, 95.75, 98.66667, 97.78947))
  

  会得到这个'数据帧'：

  #A tibble：7 x 4
  日期年度结果
  <dttm> <dbl> <dbl> <dbl>
  1 2017-11-01 2017 11 98.66667
  2 2017-12-01 2017 12 97.78947
  3 2018-01-01 2018 1 96.13636
  4 2018-02-01 2018 2 96.40000
  5 2018-03-01 2018 3 94.00000
  6 2018-04-01 2018 4 97.92857
  7 2018-05-01 2018 5 95.75000

  使您的数据值成为原子（年份在其自己的列中，月份在其自己的列中）通常会提高操作的便利性 .

  或者如果你想使用基本R日期操作而不是lubridate：

  library(tidyverse)
  
  dfYrMon_base <- 
      df1 %>% 
      mutate(date = as.Date(paste("01/", month, sep = ""), "%d/%m/%Y"),
             year = format(as.Date(date, format="%d/%m/%Y"),"%Y"),
             month = format(as.Date(date, format="%d/%m/%Y"),"%m")
            ) %>%
      arrange(year, month) %>%
      select(date, year, month, result)
  
  dfYrMon_base
  

  请注意创建的数据类型 .

  #A tibble：7 x 4
  日期年度结果
  <date> <chr> <chr> <dbl>
  1 2017-11-01 2017 11 98.66667
  2 2017-12-01 2017 12 97.78947
  3 2018-01-01 2018 01 96.13636
  4 2018-02-01 2018 02 96.40000
  5 2018-03-01 2018 03 94.00000
  6 2018-04-01 2018 04 97.92857
  7 2018-05-01 2018 05 95.75000

  回复于 2024-05-14T20:39:50+08:00
• 3

  我们可以将它转换为 yearmon 类，然后执行 order

  library(zoo)
  out <- df1[order(as.yearmon(df1$month, "%m/%Y"), df1$Result),]
  row.names(out) <- NULL
  out
  #    month   Result
  #1 11/2017 98.66667
  #2 12/2017 97.78947
  #3 01/2018 96.13636
  #4 02/2018 96.40000
  #5 03/2018 94.00000
  #6 04/2018 97.92857
  #7 05/2018 95.75000
  

  数据

  df1 <- structure(list(month = c("01/2018", "02/2018", "03/2018", "04/2018", 
  "05/2018", "11/2017", "12/2017"), Result = c(96.13636, 96.4, 
  94, 97.92857, 95.75, 98.66667, 97.78947)), .Names = c("month", 
  "Result"), class = "data.frame", 
  row.names = c("1", "2", "3", 
  "4", "5", "6", "7"))
  

  回复于 2024-05-14T20:39:50+08:00