Question

我有一个这样的数据框：

  data.frame(
        date= structure(c(15461, 15553, 15675, 15826, 15918, 16040, 16191, 
        16405, 16556, 16648, 16770, 16922, 17014, 17136), class = "Date"), 
        n= 1:14
   )

如何按学年总结n？每个学年的学期应从12月到次年8月。例如，我想在每个学年总结n。手动重构不是一种选择，因为有太多的值，有时甚至是丢失的值。

最终，重构应该如下：

  date         a.y.

"2012-05-01" 2011/2012
"2012-08-01" 2011/2012

"2012-12-01" 2012/2013
"2013-05-01" 2012/2013
"2013-08-01" 2012/2013

"2013-12-01" 2013/2014
"2014-05-01" 2013/2014

"2014-12-01" 2014/2015
"2015-05-01" 2014/2015
"2015-08-01" 2014/2015

"2015-12-01" 2015/2016
"2016-05-01" 2015/2016
"2016-08-01" 2015/2016

"2016-12-01" 2016/2017

正如您所注意到的，日期遵循类似的模式，但每个学年可能会有不同的日期。

Answer 1

如果我在看到12月份的录取后立即阅读此内容，我们就会改变学年。如果这是真的，那么以下代码将起作用。

library(data.table)
library(lubridate)
df =   data.frame(
  date= structure(c(15461, 15553, 15675, 15826, 15918, 16040, 16191, 
                    16405, 16556, 16648, 16770, 16922, 17014, 17136), class = "Date"), 
  n= 1:14
)

df$AcademicYear = cumsum(month(df$date) == 12)
setDT(df)
df[ , .(Sum = sum(n)), by = .(AcademicYear)]

  AcademicYear Sum
1:            0   3
2:            1  12
3:            2  13
4:            3  27
5:            4  36
6:            5  14

修改

重构你可以做这样的事情。它会在AcademicYear中查找一个月，然后根据月份，它知道添加或减去一年并将其粘贴在一起。然后，该列只需要重命名并如上所述求和。

df[ , "AcademicYear2" := ifelse(any(month(date) == 5), paste(year(date[month(date) == 5]) - 1,year(date[month(date) == 5]), sep = "/"), ifelse(any(month(date) == 8), paste(year(date[month(date) == 8]) - 1,year(date[month(date) == 8]), sep = "/"), paste(year(date[month(date) == 12]),year(date[month(date) == 12]) + 1, sep = "/"))), by = .(AcademicYear)] > df date n AcademicYear AcademicYear2 1: 2012-05-01 1 0 2011/2012 2: 2012-08-01 2 0 2011/2012 3: 2012-12-01 3 1 2012/2013 4: 2013-05-01 4 1 2012/2013 5: 2013-08-01 5 1 2012/2013 6: 2013-12-01 6 2 2013/2014 7: 2014-05-01 7 2 2013/2014 8: 2014-12-01 8 3 2014/2015 9: 2015-05-01 9 3 2014/2015 10: 2015-08-01 10 3 2014/2015 11: 2015-12-01 11 4 2015/2016 12: 2016-05-01 12 4 2015/2016 13: 2016-08-01 13 4 2015/2016 14: 2016-12-01 14 5 2016/2017

修改2

决定将所有代码放在一起。这可以为您提供您正在寻找的最终结果。

library(data.table) library(lubridate) df = data.frame( date= structure(c(15461, 15553, 15675, 15826, 15918, 16040, 16191, 16405, 16556, 16648, 16770, 16922, 17014, 17136), class = "Date"), n= 1:14 ) setDT(df) df$AcademicYear = cumsum(month(df$date) == 12) df[ , "AcademicYear2" := ifelse(any(month(date) == 5), paste(year(date[month(date) == 5]) - 1,year(date[month(date) == 5]), sep = "/"), ifelse(any(month(date) == 8), paste(year(date[month(date) == 8]) - 1,year(date[month(date) == 8]), sep = "/"), paste(year(date[month(date) == 12]),year(date[month(date) == 12]) + 1, sep = "/"))), by = .(AcademicYear)] df = df[ , .(Sum = sum(n)), by = .(AcademicYear = AcademicYear2)] > df AcademicYear Sum 1: 2011/2012 3 2: 2012/2013 12 3: 2013/2014 13 4: 2014/2015 27 5: 2015/2016 36 6: 2016/2017 14

Answer 2

不确定您想要哪些日期与哪些日期，但您可以使用dplyr和mutate与一系列if else语句。它很慢但是有效。

df <- data.frame(
  date= structure(c(15461, 15553, 15675, 15826, 15918, 16040, 16191, 
                    16405, 16556, 16648, 16770, 16922, 17014, 17136), class = "Date"), 
  n= 1:14
)

df <- mutate(df, term=ifelse(date >= as.Date("2012-05-01")  & date <= as.Date("2012-08-01"),  "1",
        ifelse(date >= as.Date("2012-12-01")  & date <= as.Date("2013-05-01"),  "2",
          ifelse(date >= as.Date("2013-12-01")  & date <= as.Date("2014-12-01"),  "3",
         ifelse(date >= as.Date("2015-08-01")  & date <= as.Date("2016-08-01"),  "4",
           "other")))))

如何按学年分组？

2 个答案: