R：在 R 中创建一个新列以根据两个日期确定学期

Question

我有一些数据。 ID 和日期，我正在尝试为学期创建一个新字段。

df:

id  date
1   20160822
2   20170109
3   20170828
4   20170925
5   20180108
6   20180402
7   20160711
8   20150831
9   20160111
10  20160502
11  20160829
12  20170109
13  20170501

我还有一个semester 表：

start       end         season_year
20120801    20121222    Fall-2012
20121223    20130123    Winter-2013
20130124    20130523    Spring-2013
20130524    20130805    Summer-2013
20130806    20131228    Fall-2013
20131229    20140122    Winter-2014
20140123    20140522    Spring-2014
20140523    20140804    Summer-2014
20140805    20141227    Fall-2014
20141228    20150128    Winter-2015
20150129    20150528    Spring-2015
20150529    20150803    Summer-2015
20150804    20151226    Fall-2015
20151227    20160127    Winter-2016
20160128    20160526    Spring-2016
20160527    20160801    Summer-2016
20160802    20161224    Fall-2016
20161225    20170125    Winter-2017
20170126    20170525    Spring-2017
20170526    20170807    Summer-2017
20170808    20171230    Fall-2017
20171231    20180124    Winter-2018
20180125    20180524    Spring-2018
20180525    20180806    Summer-2018
20180807    20181222    Fall-2018
20181223    20190123    Winter-2019
20190124    20190523    Spring-2019
20190524    20180804    Summer-2019

如果df$date 介于semester$start 和semester$end 之间，我想在df 中创建一个新字段，然后将各自的值semester$season_year 放入df 中

我试图看看 lubridate 包是否有帮助，但这似乎更适合计算

我看到了this question，它似乎最接近我想要的，但是，更复杂的是，并不是我们所有的学期都是六个月

Answer 1

这行得通吗？

library(lubridate)

semester$start <- ymd(semester$start)
semester$end <- ymd(semester$end)
df$date <- ymd(df$date)

LU <-  Map(`:`, semester$start, semester$end)
LU <- data.frame(value = unlist(LU),
                 index = rep(seq_along(LU), lapply(LU, length)))


df$semester <- semester$season_year[LU$index[match(df$date, LU$value)]]

Answer 2

使用non-equi更新连接使用data.table和lubridate包的解决方案可以是：

library(data.table)

setDT(df)
setDT(semester)


df[,date:=as.IDate(as.character(date), format = "%Y%m%d")]
semester[,':='(start = as.IDate(as.character(start), format = "%Y%m%d"), 
                         end=as.IDate(as.character(end), format = "%Y%m%d"))]


df[semester, on=.(date >= start, date <= end), season_year := i.season_year]

df
#    id       date season_year
# 1:  1 2016-08-22   Fall-2016
# 2:  2 2017-01-09 Winter-2017
# 3:  3 2017-08-28   Fall-2017
# 4:  4 2017-09-25   Fall-2017
# 5:  5 2018-01-08 Winter-2018
# 6:  6 2018-04-02 Spring-2018
# 7:  7 2016-07-11 Summer-2016
# 8:  8 2015-08-31   Fall-2015
# 9:  9 2016-01-11 Winter-2016
# 10: 10 2016-05-02 Spring-2016
# 11: 11 2016-08-29   Fall-2016
# 12: 12 2017-01-09 Winter-2017
# 13: 13 2017-05-01 Spring-2017

数据：

df <- read.table(text="
id  date
1   20160822
2   20170109
3   20170828
4   20170925
5   20180108
6   20180402
7   20160711
8   20150831
9   20160111
10  20160502
11  20160829
12  20170109
13  20170501",
header = TRUE, stringsAsFactors = FALSE)


semester <- read.table(text="
start       end         season_year
20120801    20121222    Fall-2012
20121223    20130123    Winter-2013
20130124    20130523    Spring-2013
20130524    20130805    Summer-2013
20130806    20131228    Fall-2013
20131229    20140122    Winter-2014
20140123    20140522    Spring-2014
20140523    20140804    Summer-2014
20140805    20141227    Fall-2014
20141228    20150128    Winter-2015
20150129    20150528    Spring-2015
20150529    20150803    Summer-2015
20150804    20151226    Fall-2015
20151227    20160127    Winter-2016
20160128    20160526    Spring-2016
20160527    20160801    Summer-2016
20160802    20161224    Fall-2016
20161225    20170125    Winter-2017
20170126    20170525    Spring-2017
20170526    20170807    Summer-2017
20170808    20171230    Fall-2017
20171231    20180124    Winter-2018
20180125    20180524    Spring-2018
20180525    20180806    Summer-2018
20180807    20181222    Fall-2018
20181223    20190123    Winter-2019
20190124    20190523    Spring-2019
20190524    20180804    Summer-2019",
header = TRUE, stringsAsFactors = FALSE)