如果落在由 R 中另一个数据集中的两个变量定义的范围内，则从一个数据集中获取变量值

Question

我有一个关于 R 中日期操作的问题。我已经环顾了好几天，但在网上找不到任何帮助。我有一个数据集，其中有 id 和两个日期，另一个数据集具有相同的 id 变量、日期和价格。例如：

x = data.frame(id = c("A","B","C","C"), 
               date1 = c("29/05/2013", "23/08/2011", "25/09/2011",  "18/11/2011"),    
               date2 = c("10/07/2013", "04/10/2011", "10/11/2011",	"15/12/2011") )
> x
  id      date1      date2
1  A 29/05/2013 10/07/2013
2  B 23/08/2011 04/10/2011
3  C 25/09/2011 10/11/2011
4  C 18/11/2011 15/12/2011

y = data.frame(id = c("A","A","A","B","B","B","B","B","B","C","C","C"),
              date = c("21/02/2013",  "19/06/2013",	"31/07/2013",	"07/10/2011",	"16/01/2012",	"10/07/2012","20/09/2012",	"29/11/2012",		"15/08/2014",	"27/09/2011",	"27/01/2012",	"09/03/2012"),
              price = c(126,109,111,14,13.8,14.1,14,	14.4,143,102,114,116))
> y
   id       date price
1   A 21/02/2013 126.0
2   A 19/06/2013 109.0
3   A 31/07/2013 111.0
4   B 07/10/2011  14.0
5   B 16/01/2012  13.8
6   B 10/07/2012  14.1
7   B 20/09/2012  14.0
8   B 29/11/2012  14.4
9   B 15/08/2014 143.0
10  C 27/09/2011 102.0
11  C 27/01/2012 114.0
12  C 09/03/2012 116.0

我想要做的是在数据集 x 中查找两个日期，如果数据集 y 中有一个日期由数据集 x 中的两个日期为相同的 id 定义，则为其选择价格值身份证和日期。如果没有它作为失踪。所以基本上我想得到一个像这样的最终数据集：

final = data.frame(id = c("A","B","C","C"), 
                   date1 = c("29/05/2013", "23/08/2011", "25/09/2011",  "18/11/2011"),    
                   date2 = c("10/07/2013", "04/10/2011", "10/11/2011",  "15/12/2011"),
                   date = c("19/06/2013",  "NA",	"27/09/2011",	"NA"),
                   price = c(109,"NA",102,"NA")  )  

> final
  id      date1      date2       date price
1  A 29/05/2013 10/07/2013 19/06/2013   109
2  B 23/08/2011 04/10/2011 20/09/2012    14
3  C 25/09/2011 10/11/2011 27/09/2011   102
4  C 18/11/2011 15/12/2011         NA    NA

任何帮助将不胜感激。

Answer 1

我将分两步进行。首先，通过 id 连接每个 df（有关连接的更多详细信息，请参见 this link），如下所示：

df <- merge(x, y, by = "id")

现在您应该拥有一个完整的数据集，其中包含比您要求的更多的条目。要根据您的标准减少它，请尝试：

df <- filter(df, date > date1, date < date2)

我相信这应该可行。

编辑：如果你真的想要那里的 NA 值而不是仅仅删除该数据，它会使它变得有点毛茸茸。在这种情况下我会做什么，而不是过滤步骤，试试这个：

df$price[date < date1] <- NA
df$price[date > date2] <- NA
df$date[date < date1] <- NA
df$date[date > date2] <- NA

Answer 2

这里的解决方案基于 data.table 包的优秀 foverlaps。

library(data.table)
## coerce characters to dates ( numeric) 
setDT(x)[,c("date1","date2"):=list(as.Date(date1,"%d/%m/%Y"),
                                   as.Date(date2,"%d/%m/%Y"))]
## and a dummy date since foverlaps looks for a start,end columns 
setDT(y)[,c("date1"):=as.Date(date,"%d/%m/%Y")][,date:=date1]
## y must be keyed
setkey(y,id,date,date1)
foverlaps(x,y,by.x=c("id","date1","date2"))[,
            list(id,i.date1,date2,date,price)]

  id    i.date1      date2       date price
1:  A 2013-05-29 2013-07-10 2013-06-19   109
2:  B 2011-08-23 2011-10-04       <NA>    NA
3:  C 2011-09-25 2011-11-10 2011-09-27   102
4:  C 2011-11-18 2011-12-15       <NA>    NA

PS：结果并不完全相同，因为您的预期输出有错误。

Answer 3

或者用lubridate和base R：

m <- merge(x, y, by='id')
d_range <- m$date1 %--% m$date2
m2 <- m[m$date %within% d_range, ]
res <- merge(x, m2, by=c('id', 'date1', 'date2'), all.x=T)

正如@Isaac 建议的那样，合并有助于加快流程。 lubridate 包中的运算符%--% 创建一个间隔。运算符%within% 测试LHS 对象是否在RHS 范围内。

  id      date1      date2       date price
1  A 2013-05-29 2013-07-10 2013-06-19   109
2  B 2011-08-23 2011-10-04       <NA>    NA
3  C 2011-09-25 2011-11-10 2011-09-27   102
4  C 2011-11-18 2011-12-15       <NA>    NA

数据

x = data.frame(id = c("A","B","C","C"), 
               date1 = c("29/05/2013", "23/08/2011", "25/09/2011",  "18/11/2011"),    
               date2 = c("10/07/2013", "04/10/2011", "10/11/2011",  "15/12/2011"))

y = data.frame(id = c("A","A","A","B","B","B","B","B","B","C","C","C"),
              date = c("21/02/2013",  "19/06/2013",  "31/07/2013",  "07/10/2011",   "16/01/2012",   "10/07/2012","20/09/2012",  "29/11/2012",       "15/08/2014",   "27/09/2011",   "27/01/2012",   "09/03/2012"),
              price = c(126,109,111,14,13.8,14.1,14,    14.4,143,102,114,116))

x[c('date1', 'date2')] <- lapply(x[c('date1', 'date2')], dmy)
y['date'] <- dmy(y[,'date'])