Question

我有一个data.table对象，我根据日期变量id和start.time按stop.time展开。下面列出的代码可行 - 我的问题是如何使我的代码更容易使用。

如何避免在声明中手动列出data.table对象的现有变量，即＆＃34; start.time＆＃34;和＆＃34; stop.time＆＃34;在test <- DT[, list(start.time, stop.time, time=seq(start.time, stop.time, by="1 month")), by=id]。如果你假设我不仅有两个变量而是几十个变量，那么手动添加变量会很烦人。目的是包括原始data.table对象中存在的所有列。

我尝试添加.SD和setdiff(names(DT), "id")而不是变量名称，但这会引发错误。

DT <- data.table(id=sample(1000), 
                 start.time=sample(seq(as.Date("01-01-2010", format="%d-%m-%Y"), as.Date("31-12-2010", format="%d-%m-%Y"), by="day"), 1000, replace=TRUE),
                 stop.time=sample(seq(as.Date("01-01-2011", format="%d-%m-%Y"), as.Date("31-12-2011", format="%d-%m-%Y"), by="day"), 1000, replace=TRUE))

# > DT
#        id start.time  stop.time
#    1: 105 2010-12-16 2011-07-30
#    2: 837 2010-01-15 2011-01-19
#    3: 155 2010-11-30 2011-10-07
#    4: 125 2010-03-31 2011-04-19
#    5: 192 2010-07-13 2011-03-27
#   ---                          
#  996: 749 2010-07-19 2011-02-18
#  997: 660 2010-06-28 2011-10-25
#  998: 808 2010-03-22 2011-12-04
#  999: 291 2010-09-03 2011-06-08
# 1000: 741 2010-01-25 2011-01-23


test <- DT[, list(start.time, stop.time, time=seq(start.time, stop.time, by="1 month")), by=id]

# > test
#         id start.time  stop.time       time
#     1: 903 2010-11-08 2011-10-22 2010-11-08
#     2: 903 2010-11-08 2011-10-22 2010-12-08
#     3: 903 2010-11-08 2011-10-22 2011-01-08
#     4: 903 2010-11-08 2011-10-22 2011-02-08
#     5: 903 2010-11-08 2011-10-22 2011-03-08
#    ---                                     
# 12346: 229 2010-09-11 2011-10-02 2011-05-11
# 12347: 229 2010-09-11 2011-10-02 2011-06-11
# 12348: 229 2010-09-11 2011-10-02 2011-07-11
# 12349: 229 2010-09-11 2011-10-02 2011-08-11
# 12350: 229 2010-09-11 2011-10-02 2011-09-1

Answer 1

您可以像这样使用.SD：

DT[, c(.SD, list(time=seq(start.time, stop.time, by="1 month"))), by=id]

按组扩展data.table对象时保留现有列

1 个答案: