将strptime函数应用于data.table的每个成员

Question

我无法将函数应用于data.table的每个成员.这是一个简化的例子:

dt <- data.table( a= c("30JAN14:23:16:00","23MAY12:02:00:00"), 
                  b=c("03AUG09:00:00:00","13JUN12:02:00:00"), 
                  c=c("31JAN14:15:19:00","23MAY12:00:00:00"))

strptime(dt[1,1,with=FALSE], "%d%B%y:%H:%M:%S")

回报 "2014-01-30 23:16:00 PST"

但是,当我尝试将其应用于data.table时,我没有得到我正在寻找的内容并收到指责性消息.

cols <- c("a","b","c")
dt[, (cols):=sapply(.SD, function(x) strptime(x, "%d%B%y:%H:%M:%S")),.SDcols=cols]

Answer 1

strptime 返回类POSIXlt,它实际上是一个列表,解释了为什么在data.table或data.frame对象中使用它会产生问题:

> dt[, (cols):=lapply(.SD, function(x) as.POSIXct(strptime(x, "%d%B%y:%H:%M:%S"))),.SDcols=cols]
> dt
                     a                   b                   c
1: 2014-01-30 23:16:00 2009-08-03 00:00:00 2014-01-31 15:19:00
2: 2012-05-23 02:00:00 2012-06-13 02:00:00 2012-05-23 00:00:00

Answer 2

您也可以使用as.IDate而as.ITime不是strptime.而且,lapply更好:

dt[, (cols):=lapply(.SD, function(x) paste(as.IDate(x, "%d%B%y:%H:%M:%S"),
                                           as.ITime(x, "%d%B%y:%H:%M:%S"),
                                           sep=" ")),
   .SDcols=cols]

这给了:

> dt
                     a                   b                   c
1: 2014-01-30 23:16:00 2009-08-03 00:00:00 2014-01-31 15:19:00
2: 2012-05-23 02:00:00 2012-06-13 02:00:00 2012-05-23 00:00:00