我需要将时间和日期与 2012-09-27 07:05:59 形式的time
列的data.frame隔离开来。
然后,我必须使用date
和time
列来提取特定日期/时间的数据。
我该怎么做呢?可能我想要反过来this one。
我尝试使用strptime
函数和lubridate
包,但无法使其正常工作。
data1 <- structure(list(event.date = structure(list(sec = c(59, 29, 59,
0, 29, 59, 29, 29, 59, 59), min = c(5L, 7L, 15L, 17L, 17L, 19L,
21L, 22L, 22L, 23L), hour = c(7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L,
7L, 7L), mday = c(27L, 27L, 27L, 27L, 27L, 27L, 27L, 27L, 27L,
27L), mon = c(8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L), year = c(112L,
112L, 112L, 112L, 112L, 112L, 112L, 112L, 112L, 112L), wday = c(4L,
4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L), yday = c(270L, 270L, 270L,
270L, 270L, 270L, 270L, 270L, 270L, 270L), isdst = c(0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L)), .Names = c("sec", "min", "hour",
"mday", "mon", "year", "wday", "yday", "isdst"), class = c("POSIXlt",
"POSIXt"), tzone = c("CST", "GMT", " ")), VE = c(45L, 55L,
45L, 50L, 55L, 35L, 30L, 45L, 55L, 30L)), .Names = c("event.date",
"VE"), row.names = c(NA, 10L), class = "data.frame")
以下内容为NA
s:
as.Date(as.character(data1$event.date),"%HH:%MM:%SS")
我也无法成功使用lubridate
...
ymd_hms(as.character(data1$event.date))
我对我应该使用哪种日期格式/包感到困惑。
答案 0 :(得分:6)
你走了:
R> data1$pt <- as.POSIXct(data1[,1]) # parse time from ISO string
R> data1$date <- as.Date(data1[,"pt"]) # transform to Date
R> data1$time <- format(data1[,"pt"], "%H:%M:%S") # transform to time string
R> data1
event.date VE pt date time
1 2012-09-27 07:05:59 45 2012-09-27 07:05:59 2012-09-27 07:05:59
2 2012-09-27 07:07:29 55 2012-09-27 07:07:29 2012-09-27 07:07:29
3 2012-09-27 07:15:59 45 2012-09-27 07:15:59 2012-09-27 07:15:59
4 2012-09-27 07:17:00 50 2012-09-27 07:17:00 2012-09-27 07:17:00
5 2012-09-27 07:17:29 55 2012-09-27 07:17:29 2012-09-27 07:17:29
6 2012-09-27 07:19:59 35 2012-09-27 07:19:59 2012-09-27 07:19:59
7 2012-09-27 07:21:29 30 2012-09-27 07:21:29 2012-09-27 07:21:29
8 2012-09-27 07:22:29 45 2012-09-27 07:22:29 2012-09-27 07:22:29
9 2012-09-27 07:22:59 55 2012-09-27 07:22:59 2012-09-27 07:22:59
10 2012-09-27 07:23:59 30 2012-09-27 07:23:59 2012-09-27 07:23:59
R>
以下是您的列类型,显示我使用第一个命令将POSIXlt
更改为POSIXct
:
R> sapply(data1, class)
$event.date
[1] "POSIXlt" "POSIXt"
$VE
[1] "integer"
$pt
[1] "POSIXct" "POSIXt"
$date
[1] "Date"
$time
[1] "character"
R>
答案 1 :(得分:1)
我对评论的理解是,您获得的数据实际上看起来像data2
。使用原始数据,我们可以获得data3,其中包含"POSIXct"
类datetime列,"Date"
类日期列和"character"
类时间列:
data2 <- transform(data1,
event.date = factor(format(event.date, "%m/%d/%Y %H:%M:%S")))
data3 <- transform(data2,
event.date = as.POSIXct(event.date, format = "%m/%d/%Y %H:%M:%S"),
date = as.Date(event.date, format = "%m/%d/%Y"),
time = sub(".* ", "", event.date),
stringsAsFactors = FALSE)
或者,如果我们想要它们全部作为角色,那么我们可以这样做:
data3a <- transform(data2,
event.date = as.character(as.POSIXct(event.date, format = "%m/%d/%Y %H:%M:%S")),
date = as.character(as.Date(event.date, format = "%m/%d/%Y")),
time = sub(".* ", "", event.date),
stringsAsFactors = FALSE)