Question

我正在尝试按以下格式减去包含日期时间信息的2个字符向量：

> dput(train2)

structure(list(time2 = c("2011-09-01 23:44:52.533", "2011-09-05 12:25:37.42", 
"2011-08-24 12:56:58.91", "2011-10-25 07:18:14.722", "2011-10-25 07:19:51.697"
), time3 = c("2011-09-01 23:43:59.752", "2011-09-05 12:25:01.187", 
"2011-08-24 12:55:13.012", "2011-10-25 07:16:51.759", "2011-10-25 07:16:51.759"
)), .Names = c("time2", "time3"), row.names = c(NA, 5L), class = "data.frame")

我一直在寻找并使用zoo，as.Date，as.POSIXct等等来尝试找到正确的代码来减去2个日期时间对象并在几秒钟内得到答案但是没有运气。

我很感激任何建议。

Answer 1

>  x1<-"2013-03-03 23:26:46.315" 
>  x2<-"2013-03-03 23:31:53.091"
>  x1 <- strptime(x1, "%Y-%m-%d %H:%M:%OS")
>  x2 <- strptime(x2, "%Y-%m-%d %H:%M:%OS")
> x1
[1] "2013-03-03 23:26:46"
> x2
[1] "2013-03-03 23:31:53"

我按照@Dirk Eddelbuettel的回答，但我正在失去精确度。如何强制R不切割第二部分？

谢天谢地（strptime的人）我自己回答了我的问题：

op <- options(digits.secs = 3)

应用此设置后，将使用精度。

http://stat.ethz.ch/R-manual/R-devel/library/base/html/strptime.html

如果您希望在几秒钟内获得差异，但在几分钟内获得，则可能会有用：

> as.numeric(x2-x1,units="secs")
[1] 306.776

Answer 2

容易腻：

R> now <- Sys.time()
R> then <- Sys.time()
R> then - now
Time difference of 5.357 secs
R> class(then - now)
[1] "difftime"
R> as.numeric(then - now)
[1] 5.357
R>

对于您的数据：

R> df
                    time2                   time3
1 2011-09-01 23:44:52.533 2011-09-01 23:43:59.752
2  2011-09-05 12:25:37.42 2011-09-05 12:25:01.187
3  2011-08-24 12:56:58.91 2011-08-24 12:55:13.012
4 2011-10-25 07:18:14.722 2011-10-25 07:16:51.759
5 2011-10-25 07:19:51.697 2011-10-25 07:16:51.759
R> df$time2 <- strptime(df$time2, "%Y-%m-%d %H:%M:%OS")
R> df$time3 <- strptime(df$time3, "%Y-%m-%d %H:%M:%OS")
R> df
                    time2                   time3
1 2011-09-01 23:44:52.533 2011-09-01 23:43:59.752
2 2011-09-05 12:25:37.420 2011-09-05 12:25:01.187
3 2011-08-24 12:56:58.910 2011-08-24 12:55:13.012
4 2011-10-25 07:18:14.722 2011-10-25 07:16:51.759
5 2011-10-25 07:19:51.697 2011-10-25 07:16:51.759
R> df$time2 - df$time3
Time differences in secs
[1]  52.781  36.233 105.898  82.963 179.938
attr(,"tzone")
[1] ""
R>

并以数字形式添加回数据框：

R> df$dt <- as.numeric(df$time2 - df$time3)
R> df
                    time2                   time3      dt
1 2011-09-01 23:44:52.533 2011-09-01 23:43:59.752  52.781
2 2011-09-05 12:25:37.420 2011-09-05 12:25:01.187  36.233
3 2011-08-24 12:56:58.910 2011-08-24 12:55:13.012 105.898
4 2011-10-25 07:18:14.722 2011-10-25 07:16:51.759  82.963
5 2011-10-25 07:19:51.697 2011-10-25 07:16:51.759 179.938
R>

R找到YYYY-MM-DD HH：MM：SS.MMM格式的时间差（秒）

2 个答案: