Question

我想减去POSIXct。我可以这样做，但取决于第一行（我猜？），差异将以秒或分钟为单位。下面你可以看到第一个差异以秒为单位，第二个差异以分钟为单位，因为我改变了第一行的时差：

#diff in seconds because 1st row time diff is small?
t1<- as.POSIXct(c("2015-02-02 20:18:03 00:00:00", "2015-02-02 20:17:02 00:00:00"),"GMT")
t2<- as.POSIXct(c("2015-02-02 20:18:02 00:00:00","2015-02-02 20:18:02 00:00:00"),"GMT")
d<-data.frame(t1= t1, t2= t2)
d$t1-d$t2


#diff in seconds because 1st row time diff is larger?
t1<- as.POSIXct(c("2015-02-02 20:13:03 00:00:00", "2015-02-02 20:17:02 00:00:00"),"GMT")
t2<- as.POSIXct(c("2015-02-02 20:18:02 00:00:00","2015-02-02 20:18:02 00:00:00"),"GMT")
d<-data.frame(t1= t1, t2= t2)
d$t1-d$t2

结果：

> #diff in seconds because 1st row time diff is small?
> t1<- as.POSIXct(c("2015-02-02 20:18:03 00:00:00", "2015-02-02 20:17:02 00:00:00"),"GMT")
> t2<- as.POSIXct(c("2015-02-02 20:18:02 00:00:00","2015-02-02 20:18:02 00:00:00"),"GMT")
> d<-data.frame(t1= t1, t2= t2)
> d$t1-d$t2
Time differences in secs
[1]   1 -60
> 
> 
> #diff in seconds because 1st row time diff is larger?
> t1<- as.POSIXct(c("2015-02-02 20:13:03 00:00:00", "2015-02-02 20:17:02 00:00:00"),"GMT")
> t2<- as.POSIXct(c("2015-02-02 20:18:02 00:00:00","2015-02-02 20:18:02 00:00:00"),"GMT")
> d<-data.frame(t1= t1, t2= t2)
> d$t1-d$t2
Time differences in mins
[1] -4.983333 -1.000000

无论第一行的差异是什么，我都希望能在几秒钟之内完成差异。有没有办法让这种情况发生？

谢谢。

Answer 1

您可以使用difftime作为建议，允许您指定测量单位，例如

difftime(t1, t2, units = "secs")

另一种方式（如同@nicola所述并出现在同一文档中）是利用-具有-.POSIXt方法的事实，并在减法操作后覆盖测量单位units<-替换方法

res <- t1 - t2
units(res) <- "secs"

减去POSIXct时如何在相同单位中产生时差

1 个答案: