r语言 - 在做POSIXct相减时，如何在相同的单位中做时差

Question

我想减去POSIXct。我可以这样做，但取决于第一行(我猜?)的差异将在秒或分钟。下面您可以看到第一个差值以秒为单位，第二个差值以分钟为单位，因为我更改了第一行的时差:

#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

我希望差异总是以秒为单位，无论第一行的差异是什么。有办法让这一切发生吗?

谢谢。

Answer 1

您可以在提案中使用difftime，它允许您指定度量单位，例如

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

另一种方法(如@nicola所提到的，并存在于同一文档中)是利用-具有-.POSIXt方法的事实，并在使用units<-替代方法进行减法操作后覆盖测量单元

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