减去 POSIXct 时如何使相同单位的时间差

我想减去 POSIXct。我可以做到这一点，但根据第一行（我猜？），差异将以秒或分钟为单位。下面您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，因为我更改了第一行中的时间差：

#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

results:

> #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

我希望差异始终以秒为单位，无论第一行差异是什么。有办法实现这一点吗？

谢谢。

您可以使用difftime例如，建议允许您指定测量单位

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

另一种方法（正如 @nicola 所提到的，并出现在同一文档中）是利用以下事实：- has a -.POSIXt方法并在减法运算后覆盖测量单位units<-更换方法

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