r - 减去POSIXct时如何以相同单位表示时差

Question

我想减去POSIXct。我可以这样做，但是取决于第一行（我猜是吗？），差异将以秒或分钟为单位。在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

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多少，我都希望以秒为单位的差异。有没有办法做到这一点？

谢谢。

Answer 1

您可以为该提议使用difftime，该提议允许您指定度量单位，例如

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

另一种方法（如@nicola所提及，并存在于同一文档中）是利用 -具有 -.POSIXt方法并在使用 units<-替换方法进行减法运算后覆盖测量单位这一事实

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