划分两个difftime对象

问题描述：

我有三个时间（ POSIXct ）对象 t1 ， t2 ， t3 ，它们指定完成任务的持续时间。

I have three time (POSIXct) objects t1, t2, t3 which specify the time duration to complete a task.

我发现 t1 ， t2 ， t3 ，执行以下操作：

I found t1, t2, t3 by doing the following:

t1 <- as.POSIXct("2016-10-30 13:53:34") - as.POSIXct("2016-10-30 13:35:34")
t2 <- as.POSIXct("2016-10-30 14:53:34") - as.POSIXct("2016-10-30 14:35:34")
t3 <- as.POSIXct("2016-10-30 15:50:34") - as.POSIXct("2016-10-30 15:40:34")

我想找到比率 t1 / t3 和 t2 / t3 。但是，出现以下错误：

I want to find the ratios t1/t3 and t2/t3. However, I get the following error:

t1/t3
# Error in `/.difftime`(t1, t3) : 
#   second argument of / cannot be a "difftime" object

无法分割两个 difftime 对象。有什么办法可以找到将两个 difftime 对象相除的结果？

I understood that two difftime objects cannot be divided. Is there any way that I could find the result of dividing two difftime objects?

答

要除以 difftime ，必须将其转换为数字。如您在评论中所述，如果您希望答案以秒为单位，则可以指定秒 $ c>单位。例如：

To divide by a difftime you must convert it to numeric. If, as you stated in a comment, you would like the answer to be expressed in seconds, you can specify the 'secs' units. For example:

t1/as.double(t3, units='secs')

如@JonathanLisic所述， as.double 通常不会占用 units 参数，这不适用于通用时间类。 difftime 的 S3 方法使用参数。

As @JonathanLisic notes, as.double does not generally take a units parameter, and this won't work for generic time classes. It is the S3 method for difftime which takes the parameter.

划分两个difftime对象

相关推荐