在使用difftime()时，如何提取时间单位？

Question

我目前正在运行一个for循环代码，在每个循环的末尾，我正在测量循环的持续时间，并打印一条消息，告诉用户这个循环花费了多长时间。

为了获得我所用的持续时间

duration <- difftime(end_time, start_time) 

我就是这样打印这份声明的

print(paste("Loop", i, "took", duration, "to run.")). 

问题是每个循环的持续时间从30秒到1个小时不等。将持续时间放入粘贴中，只会将其转化为一个没有单位的数字。

例如：

print(paste("Loop", i, "took", duration, "to run.")). 
"Loop 5 took 10.5 to run"

如何从difftime()中获得单元，以便得到类似于：“循环5”运行了10.5分钟“。

PS:有些人可能建议通过在difftime()中声明" unit“参数来标准化工期，但我希望用户能够很容易地理解工期，这就是为什么我将单元设置为默认的"auto”。

Answer 1

从units调用duration获取单元，并通过添加duration获得数值

print(paste("Loop", i, "took", round(duration[[1]], 2),  units(duration), "to run."))

下面是一个示例：

start_time <- Sys.time()

# few seconds later
end_time <- Sys.time()

duration <- difftime(end_time, start_time)

print(paste("Loop", 1, "took", round(duration[[1]], 2),  units(duration), "to run."))

结果：

[1] "Loop 1 took 6.97 secs to run."

根据持续时间的范围，该单元将是自动的。参见下面的示例：

start_time <- Sys.time()

# few days later
end_time <- as.Date("2019-01-23")

duration <- difftime(end_time, start_time)

print(paste("Loop", 1, "took", round(duration[[1]], 2),  units(duration), "to run."))

结果：

> print(paste("Loop", 1, "took", round(duration[[1]], 2),  units(duration), "to run."))
[1] "Loop 1 took 5.9 days to run."

Answer 2

一种方法是捕获来自difftime的输出，因为它使用capture.output

start_time <- as.POSIXct('2019-01-01 02:34:00')
end_time <- as.POSIXct('2019-01-01 03:14:00')

paste("Loop 5 took", capture.output(difftime(end_time, start_time)), "to run.")
#[1] "Loop 5 took Time difference of 40 mins to run."

现在您可以更改输出以使其有意义，例如从输出中删除“时间差”。

sent <- capture.output(difftime(end_time, start_time))
paste("Loop 5 took",sub("Time difference of ","", sent) , "to run.")
#[1] "Loop 5 took 40 mins to run."

用于另一个输入

start_time <- as.POSIXct('2019-01-01 02:34:00')
end_time <- as.POSIXct('2019-01-01 02:34:50')

sent <- capture.output(difftime(end_time, start_time))
paste("Loop 5 took",sub("Time difference of ","", sent) , "to run.")
#[1] "Loop 5 took 50 secs to run."

Answer 3

您应该能够通过units(duration)实现这一点。