使用 Sys.time() 计时 R 代码 [英] Timing R code with Sys.time()

问题描述

我可以使用以下代码运行一段代码 5 或 10 秒:

I can run a piece of code for 5 or 10 seconds using the following code:

period <- 10  ## minimum time (in seconds) that the loop should run for
tm <- Sys.time()  ## starting data & time
while(Sys.time() - tm < period) print(Sys.time())

代码可以正常运行 5 或 10 秒.但是当我用 60 替换周期值让它运行一分钟时，代码永远不会停止.怎么了?

The code runs just fine for 5 or 10 seconds. But when I replace the period value by 60 for it to run for a minute, the code never stops. What is wrong?

推荐答案

只要经过的时间超过 1 分钟，默认单位就会从秒变为分钟.所以你想控制单位:

As soon as elapsed time exceeds 1 minute, the default unit changes from seconds to minutes. So you want to control the unit:

while (difftime(Sys.time(), tm, units = "secs")[[1]] < period)

来自 ?difftime

 If ‘units = "auto"’, a suitable set of units is chosen, the
 largest possible (excluding ‘"weeks"’) in which all the absolute
 differences are greater than one.

 Subtraction of date-time objects gives an object of this class, by
 calling ‘difftime’ with ‘units = "auto"’.

或者使用 proc.time，它可以在几秒钟内测量自您启动 R 会话以来的各种时间(用户"、系统"、已用时间").我们想要已用"时间，即挂钟时间，因此我们检索 proc.time() 的第三个值.

Alternatively use proc.time, which measures various times ("user", "system", "elapsed") since you started your R session in seconds. We want "elapsed" time, i.e., the wall clock time, so we retrieve the 3rd value of proc.time().

period <- 10
tm <- proc.time()[[3]]
while (proc.time()[[3]] - tm < period) print(proc.time())

如果您对[[1]]和[[3]]的使用感到困惑，请咨询:

If you are confused by the use of [[1]] and [[3]], please consult:

• 如何仅从指定号码(没有名称)中提取号码?
• 如何在 R 中获取没有列名的矩阵元素?

让我添加一些用户友好的可重现示例.您在循环内带有 print 的原始代码非常烦人，因为它会在屏幕上打印数千行.我会使用 Sys.sleep.

Let me add some user-friendly reproducible examples. Your original code with print inside a loop is quite annoying as it prints thousands of lines onto the screen. I would use Sys.sleep.

test.Sys.time <- function(sleep_time_in_secs) {
  t1 <- Sys.time()
  Sys.sleep(sleep_time_in_secs)
  t2 <- Sys.time()
  ## units = "auto"
  print(t2 - t1)
  ## units = "secs"
  print(difftime(t2, t1, units = "secs"))
  ## use '[[1]]' for clean output
  print(difftime(t2, t1, units = "secs")[[1]])
  }

test.Sys.time(5)
#Time difference of 5.005247 secs
#Time difference of 5.005247 secs
#[1] 5.005247

test.Sys.time(65)
#Time difference of 1.084357 mins
#Time difference of 65.06141 secs
#[1] 65.06141

自动"单位非常聪明.如果sleep_time_in_secs = 3605(超过一个小时)，默认单位将更改为小时".

The "auto" units is very clever. If sleep_time_in_secs = 3605 (more than an hour), the default unit will change to "hours".

在使用 Sys.time 时要小心时间单位，否则您可能会在基准测试中上当.这是一个完美的例子:read.csv/fread 基准测试中的意外结果.我已经用现在删除的评论回答了它:

Be careful with time units when using Sys.time, or you may be fooled in a benchmarking. Here is a perfect example: Unexpected results in benchmark of read.csv / fread. I had answered it with a now removed comment:

您的时间单位有问题.我看到 fread 快了 20 多倍.如果 fread 需要 4 秒读取一个文件，read.csv 需要 80 秒 = 1.33 分钟.忽略单位，read.csv更快".

You got a problem with time units. I see that fread is more than 20 times faster. If fread takes 4 seconds to read a file, read.csv takes 80 seconds = 1.33 minutes. Ignoring the units, read.csv is "faster".

现在让我们测试 proc.time.

test.proc.time <- function(sleep_time_in_secs) {
  t1 <- proc.time()
  Sys.sleep(sleep_time_in_secs)
  t2 <- proc.time()
  ## print user, system, elapsed time
  print(t2 - t1)
  ## use '[[3]]' for clean output of elapsed time
  print((t2 - t1)[[3]])
  }

test.proc.time(5)
#   user  system elapsed 
#  0.000   0.000   5.005 
#[1] 5.005

test.proc.time(65)
#   user  system elapsed 
#  0.000   0.000  65.057 
#[1] 65.057

用户"时间和系统"时间为0，因为CPU和系统内核都空闲.

"user" time and "system" time are 0, because both CPU and the system kernel are idle.

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