在R中是否有衡量函数执行时间的标准化方法?
显然我可以用系统。时间，然后取它们的差异，但我想知道是否有一些标准化的方法或功能(不想发明轮子)。
我似乎记得我曾经用过如下的东西:
somesysfunction("myfunction(with,arguments)")
> Start time : 2001-01-01 00:00:00 # output of somesysfunction
> "Result" "of" "myfunction" # output of myfunction
> End time : 2001-01-01 00:00:10 # output of somesysfunction
> Total Execution time : 10 seconds # output of somesysfunction
当前回答
另一种简单但非常强大的方法是使用包profvis。它不仅测量代码的执行时间，还为您提供了执行每个函数的钻取。它也可以用于Shiny。
library(profvis)
profvis({
#your code here
})
点击这里查看一些例子。
其他回答
基于bench package网站:
bench::mark() from package bench is used to benchmark one or a series of expressions, we feel it has a number of advantages over alternatives. Always uses the highest precision APIs available for each operating system (often nanoseconds). Tracks memory allocations for each expression. Tracks the number and type of R garbage collections per expression iteration. Verifies equality of expression results by default, to avoid accidentally benchmarking inequivalent code. Has bench::press(), which allows you to easily perform and combine benchmarks across a large grid of values. Uses adaptive stopping by default, running each expression for a set amount of time rather than for a specific number of iterations. Expressions are run in batches and summary statistics are calculated after filtering out iterations with garbage collections. This allows you to isolate the performance and effects of garbage collection on running time (for more details see Neal 2014). The times and memory usage are returned as custom objects which have human readable formatting for display (e.g. 104ns) and comparisons (e.g. x$mem_alloc > "10MB"). There is also full support for plotting with ggplot2 including custom scales and formatting.
Use:
bench::mark(log10(5))
#> # A tibble: 1 × 6
#> expression min median `itr/sec` mem_alloc `gc/sec`
#> <bch:expr> <bch:tm> <bch:tm> <dbl> <bch:byt> <dbl>
#> 1 log10(5) 212ns 274ns 2334086. 0B 0
由reprex包在2021-08-18创建(v2.0.1)
正如Andrie所说，system.time()工作正常。对于短函数，我更喜欢在其中放入replication ():
system.time( replicate(10000, myfunction(with,arguments) ) )
包“tictoc”为您提供了一种非常简单的测量执行时间的方法。文档在:https://cran.fhcrc.org/web/packages/tictoc/tictoc.pdf。
install.packages("tictoc")
require(tictoc)
tic()
rnorm(1000,0,1)
toc()
要保存经过的时间到一个变量，你可以这样做:
install.packages("tictoc")
require(tictoc)
tic()
rnorm(1000,0,1)
exectime <- toc()
exectime <- exectime$toc - exectime$tic
另一种可能的方法是使用Sys.time():
start.time <- Sys.time()
...Relevent codes...
end.time <- Sys.time()
time.taken <- end.time - start.time
time.taken
与上面的答案相比，这不是最优雅的方法，但绝对是一种方法。
虽然其他解决方案对于单个函数也很有用，但我推荐使用下面的代码段，因为它更通用、更有效:
Rprof(tf <- "log.log", memory.profiling = TRUE)
# the code you want to profile must be in between
Rprof (NULL) ; print(summaryRprof(tf))
