如何获取Julia中函数的执行时间? [英] How to obtain the execution time of a function in Julia?

问题描述

我想获取Julia中某个函数的执行时间.这是一个最小的工作示例:

I want to obtain the execution time of a function in Julia. Here is a minimum working example:

function raise_to(n)
    for i in 1:n
        y = (1/7)^n
    end
end

如何获取执行raise_to(10)所需的时间?

How to obtain the time it took to execute raise_to(10) ?

推荐答案

推荐基准测试功能的方法是使用

The recommended way to benchmark a function is to use BenchmarkTools:

julia> function raise_to(n)
           y = (1/7)^n
       end
raise_to (generic function with 1 method)

julia> using BenchmarkTools

julia> @btime raise_to(10)
  1.815 ns (0 allocations: 0 bytes)

请注意，多次重复计算(如您在示例中所做的那样)是获得更准确测量结果的好主意.但是BenchmarTools为您做到了.

Note that repeating the computation numerous times (like you did in your example) is a good idea to get more accurate measurements. But BenchmarTools does it for you.

还请注意，BenchmarkTools避免了仅使用@time的许多陷阱.最值得注意的是，使用@time时，除了运行时间外，您还可能会评估编译时间.这就是为什么@time的第一次调用经常显示较大的时间/分配的原因:

Also note that BenchmarkTools avoids many pitfalls of merely using @time. Most notably with @time, you're likely to measure compilation time in addition to run time. This is why the first invocation of @time often displays larger times/allocations:

# First invocation: the method gets compiled
# Large resource consumption
julia> @time raise_to(10)
  0.007901 seconds (7.70 k allocations: 475.745 KiB)
3.5401331746414338e-9

# Subsequent invocations: stable and low timings
julia> @time raise_to(10)
  0.000003 seconds (5 allocations: 176 bytes)
3.5401331746414338e-9

julia> @time raise_to(10)
  0.000002 seconds (5 allocations: 176 bytes)
3.5401331746414338e-9

julia> @time raise_to(10)
  0.000001 seconds (5 allocations: 176 bytes)
3.5401331746414338e-9

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