如何计算几个cProfile结果的平均值?

时间:2019-01-26 03:18:22

标签: python python-3.x profiler cprofile

不是像这样只运行一次配置文件:

import cProfile

def do_heavy_lifting():
    for i in range(100):
        print('hello')

profiller = cProfile.Profile()
profiller.enable()

do_heavy_lifting()

profiller.disable()
profiller.print_stats(sort='time')

配置文件结果如下:

      502 function calls in 0.000 seconds

Ordered by: internal time

ncalls  tottime  percall  cumtime  percall filename:lineno(function)
   100    0.000    0.000    0.000    0.000 {built-in method builtins.print}
   200    0.000    0.000    0.000    0.000 cp1252.py:18(encode)
   200    0.000    0.000    0.000    0.000 {built-in method _codecs.charmap_encode}
     1    0.000    0.000    0.000    0.000 test.py:2(do_heavy_lifting)
     1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}

我想运行几次并打印平均结果,以提高准确性。

这是我想到的初始脚本食谱:

import cProfile

def do_heavy_lifting():
    for i in range(100):
        print('hello')

def best_of_profillings(target_profile_function, count):
    profile_results = []

    for index in range(count):
        profiller = cProfile.Profile()
        profiller.enable()

        target_profile_function()

        profiller.disable()
        profile_results.append(profiller)

    profile_results /= count
    return profile_results

heavy_lifting_result = best_of_profillings(do_heavy_lifting, 10)
heavy_lifting_result.print_stats(sort='time')

运行此命令后,它应该像第一个版本一样显示结果,但区别在于它们是几次运行的平均值,而不是一次运行。

草稿脚本仍然缺少profile_results /= count部分,在所有迭代之后,我将获得所有计算结果并创建平均结果,并始终将其显示在屏幕上:

      502 function calls in 0.000 seconds

Ordered by: internal time

ncalls  tottime  percall  cumtime  percall filename:lineno(function)
   100    0.000    0.000    0.000    0.000 {built-in method builtins.print}
   200    0.000    0.000    0.000    0.000 cp1252.py:18(encode)
   200    0.000    0.000    0.000    0.000 {built-in method _codecs.charmap_encode}
     1    0.000    0.000    0.000    0.000 test.py:2(do_heavy_lifting)
     1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}

1 个答案:

答案 0 :(得分:0)

我设法使用功能average()创建了以下代码。我打开了pstats的实现,发现有一个名为Stats.add()的函数,它似乎只是将结果连接到当前对象中:https://docs.python.org/3.7/library/profile.html#pstats.Stats.add

import io
import pstats
import cProfile

def do_heavy_lifting():
    for i in range(100):
        print('hello')

def average(stats, count):
    stats.total_calls /= count
    stats.prim_calls /= count
    stats.total_tt /= count

    for func, source in stats.stats.items():
        cc, nc, tt, ct, callers = source
        stats.stats[func] = ( cc/count, nc/count, tt/count, ct/count, callers )

    return stats

def best_of_profillings(target_profile_function, count):
    output_stream = io.StringIO()
    profiller_status = pstats.Stats( stream=output_stream )

    for index in range(count):
        profiller = cProfile.Profile()
        profiller.enable()

        target_profile_function()

        profiller.disable()
        profiller_status.add( profiller )

        print( 'Profiled', '%.3f' % profiller_status.total_tt, 'seconds at', index,
                'for', target_profile_function.__name__, flush=True )

    average( profiller_status, count )
    profiller_status.sort_stats( "time" )
    profiller_status.print_stats()

    return "\nProfile results for %s\n%s" % ( 
           target_profile_function.__name__, output_stream.getvalue() )

heavy_lifting_result = best_of_profillings( do_heavy_lifting, 10 )
print( heavy_lifting_result )

结果:

Profile results for do_heavy_lifting
         102.0 function calls in 0.001 seconds

   Ordered by: internal time

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
    100.0    0.001    0.000    0.001    0.000 {built-in method builtins.print}
      1.0    0.000    0.000    0.001    0.001 D:\test.py:5(do_heavy_lifting)
      1.0    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}