我正在使用列表理解vs consturrent.futures来测试一个琐碎的函数:
class Test:
@staticmethod
def something(times = 1):
return sum([1 for i in range(times)])
@staticmethod
def simulate1(function, N):
l = []
for i in range(N):
outcome = function()
l.append(outcome)
return sum(l) / N
@staticmethod
def simulate2(function, N):
import concurrent.futures
with concurrent.futures.ThreadPoolExecutor(max_workers=10) as executor:
l = [outcome for outcome in executor.map(lambda x: function(), range(N))]
return sum(l) / N
@staticmethod
def simulate3(function, N):
import concurrent.futures
l = 0
with concurrent.futures.ThreadPoolExecutor(max_workers=10) as executor:
futures = [executor.submit(function) for i in range(N)]
for future in concurrent.futures.as_completed(futures):
l += future.result()
return l / N
def simulation():
simulationRate = 100000
import datetime
s = datetime.datetime.now()
o = Test.simulate1(lambda : Test.something(10), simulationRate)
print((datetime.datetime.now() - s))
s = datetime.datetime.now()
o = Test.simulate2(lambda : Test.something(10), simulationRate)
print((datetime.datetime.now() - s))
s = datetime.datetime.now()
o = Test.simulate3(lambda : Test.something(10), simulationRate)
print((datetime.datetime.now() - s))
simulation()
测量时间,我得到了:
0:00:00.258000
0:00:10.348000
0:00:10.556000
我正在开始并发,所以我不明白什么是阻止线程更快地运行的瓶颈。
如果将任务功能更改为此,则会看到差异:
def something(n):
""" simulate doing some io based task.
"""
time.sleep(0.001)
return sum(1 for i in range(n))
在我的Mac Pro上,这给出了:
0:00:13.774700
0:00:01.591226
0:00:01.489159
conturrent.future这次更加明显。
原因是:您正在模拟基于CPU的任务,因为Python的GIL,同意。
consturrent.future为异步执行可呼叫提供了一个高级接口,您将其用于错误的场景。