python线程池(ThreadPoolExecutor)与进程池(ProcessPoolExecutor)的简单用法

python中ThreadPoolExecutor(线程池)与ProcessPoolExecutor(进程池)都是concurrent.futures模块下的,主线程(或进程)中可以获取某一个线程(进程)执行的状态或者某一个任务执行的状态及返回值。

通过submit返回的是一个future对象,它是一个未来可期的对象,通过它可以获悉线程的状态

ThreadPoolExecutor(线程池)

import time
from concurrent.futures import ThreadPoolExecutor

def get_thread_time(times):
    time.sleep(times)
    return times

# 创建线程池  指定最大容纳数量为4
executor = ThreadPoolExecutor(max_workers=4)
# 通过submit提交执行的函数到线程池中
task1 = executor.submit(get_thread_time, (1))
task2 = executor.submit(get_thread_time, (2))
task3 = executor.submit(get_thread_time, (3))
task4 = executor.submit(get_thread_time, (4))
print("task1:{} ".format(task1.done()))
print("task2:{}".format(task2.done()))
print("task3:{} ".format(task3.done()))
print("task4:{}".format(task4.done()))
time.sleep(2.5)
print('after 2.5s {}'.format('-'*20))

done_map = {
    "task1":task1.done(),
    "task2":task2.done(),
    "task3":task3.done(),
    "task4":task4.done()
}
# 2.5秒之后,线程的执行状态
for task_name,done in done_map.items():
    if done:
        print("{}:completed".format(task_name))

result:

task1:False 
task2:False
task3:False 
task4:False
after 2.5s --------------------
task1:completed
task2:completed

初始状态4个task都是未完成状态,2.5秒后task1和task2执行完成,task3和task由于是sleep(3) sleep(4)所以仍然是未完成的sleep状态

通过wait()判断线程执行的状态:

wait(fs, timeout=None, return_when=ALL_COMPLETED),wait接受3个参数,fs表示执行的task序列;timeout表示等待的最长时间,超过这个时间即使线程未执行完成也将返回;return_when表示wait返回结果的条件,默认为ALL_COMPLETED全部执行完成再返回:

import time
from concurrent.futures import (
    ThreadPoolExecutor, wait
)


def get_thread_time(times):
    time.sleep(times)
    return times


start = time.time()
executor = ThreadPoolExecutor(max_workers=4)
task_list = [executor.submit(get_thread_time, times) for times in [1, 2, 3, 4]]
i = 1
for task in task_list:
    print("task{}:{}".format(i, task))
    i += 1
print(wait(task_list, timeout=2.5))

wait在2.5秒后返回线程的状态,result:

task1:<Future at 0x7ff3c885f208 state=running>
task2:<Future at 0x7ff3c885fb00 state=running>
task3:<Future at 0x7ff3c764b2b0 state=running>
task4:<Future at 0x7ff3c764b9b0 state=running>
DoneAndNotDoneFutures(done={<Future at 0x7ff3c885f208 state=finished returned int>, <Future at 0x7ff3c885fb00 state=finished returned int>}, not_done={<Future at 0x7ff3c764b2b0 state=running>, <Future at 0x7ff3c764b9b0 state=running>})

可以看到在timeout 2.5时,task1和task2执行完毕,task3和task4仍在执行中

通过map返回线程的执行结果:

import time
from concurrent.futures import ThreadPoolExecutor


def get_thread_time(times):
    time.sleep(times)
    return times


start = time.time()
executor = ThreadPoolExecutor(max_workers=4)

i = 1
for result in executor.map(get_thread_time,[2,3,1,4]):
    print("task{}:{}".format(i, result))
    i += 1

map(fn, *iterables, timeout=None),第一个参数fn是线程执行的函数;第二个参数接受一个可迭代对象;第三个参数timeout跟wait()的timeout一样,但由于map是返回线程执行的结果,如果timeout小于线程执行时间会抛异常TimeoutError。

import time
from concurrent.futures import ThreadPoolExecutor


def get_thread_time(times):
    time.sleep(times)
    return times


start = time.time()
executor = ThreadPoolExecutor(max_workers=4)

i = 1
for result in executor.map(get_thread_time,[2,3,1,4]):
    print("task{}:{}".format(i, result))
    i += 1

map的返回是有序的,它会根据第二个参数的顺序返回执行的结果:

task1:2
task2:3
task3:1
task4:4

as_completed返回线程执行结果:

import time
from collections import OrderedDict
from concurrent.futures import (
    ThreadPoolExecutor, as_completed
)


def get_thread_time(times):
    time.sleep(times)
    return times


start = time.time()
executor = ThreadPoolExecutor(max_workers=4)
task_list = [executor.submit(get_thread_time, times) for times in [2, 3, 1, 4]]
task_to_time = OrderedDict(zip(["task1", "task2", "task3", "task4"],[2, 3, 1, 4]))
task_map = OrderedDict(zip(task_list, ["task1", "task2", "task3", "task4"]))

for result in as_completed(task_list):
    task_name = task_map.get(result)
    print("{}:{}".format(task_name,task_to_time.get(task_name)))

task1、task2、task3、task4的等待时间分别为2s、3s、1s、4s,通过as_completed返回执行完的线程结果,as_completed(fs, timeout=None)接受2个参数,第一个是执行的线程列表,第二个参数timeout与map的timeout一样,当timeout小于线程执行时间会抛异常TimeoutError。

task3:1
task1:2
task2:3
task4:4

通过执行结果可以看出,as_completed返回的顺序是线程执行结束的顺序,最先执行结束的线程最早返回。

ProcessPoolExecutor

对于频繁的cpu操作,由于GIL锁的原因,多个线程只能用一个cpu,这时多进程的执行效率要比多线程高。

线程池操作斐波拉切:

import time
from concurrent.futures import ThreadPoolExecutor


def fib(n):
    if n < 3:
        return 1
    return fib(n - 1) + fib(n - 2)


start_time = time.time()
executor = ThreadPoolExecutor(max_workers=4)
task_list = [executor.submit(fib, n) for n in range(3, 35)]
thread_results = [task.result() for task in as_completed(task_list)]
print(thread_results)
print("ThreadPoolExecutor time is: {}".format(time.time() - start_time))

result:

[8, 5, 3, 2, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 10946, 46368, 6765, 28657, 17711, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887]
ThreadPoolExecutor time is: 4.998981237411499

进程池操作斐波拉切:

import time
from concurrent.futures import ProcessPoolExecutor


def fib(n):
    if n < 3:
        return 1
    return fib(n - 1) + fib(n - 2)


start_time = time.time()
executor = ProcessPoolExecutor(max_workers=4)
task_list = [executor.submit(fib, n) for n in range(3, 35)]
process_results = [task.result() for task in as_completed(task_list)]
print(process_results)
print("ProcessPoolExecutor time is: {}".format(time.time() - start_time))

result:

[2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 75025, 28657, 46368, 196418, 121393, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887]
ProcessPoolExecutor time is: 3.3585257530212402

可以看出,对于频繁cpu操作进程是优于线程的,3.3s<4.9s

ProcessPoolExecutor在使用上和ThreadPoolExecutor大致是一样的 它们在futures中的方法也是相同的,但是对于map()方法ProcessPoolExecutor会多一个参数chunksize(ThreadPoolExecutor中这个参数没有任何作用), chunksize将迭代对象切成块,将其作为分开的任务提交给pool,对于很大的iterables,设置较大chunksize可以提高性能。