Python concurrent.futures简单实例

实例1:简单实现单个任务多线程

from concurrent.futures import ThreadPoolExecutor, as_completed, ProcessPoolExecutor


THREAD_POOL = ThreadPoolExecutor(4)



# write为函数名,'打球为参数'
def thread_write():
    all_task = []
    for _ in range(4):
        all_task.append(THREAD_POOL.submit(write, '打球'))
    for future in as_completed(all_task):  # 子线程全部完成时
        future.result()



def write(moive):
    time.sleep(3)
    print(moive)
    time.sleep(2)

实例2：多个子任务多线程实现

from concurrent.futures import ThreadPoolExecutor
import time
from datetime import datetime


# 参数times用来模拟网络请求的时间
def get_html(times):
    time.sleep(times)
    print("get page {}s finished {}".format(times,datetime.now()))
    return times


def madd(*list):
    time.sleep(2)
    return sum(list)

# wait的应用
executor = ThreadPoolExecutor(max_workers=2)
# 通过submit函数提交执行的函数到线程池中，submit函数立即返回，不阻塞
task1 = executor.submit(get_html, *(3,))  # 开一个线程去处理事务
task2 = executor.submit(madd, *(2,3,4)) # 开第二个线程去处理事务
# done方法用于判定某个任务是否完成
print(task1.done())
# cancel方法用于取消某个任务,该任务没有放入线程池中才能取消成功
print(task2.cancel())
print(datetime.now())
time.sleep(4)
print(datetime.now())
print(task1.done())
# result方法可以获取task的执行结果
print(task1.result())
print(task2.result())
wait(all_task, return_when=ALL_COMPLETED) # 等待子线程全部完成

　实例3：多线程，当单线程完成时，能补充，时刻保持四线程运行

# 线程完成时,可以重复提交
from concurrent.futures import ThreadPoolExecutor,wait,ALL_COMPLETED,FIRST_COMPLETED
import time
from datetime import datetime


# 参数times用来模拟网络请求的时间
def sleep_1():
    time.sleep(1)
    print(f'{datetime.now()}:sleep 1s')




def sleep_2():
    time.sleep(2)
    print(f'{datetime.now()}:sleep 2s')


def sleep_3():
    time.sleep(3)
    print(f'{datetime.now()}:sleep 3s')


def sleep_4():
    time.sleep(4)
    print(f'{datetime.now()}:sleep 4s')

with ThreadPoolExecutor(max_workers=4) as t:
    taskDict = {'task_1':sleep_1,'task_2':sleep_2,'task_3':sleep_3,'task_4':sleep_4}
    # all_task.extend([t.submit(value) for value in taskDict.values()])
    allTaskDict = {name:t.submit(runFunc) for name,runFunc in taskDict.items()}
    while 1:
        wait(allTaskDict.values(),return_when=FIRST_COMPLETED)
        # todo 这里判断是否需要添加
        for name,task in allTaskDict.items():
            if task.done():
                # print(task.result())
               # 重新提交线程
                allTaskDict.update({name:t.submit(taskDict[name])})

　　

使用:shutdown , 或用with来关闭线程.

pool.shutdown(wait=True)