python執行緒池(threadpool)模組使用筆記
阿新 • • 發佈:2018-12-05
https://www.cnblogs.com/xiaozi/p/6182990.html
一、安裝與簡介
pip install threadpool
pool = ThreadPool(poolsize) requests = makeRequests(some_callable, list_of_args, callback) [pool.putRequest(req) for req in requests] pool.wait()
第一行定義了一個執行緒池,表示最多可以建立poolsize這麼多執行緒;
第二行是呼叫makeRequests建立了要開啟多執行緒的函式,以及函式相關引數和回撥函式,其中回撥函式可以不寫,default是無,也就是說makeRequests只需要2個引數就可以執行;
第三行用法比較奇怪,是將所有要執行多執行緒的請求扔進執行緒池,[pool.putRequest(req) for req in requests]等同於
for req in requests:
pool.putRequest(req)
第四行是等待所有的執行緒完成工作後退出。
二、程式碼例項
import time def sayhello(str): print "Hello ",str time.sleep(2) name_list =['xiaozi','aa','bb','cc'] start_time = time.time() for i in range(len(name_list)): sayhello(name_list[i]) print '%d second'% (time.time()-start_time)
改用執行緒池程式碼,花費時間更少,更效率
import time import threadpool def sayhello(str): print "Hello ",str time.sleep(2) name_list =['xiaozi','aa','bb','cc'] start_time = time.time() pool = threadpool.ThreadPool(10) requests = threadpool.makeRequests(sayhello, name_list) [pool.putRequest(req) for req in requests] pool.wait() print '%d second'% (time.time()-start_time)
當函式有多個引數的情況,函式呼叫時第一個解包list,第二個解包dict,所以可以這樣:
def hello(m, n, o): """""" print "m = %s, n = %s, o = %s"%(m, n, o) if __name__ == '__main__': # 方法1 lst_vars_1 = ['1', '2', '3'] lst_vars_2 = ['4', '5', '6'] func_var = [(lst_vars_1, None), (lst_vars_2, None)] # 方法2 dict_vars_1 = {'m':'1', 'n':'2', 'o':'3'} dict_vars_2 = {'m':'4', 'n':'5', 'o':'6'} func_var = [(None, dict_vars_1), (None, dict_vars_2)] pool = threadpool.ThreadPool(2) requests = threadpool.makeRequests(hello, func_var) [pool.putRequest(req) for req in requests] pool.wait()
需要把所傳入的引數進行轉換,然後帶人執行緒池。
def getuserdic(): username_list=['xiaozi','administrator'] password_list=['root','','abc123!','123456','password','root'] userlist = [] for username in username_list: user =username.rstrip() for password in password_list: pwd = password.rstrip() userdic ={} userdic['user']=user userdic['pwd'] = pwd tmp=(None,userdic) userlist.append(tmp) return userlist
最後
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