我在一些资料中读到print命令不是线程安全的,解决方法是使用系统标准输出写入命令,但它仍然对我不起作用,而且对标准输出的写入不是原子的。在
下面是一个简短的示例(称为这个文件并行实验.py)公司名称:
import os
import sys
from multiprocessing import Pool
def output(msg):
msg = '%s%s' % (msg, os.linesep)
sys.stdout.write(msg)
def func(input):
output(u'pid:%d got input \"%s\"' % (os.getpid(), str(input)))
def executeFunctionInParallel(funcName, inputsList, maxParallelism):
output(u'Executing function %s on input of size %d with maximum parallelism of %d' % (
funcName.__name__, len(inputsList), maxParallelism))
parallelismPool = Pool(processes=maxParallelism)
executeBooleanResultsList = parallelismPool.map(funcName, inputsList)
parallelismPool.close()
output(u'Function %s executed on input of size %d with maximum parallelism of %d' % (
funcName.__name__, len(inputsList), maxParallelism))
# if all parallel executions executed well - the boolean results list should all be True
return all(executeBooleanResultsList)
if __name__ == "__main__":
inputsList=[str(i) for i in range(20)]
executeFunctionInParallel(func, inputsList, 4)
看看输出:
一、调用python的输出并行实验.py(注意“pid”一词在某些行中弄乱了):
^{pr2}$二。调用python的输出并行实验.py>;并行实验.log重定向到意义并行实验.log文件(注意行顺序不好,因为在调用并行调用func的executeFunctionInParallel之前和之后,应该打印一条消息):
pid:3244 got input "4"
pid:3244 got input "5"
pid:3244 got input "12"
pid:3244 got input "13"
pid:240 got input "0"
pid:240 got input "1"
pid:240 got input "8"
pid:240 got input "9"
pid:240 got input "16"
pid:240 got input "17"
pid:1268 got input "2"
pid:1268 got input "3"
pid:1268 got input "10"
pid:1268 got input "11"
pid:1268 got input "18"
pid:1268 got input "19"
pid:3332 got input "6"
pid:3332 got input "7"
pid:3332 got input "14"
pid:3332 got input "15"
Executing function func on input of size 20 with maximum parallelism of 4
Function func executed on input of size 20 with maximum parallelism of 4
如果您希望避免锁定,并且乐于使用较低级别的接口,那么可以使用
os.open
,os.write
获得POSIX OаAPPEND行为(如果您的系统支持它);然后查看Is file append atomic in UNIX?。在这是因为
multiprocessing.Pool
实际上使用子进程而不是线程。 您需要在进程之间使用显式的synchronization。注意,链接上的示例,它解决了您的问题。在相关问题 更多 >
编程相关推荐