Python namedtuple作为apply_async(..)回调的参数

Question

我正在写一个简短的程序,我想异步调用一个函数,这样它就不会阻塞调用者.为此,我正在使用Poolpython的multiprocessing模块.

在异步调用的函数中,我想返回一个namedtuple以适应我程序其余部分的逻辑,但我发现a namedtuple似乎不是从生成的进程传递给回调的受支持类型(可能是因为它不能被腌制).这是问题的最低限度.

from multiprocessing import Pool
from collections import namedtuple

logEntry = namedtuple("LogEntry", ['logLev', 'msg'])

def doSomething(x):
    # Do actual work here
    logCode = 1
    statusStr = "Message Here"
    return logEntry(logLev=logCode, msg=statusStr)

def callbackFunc(result):
    print(result.logLev)
    print(result.msg)

def userAsyncCall():
    pool = Pool()
    pool.apply_async(doSomething, [1,2], callback=callbackFunc)

if __name__ == "__main__":
    userAsyncCall() # Nothing is printed

    # If this is uncommented, the logLev and status are printed as expected:
    # y = logEntry(logLev=2, msg="Hello World")
    # callbackFunc(y)

有谁知道是否有办法namedtuple将异步进程的返回值传递给回调？对于我正在做的事情,是否有更好/更多的pythonic方法？

Answer 1

问题是 的返回值namedtuple()及其typename参数的情况不同。也就是说，命名元组的类定义与您提供的变量名称之间存在不匹配。您需要两者匹配：

LogEntry = namedtuple("LogEntry", ['logLev', 'msg'])

并相应地更新return语句doSomething()。

完整代码：

from multiprocessing import Pool
from collections import namedtuple

LogEntry = namedtuple("LogEntry", ['logLev', 'msg'])

def doSomething(x):
    # Do actual work here
    logCode = 1
    statusStr = "Message Here"
    return LogEntry(logLev=logCode, msg=statusStr)

def callbackFunc(result):
    print(result.logLev)
    print(result.msg)

def userAsyncCall():
    pool = Pool()
    return pool.apply_async(doSomething, [1], callback=callbackFunc)

if __name__ == "__main__":
    c = userAsyncCall()

    # To see whether there was an exception, you can attempt to get() the AsyncResult object.
    # print c.get()

（要查看类定义，请添加verbose=True到namedtuple().）