下面的函数foo返回一个字符串'foo'。我如何才能获得从线程的目标返回的值'foo' ?
from threading import Thread
def foo(bar):
print('hello {}'.format(bar))
return 'foo'
thread = Thread(target=foo, args=('world!',))
thread.start()
return_value = thread.join()
上面所示的“一种明显的方法”不起作用:thread.join()返回None。
当前回答
FWIW，多处理模块使用Pool类提供了一个很好的接口。如果您希望坚持使用线程而不是进程，可以直接使用multiprocessing.pool.ThreadPool类作为替代。
def foo(bar, baz):
print 'hello {0}'.format(bar)
return 'foo' + baz
from multiprocessing.pool import ThreadPool
pool = ThreadPool(processes=1)
async_result = pool.apply_async(foo, ('world', 'foo')) # tuple of args for foo
# do some other stuff in the main process
return_val = async_result.get() # get the return value from your function.
其他回答
考虑到@iman对@JakeBiesinger回答的评论，我重新组合了它，使其具有不同数量的线程:
from multiprocessing.pool import ThreadPool
def foo(bar, baz):
print 'hello {0}'.format(bar)
return 'foo' + baz
numOfThreads = 3
results = []
pool = ThreadPool(numOfThreads)
for i in range(0, numOfThreads):
results.append(pool.apply_async(foo, ('world', 'foo'))) # tuple of args for foo)
# do some other stuff in the main process
# ...
# ...
results = [r.get() for r in results]
print results
pool.close()
pool.join()
FWIW，多处理模块使用Pool类提供了一个很好的接口。如果您希望坚持使用线程而不是进程，可以直接使用multiprocessing.pool.ThreadPool类作为替代。
def foo(bar, baz):
print 'hello {0}'.format(bar)
return 'foo' + baz
from multiprocessing.pool import ThreadPool
pool = ThreadPool(processes=1)
async_result = pool.apply_async(foo, ('world', 'foo')) # tuple of args for foo
# do some other stuff in the main process
return_val = async_result.get() # get the return value from your function.
GuySoft的想法很棒，但我认为对象不一定要从Thread继承，start()可以从接口中删除:
from threading import Thread
import queue
class ThreadWithReturnValue(object):
def __init__(self, target=None, args=(), **kwargs):
self._que = queue.Queue()
self._t = Thread(target=lambda q,arg1,kwargs1: q.put(target(*arg1, **kwargs1)) ,
args=(self._que, args, kwargs), )
self._t.start()
def join(self):
self._t.join()
return self._que.get()
def foo(bar):
print('hello {0}'.format(bar))
return "foo"
twrv = ThreadWithReturnValue(target=foo, args=('world!',))
print(twrv.join()) # prints foo
我见过的一种方法是将一个可变对象(如列表或字典)传递给线程的构造函数，同时传递一个索引或其他某种类型的标识符。然后线程可以将结果存储在该对象的专用槽中。例如:
def foo(bar, result, index):
print 'hello {0}'.format(bar)
result[index] = "foo"
from threading import Thread
threads = [None] * 10
results = [None] * 10
for i in range(len(threads)):
threads[i] = Thread(target=foo, args=('world!', results, i))
threads[i].start()
# do some other stuff
for i in range(len(threads)):
threads[i].join()
print " ".join(results) # what sound does a metasyntactic locomotive make?
如果你真的想要join()返回被调用函数的返回值，你可以用Thread子类来实现，如下所示:
from threading import Thread
def foo(bar):
print 'hello {0}'.format(bar)
return "foo"
class ThreadWithReturnValue(Thread):
def __init__(self, group=None, target=None, name=None,
args=(), kwargs={}, Verbose=None):
Thread.__init__(self, group, target, name, args, kwargs, Verbose)
self._return = None
def run(self):
if self._Thread__target is not None:
self._return = self._Thread__target(*self._Thread__args,
**self._Thread__kwargs)
def join(self):
Thread.join(self)
return self._return
twrv = ThreadWithReturnValue(target=foo, args=('world!',))
twrv.start()
print twrv.join() # prints foo
这有点麻烦，因为一些名称混乱，它访问特定于线程实现的“私有”数据结构……但它确实有效。
对于Python 3:
class ThreadWithReturnValue(Thread):
def __init__(self, group=None, target=None, name=None,
args=(), kwargs={}, Verbose=None):
Thread.__init__(self, group, target, name, args, kwargs)
self._return = None
def run(self):
if self._target is not None:
self._return = self._target(*self._args,
**self._kwargs)
def join(self, *args):
Thread.join(self, *args)
return self._return
您可以在线程函数的作用域之上定义一个可变变量，并将结果添加到该变量中。(我还修改了代码，使其与python3兼容)
returns = {}
def foo(bar):
print('hello {0}'.format(bar))
returns[bar] = 'foo'
from threading import Thread
t = Thread(target=foo, args=('world!',))
t.start()
t.join()
print(returns)
返回{'world!”:“foo”}
如果使用函数input作为结果字典的键，则保证每个惟一的输入都在结果中给出一个条目
