我有一个使用Python 3.4.2的非常简单的线程示例。在此示例中，我创建了五个线程，这些线程仅返回字符串“ Result”并将其附加到名为线程的数组中。在另一个for循环中，迭代五次将线程连接到项x。我正在尝试打印结果x，它应该产生一个看起来像['Resut'，'Result'，'Result'，'Result'，'Result']的列表，但print命令只产生线程的标题和它是封闭的事实。我显然误会了如何在python中使用线程。如果有人可以提供如何充分完成此测试用例的示例，我将不胜感激。
import threading
def Thread_Test():
return ("Result")
number = 5
threads = []
for i in range(number):
Result = threading.Thread(target=Thread_Test)
threads.append(Result)
Result.start()
for x in threads:
x.join()
print (x)
请找到以下有关队列和线程的简单示例，
import threading
import Queue
import timeit
q = Queue.Queue()
number = 5
t1 = timeit.default_timer()
# Step1: For example, we are running multiple functions normally
result = []
def fun(x):
result.append(x)
return x
for i in range(number):
fun(i)
print result ," # normal result"
print (timeit.default_timer() - t1)
t2 = timeit.default_timer()
#Step2: by using threads and queue
def fun_thrd(x,q):
q.put(x)
return
for i in range(number):
t1 = threading.Thread(target = fun_thrd, args=(i,q))
t1.start()
t1.join()
thrd_result = []
while True:
if not q.empty():
thrd_result.append(q.get())
else:
break
print thrd_result , "# result with threads involved"
print (timeit.default_timer() - t2)
t3 = timeit.default_timer()
#step :3 if you want thread to be run without depending on the previous thread
threads = []
def fun_thrd_independent(x,q):
q.put(x)
return
def thread_indep(number):
for i in range(number):
t = threading.Thread(target = fun_thrd_independent, args=(i,q))
t.start()
threads.append(t)
thread_indep(5)
for j in threads:
j.join()
thread_indep_result = []
while True:
if not q.empty():
thread_indep_result.append(q.get())
else:
break
print thread_indep_result # result when threads are independent on each other
print (timeit.default_timer() - t3)
[0, 1, 2, 3, 4] # normal result
3.50475311279e-05
[0, 1, 2, 3, 4] # result with threads involved
0.000977039337158
[0, 1, 2, 3, 4] result when threads are independent on each other
0.000933170318604
根据数据规模的不同会有很大的不同
希望这会有所帮助，谢谢
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我来说两句