附加以列出线程的结果

Question

这是我的代码：

import threading

def worker(num):
    """thread worker function"""
    print '{Worker: %s}' % num
    return

threads = []
for i in range(5):
    t = threading.Thread(target=worker, args=(i,))
    threads.append(t)
    t.start()
print(threads)

这是上面代码的结果：

{Worker: 0}
{Worker: 1}
{Worker: 2}
{Worker: 3}
{Worker: 4}
[<Thread(Thread-1, stopped 17204)>, <Thread(Thread-2, stopped 852)>, <Thread(Thread-3, stopped 17092)>, <Thread(Thread-4, stopped 10632)>, <Thread(Thread-5, stopped 2396)>]

我如何将这些结果附加到列表中并创建一个类似这样的列表：

[[{Worker: 0}],[{Worker: 1}],[{Worker: 2}],[{Worker: 3}],[{Worker: 4}]]

edited: 实际上我的 def 包含一个请求 REST 方法并且它消耗了一些秒并且经过一些尝试我发现线程不等待获取结果而不是填充数组列表，然后我将我的代码改造成一个字符串变量来为你模拟，在这种情况下，我在 def 的第一行放了一个 time.sleep(1) 然后是运行线程，它有一个空列表....我将睡眠时间减少到 0.001，直到得到结果并且线程工作和数组已填充。

import threading

def worker(txt):
    """thread worker function"""
    if txt != '[{}]':
        time.sleep(0.01) # this time has empty result till replace with 0.001 or lesser
        lst.append(['{Worker: %s}' % txt])
        # time.sleep(5)
        return


xx = ['abc','def','ghi','jkl','mno','pqr','stw','xyz']

threads, lst = [], []
for i in xx:
    t = threading.Thread(target=worker, args=(i,))
    threads.append(t)
    t.start()
    # t.join()
print(lst) #[['{Worker: abc}'], ['{Worker: jkl}'], ['{Worker: ghi}'], ['{Worker: def}']] for 0.001 sleep time
# print(threads)

Answer 1

你的意思是这样的吗？

import threading

def worker(num):
    """thread worker function"""
    lst.append(['{Worker: %s}' % num])
    return

threads, lst = [], []
for i in range(5):
    t = threading.Thread(target=worker, args=(i,))
    threads.append(t)
    t.start()
print(lst)
#print(threads)

# [['{Worker: 0}'], ['{Worker: 1}'], ['{Worker: 2}'], ['{Worker: 3}'], ['{Worker: 4}']]

附加以列出线程的结果

append to list the result of a thread

python

python-multithreading