将字典传递给Python中的线程函数

Question

所以,我有一个函数explain(item),它接受1个参数.该参数旨在成为具有8-9个键的字典.我打电话的时候explain(item)一切都很好.但是当我打电话时(物品是同一个变量)

threads.append(threading.Thread(target = explain, args=(item)))
threads[i].start()
threads[i].join()

我得到像这样的错误:

Exception in thread Thread-72:
Traceback (most recent call last):
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/threading.py", line 810, in __bootstrap_inner
    self.run()
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/threading.py", line 763, in run
    self.__target(*self.__args, **self.__kwargs)
TypeError: explain() takes exactly 1 argument (10 given)

我究竟做错了什么？

Answer 1

看起来你打算传递一个项目元组作为args参数,threading.Thread()但使用args=(item)相当于args=item.你需要添加一个逗号来创建一个元组,所以它将是args=(item,):

threads.append(threading.Thread(target = explain, args=(item,))).

如果没有尾随逗号,则括号只是一种分组表达式的方法,例如:

>>> (100)  # this is just the value 100
100
>>> (100,) # this is a one-element tuple that contains 100
(100,)