How to pass list to Python worker thread?

Question

How to pass a list to Python threading.Thread? In the beflow example, I am converting the list to a dictionary. Is there a way to pass list as such to doWork method?

def launcher():
    #.... 
   list = ['name1', 'name2', 'name3','name4']
   nameList = { 'names' : list }
   workThread = threading.Thread(target=self.doWork, kwargs=nameList)
   workThread.start()

def doWork(self, **kwargs):
    for i, name in enumerate(kwargs['nameList']):
        print i, name

Answer 1

The reason being that workThread = threading.Thread(target=doWork, args=lst) takes the second argument as a list and then maps the contents of the lst to the various arguments required by the method or function, Now since you want to pass only a single argument to the function which is a list, So if you see the common Syntax it is written some thing like: args = (param1, param2,...)

Now since you have only 1 argument so this reduces to args = (param1) Now replacing the param1 with a list you will get args = ([1, 2, 3...]), So actually the point here is that you need a nested list to pass as args because the arguments are enclosed inside the list or a tuple so you need to wrap your list inside a list or tuple. Any content under that wrapper list would be mapped with the arguments of the function or method being called.

To make the scenario clear:

If you pass args = [1, 2, 3, 4] then you mean that you are passing these 4 values as the arguments to that function or method.

On the other side if you pass args = [[1, 2, 3, 4]] or args = ([1, 2, 3, 4], ) or args = ((1, 2, 3, 4), ) Then only the whole list or tuple would be passed as a parameter to the function or method called.

I have simplified your code snippet to look like :

import threading

lst = ['name1', 'name2', 'name3','name4', "a"]

def doWork(param):
    for i, name in enumerate(param):
        print i, name

workThread = threading.Thread(target=doWork, args=(lst,))
workThread.start()