将List作为参数传递给另一个方法，但是该方法在Python中将该参数接受为Tuple吗？

Question

I am passing a List as an argument to another method. But the method accepting the parameter as Tuple.

class A():
   def method(self):
      lst = [{'CITY': 'Boston', 'STATE': 'MA'},
           {'CITY': 'New York', 'STATE': 'NY'}]
      self.method1(self,*lst)

   def method1(self,*param):
      print(type(param))
      print(param)

b =A()
b.method() # Printing type as Tuple

I want to get the List not the Tuple. Any suggestion!

Answer 1

When you do *lst , you're unpacking the contents of lst as arguments to the function method1 . The function method1 is actually taking multiple args because you've done *param - it takes as many parameters as you pass it.

When you unpack lst and pass it to method1 , it takes it in as a tuple by definition.

Observe:

>>> L = ['hi', 'hello']
>>> def A(*args):
...     print(args)
... 
>>> A(*L)
('hi', 'hello')

---

To fix your issue, remove both your asterisks and take in a list object - don't unpack your list.