CODEWITHSUNDEEP
pythonarrayspython-2.7args
Daniel Watson
Daniel Watson

Reputation: 473

Python converting *args to list

This is what I'm looking for:

def __init__(self, *args):
  list_of_args = #magic
  Parent.__init__(self, list_of_args)

I need to pass *args to a single array, so that:

MyClass.__init__(a, b, c) == Parent.__init__([a, b, c])

Upvotes: 27

Views: 44402

Answers (4)

tobeistodo123
tobeistodo123

Reputation: 11

Try this:

listname = list(args)

Upvotes: 1

arunkumaraqm
arunkumaraqm

Reputation: 367

If you're looking for something in the same direction as @simon 's solution, then:

def test_args(*args): lists = [*args] print(lists) test_args([7],'eight',[[9]])

Result:

[[7], 'eight', [[9]]]

Upvotes: 0

Simon
Simon

Reputation: 19938

There is this piece of code that I picked up in sentdex tutorials that deals with this:

https://www.youtube.com/watch?v=zPp80YM2v7k&index=11&list=PLQVvvaa0QuDcOdF96TBtRtuQksErCEBYZ

Try this:

def test_args(*args):
    lists = [item for item in args]
    print lists

test_args('Sun','Rain','Storm','Wind')

Result:

['Sun', 'Rain', 'Storm', 'Wind']

Upvotes: 11

Andrew Clark
Andrew Clark

Reputation: 208475

Nothing too magic:

def __init__(self, *args):
  Parent.__init__(self, list(args))

Inside of __init__, the variable args is just a tuple with any arguments that were passed in. In fact you can probably just use Parent.__init__(self, args) unless you really need it to be a list.

As a side note, using super() is preferable to Parent.__init__().

Upvotes: 34

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