Passing every element of a list as argument to a function

[Prasad, Ramit]

-----Original Message-----
From: python-list-bounces+ramit.prasad=jpmorgan.com at python.org [mailto:python-list-bounces+ramit.prasad=jpmorgan.com at python.org] On Behalf Of Antonio Vera
Sent: Tuesday, August 09, 2011 12:02 PM
To: python-list at python.org
Subject: Passing every element of a list as argument to a function

Hi!,
I have a very simple syntax question. I want to evaluate a library
function f receiving an arbitrary number of arguments (like
itertools.product), on the elements of a list l. This means that I
want to compute f(l[0],l[1],...,l[len(l)-1]).

Is there any operation "op" such that f(op(l)) will give the sequence
of elements of l as arguments to f?

Thanks for your time.
Best,
Antonio
-- 
http://mail.python.org/mailman/listinfo/python-list

op(*l) for a list (or positional arguments).

If you are trying to pass named keyword arguments then you must pass it a dictionary { 'keywordName' : 'value' }
Example:
>>>def F(name=None):
      pass
>>>F(**{'name':'boo'})


Ramit


Ramit Prasad | JPMorgan Chase Investment Bank | Currencies Technology
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