On Oct 2, 4:18*am, Terrence Brannon <metap...@gmail.comwrote:
Quote:
Ok, here is some code:
>
def calc_profit(std_clicks, vip_clicks, ad_rate=200,
upline_status=None):
* * payout = {}
* * payout_std = std_clicks * rates['std'].per_click
* * payout_vip = vip_clicks * rates['vip'].per_click
>
... now note that std_clicks and vip_clicks are passed to the
function.
>
Now, I improved this function this way:
>
def calc_profit(std_clicks, vip_clicks, ad_rate=200,
upline_status=None):
* * clicks = {}
* * clicks['std'] = std_clicks
* * clicks['vip'] = vip_clicks
>
* * payout = {}
* * for member_type in rates:
* * * * payout[member_type] = clicks[member_type] *
rates[member_type].per_click
>
But it seems wasteful to have to re-bind the passed-in function args
to a dictionary in the function. I think there must be some way to
improve this code and get the dictionary built without me manually
doing it...
>
I know there is something like *args, or **args, but since
docs.python.org is down, I cant check.
*args is for variable-length parameters, **args is for keyword
parameters.
Quote:
Quote:
Quote:
>>def f( **kwar ):
.... print kwar
....
{'a': 2, 'b': 3}
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: f() takes exactly 0 arguments (1 given)
{'a': 'abc'}
