Python: Using defined arguments with *args

Question

Rewritten to make more clear the use-case and answer better Anentropic's question.

def use_all (step_todo, wait_complete=True, *args):
    execute_step1 (step-todo)
    handle_args (*args)
    if not wait_complete: 
       do_somehing ()
       return
    execute_stepN ()

@decorate
def use_from (step_todo, *args):
    use_all (step_todo, args)

@decorate
def use_many ():
    use_all (step_todo1, wait_complete=False)
    use_all (step_todo2, args2)
    use_all (step_todo3)

The use_all is the main "executive" to process the steps (exactlypxssh for installation). It shall not be decorated with start/stop comments as may be called several times from a procedure (e.g. step_many which is reboot - reason for no wait_complete), but single step shall be.

As the use-case is specific, I may see solution to handle the *args as _single named variable containing tuple, e.g.

def use_all (step_todo, wait_complete=True, args_list=()):

Is this correct (and recommended) solution?

This is somehow linked to questions python function *args and **kwargs with other specified keyword arguments or Using default arguments before positional arguments . Is it possible not to parse kwargs and keep Python R2.7?

Thanks Jan

Anentropic · Accepted Answer · 2014-11-19 16:03:11Z

0

You need to do this:

def use_from(step_todo, *args):
    use_all(step_todo, *args)

...otherwise you are calling use_all with a single arg containing a list of values, instead of calling it with the multiple args

also, don't put a space between your function and the parentheses, it's bad style :)

to get around the problem of wait_complete taking value of your first arg you need to pass it explicitly, eg:

def use_from(step_todo, *args):
    use_all(step_todo, True, *args)

edited Nov 19, 2014 at 16:03

answered Nov 19, 2014 at 15:56

Anentropic

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5 Comments

grasshopper Over a year ago

Good points, both of them, but this still doesn't fix the problem of wait_complete taking the first args value.

Anentropic Over a year ago

yes, because of this you will always have to explicitly pass a value for wait_complete

Rusty Over a year ago

wait_complete eats up the first arg, you could call it like this use_all(step_todo, True, *args), but that is just plain ugly, I think use_all()'s parameter list needs to be reworked.

grasshopper Over a year ago

yep, so essentially the answer to the OP is "it's not possible to do that in python 2.7"

Anentropic Over a year ago

I don't see the point of the use_all function, perhaps with a more complete example we could suggest a useful alternative

Raydel Miranda · Accepted Answer · 2014-11-19 16:04:35Z

0

That is not possible to do in Python 2.7, however in python 3.x you could:

def use_all (step_todo, *args, wait_complete=True):
    print('W_C: {0}'.format(wait_complete))
    print('Args: {0}'.format(args))

def use_from (step_todo, *args, **kwargs):
    use_all (step_todo, *args, **kwargs)

use_from ('Step1')
use_from ('Step3', 'aa:bb:cc')

Ouput:

>>> W_C: True
>>> Args: ()
>>> W_C: True
>>> Args: ('aa:bb:cc',)

edited Nov 19, 2014 at 16:04

answered Nov 19, 2014 at 15:57

Raydel Miranda

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