[Python-ideas] Method chaining notation

[Masklinn]

On 2014-02-26, at 16:28 , Ron Adam <ron3200 at gmail.com> wrote:

> 
> 
> On 02/26/2014 01:19 AM, Stephen J. Turnbull wrote:
>> Ron Adam writes:
>> 
>>  > >   >     def names(defaults, pos_names, pos_args, kwds):
>>  > >   >         return  {}.=update(defaults) \
>>  > >   >                   .=update(zip(pos_names, pos_args) \
>>  > >   >                   .=update(kwds)
>> 
>>  > >      def names(defaults, pos_names, pos_args, kwds):
>>  > >          for dct in pos_names, pos_args, kwds:
>>  > >              defaults.update(dct)
>>  > >          return defaults
>>  >
>>  > Not quite the same but close.  I just tried to come up with a more
>>  > realistic example without having to look up a lot code.  How does
>>  > pos_args in your example get paired with names?
>> 
>> Sorry, I knew that before dinner but forgot after dinner.  Same way as
>> in yours:
>> 
>>     def names(defaults, pos_names, pos_args, kwds):
>>         for dct in zip(pos_names, pos_args), kwds:
>>             defaults.update(dct)
>>         return defaults
> 
> Yes, ok.
> 
>> If that doesn't work in my version (I've never used zip that way), how
>> does it work in yours?  BTW, I'd actually be more likely to write that
>> now as
>> 
>>     def names(defaults, *updates):
>>         for update in updates:
>>             defaults.update(update)
>>         return defaults
>> 
>> and call it with "names(zip(pos_names, pos_args), kwds)".
> 
> The main difference between this and the one I posted is in this, defaults is mutated in your version.  I'd prefer it not be.
> 
> Dictionaries are pretty flexible on how they are initiated, so it's surprising we can't do this...
> 
>        D = dict(keys=names, values=args)

You can. It may not do what you want, but you definitely can do this:

    >>> dict(keys=names, values=args)
    {'keys': ['a', 'b', 'c', 'd'], 'values': [0, 1, 2]}

Although you're probably looking for:

    >>> dict(zip(names, args))
    {'a': 0, 'c': 2, 'b': 1}

and if you want to do a fill because you don't have enough args:

    >>> dict(izip_longest(names, args, fillvalue=None))
    {'a': 0, 'c': 2, 'b': 1, 'd': None}

(itertools is like friendship, it's bloody magic)

> The .fromkeys() method is almost that, but sets all the values to a single value.  I think I would have written that a bit different.
> 
>       def fromkeys(self, keys, values=None, default=None):
>             D = {}
>             if D is not None:
>                 D.update(zip(keys, values)]
>             for k in keys[len(vaues):]:
>                 D[k] = default
>             return D
> 
> And probably named it withkeys instead of fromkeys.  <shrug>  It's what I expected fromkeys to do.
> 
> cheers,
>   Ron
>