Hey Kirby!! This is the best explanation I've seen of Python's argument passing subtleties. If you don't mind I would like to post it on our PyKata website in a section "Tips from the Masters", with you as author, of course. We could also put a link to a page on your website, if you prefer.

-- Dave

************************************************************     *
* David MacQuigg, PhD    email: macquigg at ece.arizona.edu   *  *
* Research Associate                phone: USA 520-721-4583   *  *  *
* ECE Department, University of Arizona                       *  *  *
*                                 9320 East Mikelyn Lane       * * *
* http://purl.net/macquigg        Tucson, Arizona 85710          *
************************************************************     *



kirby urner wrote:
...

Even functions with purely positional
parameters may be passed a dict, in
which case the positional parameters
will be treated like named ones, i.e.
the dict will map to them.

I'll use the new formating protocol:

from __future__ import print_function

def f(a, b, c):
        print("a={0}, b={1}, c={2}".format(a,b,c))

        
f(1,2,3)
a=1, b=2, c=3

Here's what I'm illustrating:

thedict = dict(c=5,a=10,b='cat')

f(**thedict)
a=10, b=cat, c=5

Now what if we try to sneak in an argument
that doesn't map:

thedict = dict(c=5,a=10,b='cat', d='dog')
f(**thedict)

Traceback (most recent call last):
  File "<pyshell#11>", line 1, in <module>
    f(**thedict)
TypeError: f() got an unexpected keyword argument 'd'

The d is caught, an exception is raised.

So here we might add the all purpose "pooper scooper"
(yeah, scatological -- sometimes effective pedagogy):

def f(a, b, c, **kwargs):
        print("a={0}, b={1}, c={2}".format(a,b,c))

The d now gets through, though nothing echoes:
        
f(**thedict)
a=10, b=cat, c=5

Lets make sure we see what's in "overflow":

def f(a, b, c, **kwargs):
        print("a={0}, b={1}, c={2}".format(a,b,c))
        print(kwargs)

        
f(**thedict)
a=10, b=cat, c=5
{'d': 'dog'}

Is an all purpose pooper scooper allowed to have
defaults?  No, default arguments are what turn
parameters into named parameters.  *args and
**kwargs (often so named by convention) are
not named parameters so much as "collectors"
of "overflow" arguments (made-up terminology).

def f(a, b, c, d='dog', *e):
        print("a={0}, b={1}, c={2} d={3} e={4}".format(a,b,c,d,e))

        
f(**thedict)
a=10, b=cat, c=5 d=dog e=()
Speaking of made-up terminology, when I pass
**thedict as an argument, I tend to think of the
** as "explode".

If you don't double-star it, it goes through as just
the one argument and an exception gets raised:

f(thedict)

Traceback (most recent call last):
  File "<pyshell#31>", line 1, in <module>
    f(thedict)
TypeError: f() takes at least 3 arguments (1 given)

The same "explode" operator works for tuples in
that it "unpacks" the one argument into len(tuple)
arguments.  The typical use is to make (x,y,z)-tuples
map to some function that expects x, y and z as
separate arguments.

The error messages remind us when we fail to
"explode" one argument into many:

def f(x, y, z):
        return x**2 + y**2 + z**2

f(1,2,3)
14
coords = (1,2,3)
f(coords)

Traceback (most recent call last):
  File "<pyshell#38>", line 1, in <module>
    f(coords)
TypeError: f() takes exactly 3 arguments (1 given)
f(*coords)
14
f(dict(y=2,x=1,z=3))

Traceback (most recent call last):
  File "<pyshell#40>", line 1, in <module>
    f(dict(y=2,x=1,z=3))
TypeError: f() takes exactly 3 arguments (1 given)
f(**dict(y=2,x=1,z=3))
14

Kirby
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