if does not evaluate

[Robert Brewer]

Jim Newton wrote:
> A question that has bothered me ever since looking at python
> the first time is that not everything evaluates to something.

Are you only bothered by it because there's no ifelse operator? ;)

> I am wondering what is the reason for this.  Would making
> everying evaluate have caused the langugage to be less
> efficient execution-wise?  or was it a choice to enforce some
> sort of standard?

Only Zen standards:

1. Beautiful is better than ugly.
2. Explicit is better than implicit.
3. Simple is better than complex.
6. Sparse is better than dense.
7. Readability counts.

> I've read a few discussions about the fact that there is
> no if/then/else operartor.
> 
> Wouldn't this problem be easily solved by making the built
> in keywords actually be evaluatable.
> 
> I.e.,  x = if something:
>                expr1
>             else:
>                expr2
> 
> parentheses would of course be optional as they are for
> all expressions.

You'll first have to convince many people that this is a "problem" to be
"solved". Why is your solution "better than":

if something:
    x = expr1
else:
    x = expr2

Then you need to show why the alternatives aren't good enough:

If your main desire is to code in some other language while still using
Python, write your own VB-style IIf function:

>>> def IIf(expression, truepart, falsepart):
... 	if expression:
... 		return truepart
... 	return falsepart
... 
>>> x = IIf('something', 'expr1', 'expr2')
>>> x
'expr1'

If your main desire is to provide a one-liner, use a tuple (which is
still ugly):

>>> x = ('expr2', 'expr1')[bool('something')]
>>> x
'expr1'

Or a mapping (which is prettier, but still uglier than the regular if:
else:

>>> x = {True: 'expr1',
...      False: 'expr2'}[bool('something')]
>>> x
'expr1'

Or a more generic morphing function:

>>> def surjection(input, mappings):
... 	mappings = dict(mappings)
... 	return mappings[input]
... 
>>> x = surjection(bool('something'), {True: 'expr1', False: 'expr2'})
>>> x
'expr1'
>>> x = surjection(bool('something'), [(True, 'expr1'), (False,
'expr2')])
>>> x
'expr1'
>>> x = surjection('something', [('something', 'expr1'), ('other',
'expr2'), ('than', 'expr2')])
>>> x
'expr1'



Robert Brewer
MIS
Amor Ministries
fumanchu at amor.org