undo a dictionary

[Terry Reedy]

mmm wrote:
> I found code to undo a dictionary association.
> 
> def undict(dd, name_space=globals()):
>     for key, value in dd.items():
>         exec "%s = %s" % (key, repr(value)) in name_space

You are not undoing anything.  You are updating globals() from another 
dict.  But why repr(value)?  Without that, globals().update(dd) would 
work.  In 2.6?/3.0, replace 'dd' with '{a:b for a,b in dd.items()}

dd = { 'a':1, 'b': 'B'}
globals().update({a:b for a,b in dd.items()})
print(a,b)
# 1,B

>>>> dx= { 'a':1, 'b': 'B'}
>>>> undict(dx)
> 
> I get
>>>> print A, B
> 1 B
> 
> Here,  a=1 and b='B'

Don't fake interactive output.  You would have to "print a,b".  Above 
gives a NameError.

> This works well enough for simple tasks and I understand the role of
> globals() as the default names space, but creating local variables is
> a problem.

Within functions, yes. Just access the values in the dict.

 > Also having no output arguemtns to undict() seems
> counterintuitive.

In Python, this is standard for functions that mutate.

 > Also, the function fails if the key has spaces or
> operand characters (-,$,/,%).

Exec is tricky.  Most people hardly ever use it.

 > Finally I know I will have cases where
> not clearing (del(a,b)) each key-value pair might create problems in a
> loop.

You cannot mutate a dict while iterating through it.

> So I wonder if anyone has more elegant code to do the task that is
> basically the opposite of creating a dictionary from a set of
> globally assigned variables.

See above.

> And for that matter a way to create a
> dictionary from a set of variables (local or global).

You have to be more specific: there are {} displays and dict(args) call 
and other methods.  Read the manual.

tjr