On Tue, 23 Dec 2008 10:20:59 -0500 Steve Holden <st...@holdenweb.com> wrote: > D'Arcy J.M. Cain wrote: > > Well, if all you want is a loop: > > > > for v in vars: > > locals()[v] = [] > > > Note that this isn't guaranteed to work. While locals() will return a > dict containing the names and values from the local namespace, you won't > affect the local namespace by assigning values to the appropriate keys: > > >>> def f(): > ... a = "hello" > ... locals()["a"] = "goodbye" > ... print a
This was my test: >>> locals()['x'] = "hello" >>> x 'hello' >>> locals()['x'] = "goodbye" >>> x 'goodbye' Just didn't want people to think that I post without testing. In any case, even if that worked as expected I am pretty sure that it is the wrong solution but without knowing more about what the OP is doing it is impossible to know what the right answer is. -- D'Arcy J.M. Cain <da...@druid.net> | Democracy is three wolves http://www.druid.net/darcy/ | and a sheep voting on +1 416 425 1212 (DoD#0082) (eNTP) | what's for dinner. -- http://mail.python.org/mailman/listinfo/python-list
