Dictionary assignment
Christopher Boomer (don't dash!)
c-b-o-o-m-e-r at tiscali.co.uk
Mon Aug 18 11:53:11 EDT 2003
"Mark Daley" <mark at diversiform.com> wrote in message
news:mailman.1060979551.16837.python-list at python.org...
Start by making your code a bit more robust.
for key in current.keys():
lookup=self.formats.get()
if not format.has_key(lookup):
format[lookup]={}
print "No such lookup:",lookup # a debugging warning
format[self.formats.get()][key] = current[key]
You can work from there, depending on what outcome you want in the event of
an error...
You also have to be sure that the result of get() is a valid key (string,
tuple) not, say, a list, which is why you should avoid statements like
mylist=tuple(temp). There's a list() like the tuple() you use, so to save
more headahes, avoid list= altogether
And don't mind him. We don't solve problems, but we do tell you how to. ;-)
Christopher Boomer.
More information about the Python-list
mailing list