Newbie Dictionary Question

[Gary Walker]

Thanks!! I'd put sample code but there's no need. You've more than answered
the question, and brilliantly too, I might add. I'm actually a Delphi/Kylix
programmer (that is to say, I get paid to do that), and I'm learning Python
to amuse myself. Your answer made total sense, in a way that the online
tutorials and O'Reilly books have not.

I really appreciate the nitpicking part, too! That was the detail that I was
missing. In fact, within two minutes of reading your message, I had my
little program working!!

Thanks again!

Gary


Quinn Dunkan wrote in message ...
>On Sat, 31 Mar 2001 14:57:19 GMT, Gary Walker <borealis3 at home.com> wrote:
>>I *think* I want to do the following:
>>
>>I want to create a list (this I'm able to do)
>>Then I want to dynamically create a dictionary for each item in the list,
>>add some keys (and values), and then put the dictionary in the list.
>>
>>(This part is frustrating me, because my newbie python code is referencing
>>the SAME dictionary for each of the list indexes, this NOT what I want. I
>>want a new instance of a dictionary for each pass thru the list as I loop)
>
>list = []
>for elt in thingsies:
>    list.append({'key': elt})
>
>>I want to do this because this way I can have an ordered list of key/value
>>pairs... at least that was my plan...
>
>Well, then a dictionary isn't going to help you, it's not ordered.  You
might
>want a list of pairs:
>
>a = [
>    ('foo', 'a'),
>    ('bar', 'q'),
>    ('baz', 'j'),
>    ('faz', 't'),
>]
>
>Or, if you mean 'sorted' when you say 'ordered', you could sort a
dictionary's
>keys and iterate over that:
>
>d = { ... }
>ks = d.keys()
>ks.sort()
>for k in ks:
>    print d[k]
>
>>In other words:
>>How do I dynamically create a dictionary whose name is not known until
>>runtime?
>
>Use braces:
>
>{'foo': bar}
>
>Nitpick: dictionaries don't have names.  No objects have names.
Dictionaries
>might be bound to one, two, three, a million, or zero names, but they
>themselves don't have names.  So "{'foo': bar}" creates a dictionary
without a
>name.  "d = {'foo': bar}" creates a dictionary without a name *and* a name
>in the current environment (global or local) that refers to that
dictionary.
>"def f(x): ..." creates a function without a name and a name in the current
>environment that refers to that function.  "a = b = {}" creates a function
>without a name, and two names in the current environment that refer to it.
>"a[0] = {}" creates a dictionary without a name and asks 'a' to store a
>reference to it in its 0th element.
>
>>Actually, they don't have to be ordered, so I suppose (thinking aloud
here)
>>that I could use a dictionary of dictionaries, but again - it seems to me
>>that I'd need to dynamically create a dictionary whose name isn't known
>>until runtime...
>>
>>Or am I missing the obvious? This is the most likely scenario :)
>
>You should post some example code.  I'm not sure I know what you're talking
>about.