Thank you all, this is exactly what I was trying to do and the syntax is
beautiful... :-)
--
Pablo Englebienne
"Progress is made by lazy men looking for easier ways to do things." - Robert
A. Heinlein
Kent Johnson wrote:
On Thu, Nov 13, 2008 at 10:01 AM, A.T.Hofkamp <[EMAIL PROTECTED]> wro
Kent Johnson wrote:
On Thu, Nov 13, 2008 at 10:01 AM, A.T.Hofkamp <[EMAIL PROTECTED]> wrote:
d3 = dict(( (rv, dict.fromkeys(c)) for rv in r ))
You don't need the double parentheses, this works just as well:
d3 = dict( (rv, dict.fromkeys(c)) for rv in r )
A generator expression just has to be
On Thu, Nov 13, 2008 at 10:01 AM, A.T.Hofkamp <[EMAIL PROTECTED]> wrote:
d3 = dict(( (rv, dict.fromkeys(c)) for rv in r ))
You don't need the double parentheses, this works just as well:
d3 = dict( (rv, dict.fromkeys(c)) for rv in r )
A generator expression just has to be in parentheses, it'
On Thu, Nov 13, 2008 at 8:11 AM, Pablo Englebienne
<[EMAIL PROTECTED]> wrote:
> Hi, I'm trying to work with a dictionary of dictionaries and I'm having
> trouble accessing a specific element of it:
>
> $ python
> Python 2.6 (trunk:66714:66715M, Oct 1 2008, 18:36:04)
> [GCC 4.0.1 (Apple Computer, I
Pablo Englebienne wrote:
Hi, I'm trying to work with a dictionary of dictionaries and I'm having
trouble accessing a specific element of it:
$ python
Python 2.6 (trunk:66714:66715M, Oct 1 2008, 18:36:04)
[GCC 4.0.1 (Apple Computer, Inc. build 5370)] on darwin
Type "help", "copyright", "credits
Hi, I'm trying to work with a dictionary of dictionaries and I'm
having trouble accessing a specific element of it:
$ python
Python 2.6 (trunk:66714:66715M, Oct 1 2008, 18:36:04)
[GCC 4.0.1 (Apple Computer, Inc. build 5370)] on darwin
Type "help", "copyright", "credits" or "license" for more i
