On Sat, 2005-12-31 at 09:33 -0500, Kent Johnson wrote:
> Could be
>self.results[key] = [0*24]
[0]*24
Excellent points and advice, just noting a typo.
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Lloyd Kvam
Venix Corp
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Paul Kraus wrote:
> So now for all my reports all I have to write are little 5 or 6 line scripts
> that take a text file split the fields and format them before basing them off
> into my custom object. Very slick and this is my first python project. Its
> cluttered and messy but for about 1 hour
[snip]
Paul Kraus wrote:
> Now I have to find a way to take the output at the end and pipe it
> out to an external Perl program that creates an excel spreadsheet (
> no real clean easy way to do this in python but hey each tool has its
> usefullness). I wish I could hide this in the object though
> That is the approach Paul took originally (see the other fork of this
> thread). He is accumulating a sparse 3d matrix where the keys are year,
> field6 and month. (He hasn't said what field6 represents.) The problem
> is that he wants to print out counts corresponding to all the existing
> year
Danny Yoo wrote:
> This being said, what data are you modeling? It almost sounds like you're
> implementing some kind of 3d matrix, even a sparse one. More information
> about the data you're modelling might lead to a different representation
>
> For example, would using a dictionary whose keys
> ok so assuming I had a dictionary with 1key that contained a list like
> so... dictionary[key1][0]
>
> How would I increment it or assign it if it didn't exist. I assumed like
> this. dict[key1][0] = dictionary.get(key1[0],0) + X
Hi Paul,
Given a dictionary d and some arbitrary key k, let's ass
On Wednesday 28 December 2005 11:30 am, Kent Johnson wrote:
> Python lists don't create new elements on assignment (I think Perl lists
> do this?) so for example
> dictionary[(key1, key2)][10] = X
ok so assuming I had a dictionary with 1key that contained a list like so...
dictionary[key1][0]
How
Paul Kraus wrote:
> Here is the code that I used. Its functional and it works but there has got
> to
> be some better ways to do a lot of this. Transversing the data structure
> still seems like I have to be doing it the hard way.
>
> The input data file has fixed width fields that are delimite
Paul Kraus wrote:
> I am trying to build a data structure that would be a dictionary of a
> dictionary of a list.
>
> In Perl I would build the structure like so $dictionary{key1}{key2}[0] = X
> I would iterate like so ...
> foreach my $key1 ( sort keys %dictionary ) {
> foreach my $key2 (
On Wednesday 28 December 2005 10:18 am, Paul Kraus wrote:
> I am trying to build a data structure that would be a dictionary of a
> dictionary of a list.
>
> In Perl I would build the structure like so $dictionary{key1}{key2}[0] = X
> I would iterate like so ...
> foreach my $key1 ( sort keys %dict
I am trying to build a data structure that would be a dictionary of a
dictionary of a list.
In Perl I would build the structure like so $dictionary{key1}{key2}[0] = X
I would iterate like so ...
foreach my $key1 ( sort keys %dictionary ) {
foreach my $key2 ( sort keys %{$dictionary{$key1}
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