On Aug 15, 11:45 am, [EMAIL PROTECTED] wrote:
> by the way, iterating over bar will throw KeyError if that key does not exist
> in foo. to see that in action, simply set another key in bar after
> copy.deepcopy stmt in this example..
> bar['xtra'] = 0
> and re-run
My first posting in this th
ari, Edwin
Sent: Thursday, August 14, 2008 9:24 PM
To: 'Brandon'; python-list@python.org
Subject: RE: updating dictionaries from/to dictionaries
if the values for any of the keys are None, both do not work as can be seen
below!!.
since both loops are iterating over keys(), 'get&#
On Aug 15, 11:23 am, [EMAIL PROTECTED] wrote:
> if the values for any of the keys are None, both do not work as can be seen
> below!!.
>
If any of the values are None, nothing can work. The OP justifiably
(given his problem domain (bigram frequencies)) expects that all
values will be integers. If
x27;NoneType'",))
hope that helps to clarify both point of views.
thanks Edwin
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED]
On Behalf Of Brandon
Sent: Thursday, August 14, 2008 8:35 PM
To: python-list@python.org
Subject: Re: updating dictionaries from/to
> (1) iterating over foo:
> for key in foo:
>foo[key] += bar.get(key, 0)
>
> (2) iterating over bar:
> for key in bar:
>foo[key] += bar[key]
>
> I (again) challenge you to say *why* you feel that the "iterating over
> bar" solution will not work.
Well if you're going to be clever enough
On Aug 13, 5:33 am, Brandon <[EMAIL PROTECTED]> wrote:
> On Aug 12, 7:26 am, John Machin <[EMAIL PROTECTED]> wrote:
>
>
>
> > On Aug 12, 12:26 pm, Brandon <[EMAIL PROTECTED]> wrote:
>
> > > You are very correct about the Laplace adjustment. However, a more
> > > precise statement of my overall pro
On Aug 12, 7:26 am, John Machin <[EMAIL PROTECTED]> wrote:
> On Aug 12, 12:26 pm, Brandon <[EMAIL PROTECTED]> wrote:
>
>
>
> > You are very correct about the Laplace adjustment. However, a more
> > precise statement of my overall problem would involve training and
> > testing which utilizes bigram
On Aug 12, 12:26 pm, Brandon <[EMAIL PROTECTED]> wrote:
>
> You are very correct about the Laplace adjustment. However, a more
> precise statement of my overall problem would involve training and
> testing which utilizes bigram probabilities derived in part from the
> Laplace adjustment; as I und
John:
> "append"? Don't you mean "add"???
Yes, that is what I meant, my apologies.
> What you need to do is practice translating from your
> requirements into Python, and it's not all that hard:
>
> "run a loop through foo" -> for key in foo:
> "match any of its keys that also exist in bar" -> i
On Aug 12, 9:14 am, Brandon <[EMAIL PROTECTED]> wrote:
> I wasn't sure about the update method either, since AFAICT (not far)
> the values would in fact update, not append as I needed them to.
"append"? Don't you mean "add"???
> But
> the iteritems and get combo definitely worked for me.
Under s
I wasn't sure about the update method either, since AFAICT (not far)
the values would in fact update, not append as I needed them to. But
the iteritems and get combo definitely worked for me.
Thank you for the suggested link. I'm familiar with that page, but my
skill level isn't so far along yet
On Aug 12, 2:52 am, Steven D'Aprano <[EMAIL PROTECTED]
cybersource.com.au> wrote:
> On Mon, 11 Aug 2008 00:27:46 -0700, Brandon wrote:
> > This should be pretty simple: I have two dictionaries, foo and bar. I
> > am certain that all keys in bar belong to foo as well, but I also know
> > that not a
"Harder to say what you want to do than to just do it."
The truly terrible thing is when you know that's the case even as
you're saying it. Thanks for the help, all!
--
http://mail.python.org/mailman/listinfo/python-list
On Mon, 11 Aug 2008 00:27:46 -0700, Brandon wrote:
> This should be pretty simple: I have two dictionaries, foo and bar. I
> am certain that all keys in bar belong to foo as well, but I also know
> that not all keys in foo exist in bar. All the keys in both foo and bar
> are tuples (in the bigra
On Aug 11, 6:24 pm, "Calvin Spealman" <[EMAIL PROTECTED]> wrote:
> for k in foo:
> foo[k] += bar.get(k, 0)
An alternative:
for k in bar:
foo[k] += bar[k]
The OP asserts that foo keys are a superset of bar keys. If that
assertion is not true (i.e. there are keys in bar that are not in foo,
y
for k in foo:
foo[k] += bar.get(k, 0)
On Mon, Aug 11, 2008 at 3:27 AM, Brandon <[EMAIL PROTECTED]> wrote:
> Hi all,
>
> I am not altogether experienced in Python, but I haven't been able to
> find a good example of the syntax that I'm looking for in any tutorial
> that I've seen. Hope somebody
Hi all,
I am not altogether experienced in Python, but I haven't been able to
find a good example of the syntax that I'm looking for in any tutorial
that I've seen. Hope somebody can point me in the right direction.
This should be pretty simple: I have two dictionaries, foo and bar.
I am certai
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