Kent,
On Wed, 14 May 2008, Kent Johnson wrote:
I understand about removing elements from a container you're iterating. Is
data.remove(x) problematic in this context?
Yes. It can cause the iteration to skip elements ofthe list. Better to
post-process the list with a list comprehension:
evts = [
On Wed, May 14, 2008 at 9:06 PM, Kent Johnson <[EMAIL PROTECTED]> wrote:
>> I understand about removing elements from a container you're iterating. Is
>> data.remove(x) problematic in this context?
>
> Yes. It can cause the iteration to skip elements of the list.
For example:
In [1]: l=range(5)
I
On Wed, May 14, 2008 at 6:03 PM, Jon Crump <[EMAIL PROTECTED]> wrote:
> def events(data):
> evts = []
> for x in data:
>for y in data:
> if (x['placename'] == y['placename']) and (x['end'].month + 1 ==
> y['start'].month) and (y['start'] - x['end'] == datetime.timedelta(1)):
>x['
Bob, and Kent, Many thanks!
Sounds like the key 'processed' is created by the assignment x['processed'] =
True. So those dictionaries that have not experienced this assignment have no
such key. You should instead use: if 'processed' in x:
Doh! Now that WAS obvious
Try lst.remove(x)
Now t
On Wed, May 14, 2008 at 3:02 PM, Jon Crump <[EMAIL PROTECTED]> wrote:
> Something basic about lists and loops that I'm not getting here.
> I have a function that searches through them to find pairs of dictionaries
> that satisfy certain criteria. When the nested loops find such a pair, I
> need to
Jon Crump wrote:
Something basic about lists and loops that I'm not getting here. I've
got a long list of dictionaries that looks like this:
lst = [{'placename': u'Stow, Lincolnshire', 'long-name': u'Stow,
Lincolnshire.', 'end': datetime.date(1216, 9, 28), 'start':
datetime.date(1216, 9, 26)}
Something basic about lists and loops that I'm not getting here. I've got
a long list of dictionaries that looks like this:
lst = [{'placename': u'Stow, Lincolnshire', 'long-name': u'Stow,
Lincolnshire.', 'end': datetime.date(1216, 9, 28), 'start':
datetime.date(1216, 9, 26)},
{'placename': u'
