Spir sent this solely to me by accident, I think.
-- Forwarded message --
From: spir ☣
Date: 2010/4/19
Subject: Re: [Tutor] List index usage: is there a more pythonesque way?
To: cmca...@googlemail.com
On Mon, 19 Apr 2010 12:59:40 +0100
C M Caine wrote:
> That's t
On Mon, Apr 19, 2010 at 9:23 AM, Alan Gauld wrote:
>
> "C M Caine" wrote
>
>> That's the first I've read of iterating through dictionaries, I'd
>>
>> assumed it was impossible because they're unordered.
>>
>
> Iteration doesn't require order, only to get each item once.
> Even in very old Python
"C M Caine" wrote
That's the first I've read of iterating through dictionaries, I'd
assumed it was impossible because they're unordered.
Iteration doesn't require order, only to get each item once.
Even in very old Python versions you could iterate a dictionary
via the keys() method. More
That's the first I've read of iterating through dictionaries, I'd
assumed it was impossible because they're unordered. Your explanation
for defining your own iterables is much easier to understand than the
one I read before as well.
Thanks again.
2010/4/19 spir ☣ :
> On Mon, 19 Apr 2010 00:37:11
On Mon, 19 Apr 2010 00:37:11 +0100
C M Caine wrote:
> That's two new things I've learnt. I didn't realise that for loops
> could be used like that (with more than one... key?).
Consider considering things differently: a for loop always iterates over items
of a collection you indicate:
l = [1,2
> That's two new things I've learnt. I didn't realise that for loops
> could be used like that (with more than one... key?).
Technically its still one key but enumerate returns a tuple
of index and value and we use tuple unpacking to assign
the values to the loop variables. That is we could writ
> Something is wrong in the following if statement, as both paths execute the
> same code.
>
>> if spaceDict['1st'+key] == 0:
>> spaceDict['nth'+key] = i
>> else:
>> spaceDict['nth'+key] = i
>> for form in forms:
>> adjuste
That's two new things I've learnt. I didn't realise that for loops
could be used like that (with more than one... key?).
Thanks, I'm changing my code even now!
On 19 April 2010 00:09, Alan Gauld wrote:
>
> "C M Caine" wrote
>
>> for i in range(len(timetable)):
>> numDict[timetab
On 4/18/2010 6:53 PM, C M Caine wrote:
# Example data for forms and timetable:
forms = ["P7", "P8", "P9", "P10", "P11", "S7", "S8", "S9", "S10",
"S11", "IMA", "CAT", "FOR", "RLS", "EMPTY"]
timetable = ['CAT', 'P10', 'P8', 'EMPTY', 'EMPTY', 'EMPTY', 'S10',
'S8', 'IMA', 'EMPTY', 'S7', 'S10', 'P9',
"C M Caine" wrote
for i in range(len(timetable)):
numDict[timetable[i]] += 1
if spaceDict['1st'+timetable[i]] == 0:
spaceDict['nth'+timetable[i]] = i
for index, item in enumerate(timetable):
numDict[item] += 1
if spaceDict[
# Example data for forms and timetable:
forms = ["P7", "P8", "P9", "P10", "P11", "S7", "S8", "S9", "S10",
"S11", "IMA", "CAT", "FOR", "RLS", "EMPTY"]
timetable = ['CAT', 'P10', 'P8', 'EMPTY', 'EMPTY', 'EMPTY', 'S10',
'S8', 'IMA', 'EMPTY', 'S7', 'S10', 'P9', 'EMPTY', 'EMPTY', 'EMPTY',
'S7', 'EMPTY'
11 matches
Mail list logo