David Isaac wrote:
"Nick Coghlan" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
A bug report on Sourceforge would help in getting the problem fixed for
the 2.5
docs
Done.
Bug 1121416, for anyone else interested. Looks Raymond agrees with me about the
left-to-right evaluation of iter
"Nick Coghlan" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> A bug report on Sourceforge would help in getting the problem fixed for
the 2.5
> docs
Done.
> For the 'left-to-right' evaluation thing, that's technically an
implementation
> artifact of the CPython implementation, si
David Isaac wrote:
If that is right, I still cannot extract it from the doc cited above.
So where should I have looked?
Ouch. The terminology's evolved, and it looks to me like the docs for the older
builtins haven't been updated to track it.
The terminology has pretty much settled to 'iterable'
> Alan Isaac wrote:
> > I see that [iter(l)]*N produces an N element list with each element
being
> > the same iterator object, but after that
> > http://www.python.org/doc/2.3.5/lib/built-in-funcs.html
> > just didn't get me there.
"Nick Coghlan" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL P
David Isaac wrote:
"Nick Coghlan" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
Using zip(*[iter(l)]*N) or zip(*(iter(l),)*N) simply extends the above to
the
general case.
Clearly true.
But can you please go into much more detail for a newbie?
I see that [iter(l)]*N produces an N ele
"Nick Coghlan" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> Using zip(*[iter(l)]*N) or zip(*(iter(l),)*N) simply extends the above to
the
> general case.
Clearly true.
But can you please go into much more detail for a newbie?
I see that [iter(l)]*N produces an N element list with
Steven Bethard wrote:
I think you can write that second one so that it works for iterables
without a __len__:
py> def padded_partition(iterable, part_len, pad_val=None):
... itr = itertools.chain(
... iter(iterable), itertools.repeat(pad_val, part_len - 1))
... return itertools.iz
Chris Wright wrote:
> Hi,
>
> 1) I want to iterate over a list "N at a time"
> sort of like:
>
> # Two at a time... won't work, obviously
>
> >>> for a, b in [1,2,3,4]:
> ... print a,b
> ...
> Traceback (most recent call last):
>File "", line 1, in ?
> TypeError: unpack non-sequence
> >>>
Nick Coghlan wrote:
I'd definitely recommend hiding this trick inside a function. Perhaps
something like (using Michael's function name):
from itertools import izip, repeat, chain
def partition(seq, part_len):
return izip(*((iter(seq),) * part_len))
def padded_partition(seq, part_len, pad_val=N
cookbook's not an easy grep but:
http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/303060
http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/303279
http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/347689
--
http://mail.python.org/mailman/listinfo/python-list
Duncan Booth wrote:
Try this:
l = [1, 2, 3, 4]
for a, b in zip(*[iter(l)]*2):
print a, b
zip(*[iter(seq)]*N) will group by N (but if there are any odd items at the
end it will ignore them).
map(None, *[iter(seq)]*N) will group by N padding the last item with None
if it needs to.
For anyone el
For problems like this I use a partition function defined in a utils.py
file that I use (based on Peter Norvig's utils file at
http://aima.cs.berkeley.edu/python/utils.py). Using partition, the
problem you posed can be solved by writing
#for a, b in partition([1, 2, 3, 4], 2):
#print a, b
Th
Chris Wright wrote:
> 1) I want to iterate over a list "N at a time"
> sort of like:
>
> # Two at a time... won't work, obviously
>
> >>> for a, b in [1,2,3,4]:
> ... print a,b
> ...
Try this:
l = [1, 2, 3, 4]
for a, b in zip(*[iter(l)]*2):
print a, b
zip(*[iter(seq)]*N) will gr
Le Fri, 28 Jan 2005 13:59:45 GMT, Chris Wright a écrit :
> Hi,
>
> 1) I want to iterate over a list "N at a time"
>
>
> Is there a nifty way to do with with list comprehensions,
> or do I just have to loop over the list ?
>
> cheers and thanks
seq = xrange(1, 9) # an iterable [1, 2, ... 8]
N
In article <[EMAIL PROTECTED]>,
Chris Wright <[EMAIL PROTECTED]> wrote:
> Hi,
>
> 1) I want to iterate over a list "N at a time"
You could do it with slicing and zip:
>>> l = [1, 2, 3, 4, 5, 6, 7, 8]
>>> zip (l[::2], l[1::2])
[(1, 2), (3, 4), (5, 6), (7, 8)]
To my eyes, that's a bit cryptic,
l = [1,2,3,4]
for a, b in zip(l[::2], l[1::2]):
print a,b
--
Regards,
Diez B. Roggisch
--
http://mail.python.org/mailman/listinfo/python-list
Hi,
1) I want to iterate over a list "N at a time"
sort of like:
# Two at a time... won't work, obviously
>>> for a, b in [1,2,3,4]:
... print a,b
...
Traceback (most recent call last):
File "", line 1, in ?
TypeError: unpack non-sequence
>>>
Is there a nifty way to do with with list com
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