Using namedtuples field names for column indices in a list of lists
Peter Otten
__peter__ at web.de
Mon Jan 9 19:09:46 EST 2017
breamoreboy at gmail.com wrote:
> On Monday, January 9, 2017 at 5:34:12 PM UTC, Tim Chase wrote:
>> On 2017-01-09 08:31, breamoreboy wrote:
>> > On Monday, January 9, 2017 at 2:22:19 PM UTC, Tim Chase wrote:
>> > > I usually wrap the iterable in something like
>> > >
>> > > def pairwise(it):
>> > > prev = next(it)
>> > > for thing in it:
>> > > yield prev, thing
>> > > prev = thing
>> >
>> > Or from
>> > https://docs.python.org/3/library/itertools.html#itertools-recipes:->> >
>> > def pairwise(iterable):
>> > "s -> (s0,s1), (s1,s2), (s2, s3), ..."
>> > a, b = tee(iterable)
>> > next(b, None)
>> > return zip(a, b)
>> >
>> > This and many other recipes are available in the more-itertools
>> > module which is on pypi.
>>
>> Ah, helpful to not have to do it from scratch each time. Also, I see
>> several others that I've coded up from scratch (particularly the
>> partition() and first_true() functions).
>>
>> I usually want to make sure it's tailored for my use cases. The above
>> pairwise() is my most common use case, but I occasionally want N-wise
>> pairing
>
> def ntuplewise(iterable, n=2):
> args = tee(iterable, n)
> loops = n - 1
> while loops:
> for _ in range(loops):
> next(args[loops], None)
> loops -= 1
> return zip(*args)
>
>>
>> s -> (s0,s1,…sN), (s1,s2,…S{N+1}), (s2,s3,…s{N+2}), …
>>
>> or to pad them out so either the leader/follower gets *all* of the
>> values, with subsequent values being a padding value:
>>
>> # lst = [s0, s1, s2]
>> (s0,s1), (s1, s2), (s2, PADDING)
>
> Use zip_longest instead of zip in the example code above.
>
>> # or
>> (PADDING, s0), (s0, s1), (s1, s2)
>
> Haven't a clue off of the top of my head and I'm too darn tired to think
> about it :)
In both cases modify the iterable before feeding it to ntuplewise():
>>> PADDING = None
>>> N = 3
>>> items = range(5)
>>> list(ntuplewise(chain(repeat(PADDING, N-1), items), N))
[(None, None, 0), (None, 0, 1), (0, 1, 2), (1, 2, 3), (2, 3, 4)]
>>> list(ntuplewise(chain(items, repeat(PADDING, N-1)), N))
[(0, 1, 2), (1, 2, 3), (2, 3, 4), (3, 4, None), (4, None, None)]
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