[Python-ideas] `OrderedDict.items().__getitem__`

[Mathias Panzenböck]

Why not:

	get_first = lambda d: next(iter(d.items()))

No need for a full copy of the dict.

On 01/11/2014 09:51 PM, Ryan Gonzalez wrote:
> Based on your popitem idea:
>
> get_first = lambda d: d.copy().popitem()
> get_last = lambda d: d.copy().popitem(last=True)
>
>
>
> On Sat, Jan 11, 2014 at 8:36 AM, Chris Angelico <rosuav at gmail.com <mailto:rosuav at gmail.com>> wrote:
>
>     On Sun, Jan 12, 2014 at 1:18 AM, Ram Rachum <ram.rachum at gmail.com <mailto:ram.rachum at gmail.com>> wrote:
>      > I think that `OrderedDict.items().__getitem__` should be implemented, to
>      > solve this ugliness:
>      >
>      > http://stackoverflow.com/questions/21062781/shortest-way-to-get-first-item-of-ordereddict-in-python-3
>      >
>      > What do you think?
>
>     Well, the first problem with that is that __getitem__ already exists,
>     and it's dict-style :) So you can't fetch out an item by its position
>     that way. But suppose you create a method that returns the Nth
>     element.
>
>     The implementation in CPython 3.4 is a linked list, so getting an
>     arbitrary element by index would be quite inefficient. Getting
>     specifically the first can be done either with what you see in that
>     link (it could be made a tiny bit shorter, but not much), but anything
>     else would effectively entail iterating over the whole thing until you
>     get to that position, so you may as well do that explicitly.
>     Alternatively, if you're okay with it being a destructive operation,
>     you can use popitem() to snag the first (or last, if you wish)
>     key/value pair.
>
>     ChrisA
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> --
> Ryan
> When your hammer is C++, everything begins to look like a thumb.
>
>
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