Looking up a dictionary _key_ by key?
Laura Creighton
lac at openend.se
Sun Jul 5 02:25:38 EDT 2015
In a message of Tue, 23 Jun 2015 18:06:45 -0700, Paul Rubin writes:
>Chris Angelico <rosuav at gmail.com> writes:
>>> Would I have to do an O(n) search to find my key?
>> Iterate over it - it's an iterable view in Py3 - and compare.
>
>I think the question was whether the O(n) search could be avoided, not
>how to do it. I don't see a way to avoid it. There is fundamental
>brokenness in having unequal objects compare as equal, and the breakage
>messes up the dictionary when those objects are used as keys.
>
>Solution is to either fix the object equality test, or wrap them in
>something (maybe a tuple containing the objects and the distinguishing
>fields that are missing from the original object's equality method) that
>treats unequal objects as unequal.
>--
>https://mail.python.org/mailman/listinfo/python-list
This just showed up in my mailbox:
Subject: [ANN] pyskiplist-1.0.0
From: Geert Jansen <geertj at gmail.com>
PySkipList is a fast, pure Python implementation of an indexable
skiplist. It implements a SkipList data structure that provides an
always sorted, list-like data structure for (key, value) pairs.
... more details including timing. For the full text see
https://github.com/geertj/pyskiplist
It's also available on PyPI.
Looks to me as if he's fixed the 0(n) problem ....
Laura
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