Looking up a dictionary _key_ by key?

Laura Creighton lac at openend.se
Sun Jul 5 08:25:38 CEST 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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