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Re: how to convert code that uses cmp to python3

Started byIan Kelly <ian.g.kelly@gmail.com>
First post2016-04-08 08:22 -0600
Last post2016-04-08 08:22 -0600
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  Re: how to convert code that uses cmp to python3 Ian Kelly <ian.g.kelly@gmail.com> - 2016-04-08 08:22 -0600

#106685 — Re: how to convert code that uses cmp to python3

FromIan Kelly <ian.g.kelly@gmail.com>
Date2016-04-08 08:22 -0600
SubjectRe: how to convert code that uses cmp to python3
Message-ID<mailman.88.1460125409.2253.python-list@python.org>
On Fri, Apr 8, 2016 at 8:08 AM, Chris Angelico <rosuav@gmail.com> wrote:
> On Fri, Apr 8, 2016 at 11:31 PM, Antoon Pardon
> <antoon.pardon@rece.vub.ac.be> wrote:
>> Doing it as follows:
>>     seq1 < seq2
>>     seq2 < seq1
>>
>> takes about 110 seconds.
>>
>>
>> Doing it like this:
>>     delta = cmp(seq1, seq2)
>>     delta < 0
>>     delta > 0
>>
>> takes about 50 seconds.
>
> Why are you comparing in both directions, though? cmp() is more
> equivalent to this:
>
> seq1 == seq2
> seq1 < seq2
>
> You only need ONE comparison, and the other is presumed to be its
> opposite. When, in the Python 3 version, would you need to compare
> twice?

When there are three possible code paths depending on the result.

def search(key, node):
    if node is None:
        raise KeyError(key)
    if key < node.key:
        return search(key, node.left)
    elif key == node.key:
        return node
    else:
        return search(key, node.right)

How would you implement this with only one comparison?

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