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From: Victor Eijkhout on 31 Mar 2010 14:58
Robert Kern <robert.kern(a)gmail.com> wrote:

> second[first.argsort()]

Really cool. Thanks.

> Ask numpy questions on the numpy mailing list.

I will. I thought that this question would have an answer in a generic
python idiom.

Victor.
--
Victor Eijkhout -- eijkhout at tacc utexas edu
From: Robert Kern on 31 Mar 2010 15:13
On 2010-03-31 13:58 PM, Victor Eijkhout wrote:
> Robert Kern<robert.kern(a)gmail.com> wrote:
>
>> second[first.argsort()]
>
> Really cool. Thanks.
>
>> Ask numpy questions on the numpy mailing list.
>
> I will. I thought that this question would have an answer in a generic
> python idiom.

When dealing with numpy arrays, the generic Python idiom is often much slower.

--
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma
that is made terrible by our own mad attempt to interpret it as though it had
an underlying truth."
-- Umberto Eco

From: Steve Holden on 31 Mar 2010 15:22
Victor Eijkhout wrote:
> Robert Kern <robert.kern(a)gmail.com> wrote:
>
>> second[first.argsort()]
>
> Really cool. Thanks.
>
>> Ask numpy questions on the numpy mailing list.
>
> I will. I thought that this question would have an answer in a generic
> python idiom.
>
> Victor.

Not an unreasonable assumption, but it turns out that for most Python
users (estimate PFTA: 97%) numpy/scipt is esoteric knowledge.

regards
Steve
--
Steve Holden +1 571 484 6266 +1 800 494 3119
See PyCon Talks from Atlanta 2010 http://pycon.blip.tv/
Holden Web LLC http://www.holdenweb.com/
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From: Raymond Hettinger on 31 Mar 2010 16:09
On Mar 30, 4:25 pm, s...(a)sig.for.address (Victor Eijkhout) wrote:
> I have two arrays, made with numpy. The first one has values that I want
> to use as sorting keys; the second one needs to be sorted by those keys.
> Obviously I could turn them into a dictionary  of pairs and sort by the
> first member, but I think that's not very efficient, at least in space,
> and this needs to be done as efficiently as possible.

Alf's recommendation is clean and correct. Just make a list of
tuples.

FWIW, here's a little hack that does the work for you:

>>> values = ['A', 'B', 'C', 'D', 'E']
>>> keys = [50, 20, 40, 10, 30]
>>> keyiter = iter(keys)
>>> sorted(values, key=lambda k: next(keyiter))
['D', 'B', 'E', 'C', 'A']


Raymond
From: Steve Howell on 1 Apr 2010 10:59
On Mar 31, 1:09 pm, Raymond Hettinger <pyt...(a)rcn.com> wrote:
> On Mar 30, 4:25 pm, s...(a)sig.for.address (Victor Eijkhout) wrote:
>
> > I have two arrays, made with numpy. The first one has values that I want
> > to use as sorting keys; the second one needs to be sorted by those keys..
> > Obviously I could turn them into a dictionary  of pairs and sort by the
> > first member, but I think that's not very efficient, at least in space,
> > and this needs to be done as efficiently as possible.
>
> Alf's recommendation is clean and correct.  Just make a list of
> tuples.
>
> FWIW, here's a little hack that does the work for you:
>
> >>> values = ['A', 'B', 'C', 'D', 'E']
> >>> keys = [50, 20, 40, 10, 30]
> >>> keyiter = iter(keys)
> >>> sorted(values, key=lambda k: next(keyiter))
>
> ['D', 'B', 'E', 'C', 'A']
>

Another option:

[values[i] for i in sorted(range(len(keys)), key=lambda i: keys[i])]

Sort the indexes according to keys values, then use indexes to get the
values.

It might read more clearly when broken out into two lines:

>>> sorted_indexes = sorted(range(len(keys)), key = lambda i: keys[i])
>>> sorted_indexes
[3, 1, 4, 2, 0]
>>> [values[i] for i in sorted_indexes]
['D', 'B', 'E', 'C', 'A']

The advantage of Raymond's solution is that he only creates one new
Python list, whereas my solutions create an intermediate Python list
of integers. I don't think my solution really is that space-wasteful,
though, since by the time the second list gets created, any internal
intermediate lists from CPython's sorted() implementation will
probably have been cleaned up.



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