sort problem

[Michele Petrazzo]

I have a list of lists (a grid table) that can have about 15000 - 20000
"rows" and 10 "cols", so:

1 [ [ 'aaa', 'vv', 'cc', 23, ... ],
2 [ 'aav', 'vv', 'cc', 45, ... ],
....
15000 [ 'sad', 'ad', 'es', 123, ... ], ]

I need to sort this list, but I need to specify two things: the "column"
and its type (string or int), so for example in this list, I want to
sort the fourth column that has int values. The type because that I want
that 1, 2, 12 will be sort in this order, not 1, 12, 2 like strings.

I have already tried to modify some code found on aspn, but the results
are always too slow for me, about 30-40 sec.
Can someone has some code or some point where can I start for speedup my
code?

Thanks,
Michele

[Lasse Vågsæther Karlsen]

Re: sort problem

How about:

list.sort(key=l ambda x: x[3])

Does that work?

[Michele Petrazzo]

Re: sort problem

Lasse Vågsæther Karlsen wrote:[color=blue]
> How about:
>
> list.sort(key=l ambda x: x[3])
>
> Does that work?
>[/color]

Yes, on my linux-test-box it work, but I my developer pc I don't have
the 2.4 yet. I think that this is a good reason for update :)

Thanks,
Michele

[Kent Johnson]

Re: sort problem

Michele Petrazzo wrote:[color=blue]
> Lasse Vågsæther Karlsen wrote:
>[color=green]
>> How about:
>>
>> list.sort(key=l ambda x: x[3])[/color][/color]

Better to use key=operator.it emgetter(3)
[color=blue]
> Yes, on my linux-test-box it work, but I my developer pc I don't have
> the 2.4 yet. I think that this is a good reason for update :)[/color]

or learn about decorate-sort-undecorate:

lst = [ ...whatever ]
lst = [ x[3], i, x for i, x in enumerate(lst) ]
lst.sort()
lst = [ x for _, _, x in lst ]

Kent
[color=blue]
>
> Thanks,
> Michele[/color]

[Michele Petrazzo]

Re: sort problem

Kent Johnson wrote:[color=blue]
> or learn about decorate-sort-undecorate:
>
> lst = [ ...whatever ] lst = [ x[3], i, x for i, x in enumerate(lst) ]
>[/color]
I think that here the code must be changed (for the future):
lst = [ (x[3], i, x) for i, x in enumerate(lst) ]
[color=blue]
> lst.sort() lst = [ x for _, _, x in lst ][/color]


Wow, this work with my py 2.3!
[color=blue]
>
> Kent
>[/color]

Thanks,
Michele

[Alex Martelli]

Re: sort problem

Michele Petrazzo <michele.petraz zo@TOGLIunipex. it> wrote:
[color=blue]
> Lasse Vågsæther Karlsen wrote:[color=green]
> > How about:
> >
> > list.sort(key=l ambda x: x[3])
> >
> > Does that work?[/color]
>
> Yes, on my linux-test-box it work, but I my developer pc I don't have
> the 2.4 yet. I think that this is a good reason for update :)[/color]

Updating is a good idea, and will let you get even faster by avoiding
the lambda:

import operator

thelist.sort(ke y=operator.item getter(3))

However, until you can upgrade you might be happy enough with a direct
implementation of the decorate-sort-undecorate (DSU) idiom which they
new "key=" named argument to sort implements. To wit:

aux = [ (x[3], x) for x in thelist ]
aux.sort()
thelist[:] = [ x[-1] for x in aux ]

Note that the "decoration " can include as many "columns" as you want,
transformations obtained by calling int(...) or str(...) on some of the
columns, and so on. This applies to "key=" in 2.4 just as well as to
the (slightly slower) direct implementation in 2.3 and earlier.


Alex