> > Hi all,
>
> >
>
> > I would like to sort according to this order:
>
> >
>
> > (' ', '.', '\'', '-', '0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'a',
>
> > 'A', '?', '?', '?', '?', '?', '?', '?', '?', '?', '?', 'b', 'B', 'c', 'C',
>
> > '?', '?', 'd', 'D', 'e', 'E', '?', '?', '?', '?', '?', '?', '?', '?', 'f',
>
> > 'F', 'g', 'G', 'h', 'H', 'i', 'I', '?', '?', '?', '?', '?', '?', '?', '?',
>
> > 'j', 'J', 'k', 'K', 'l', 'L', 'm', 'M', 'n', '?', 'N', '?', 'o', 'O', '?',
>
> > '?', '?', '?', '?', '?', '?', '?', '?', '?', 'p', 'P', 'q', 'Q', 'r', 'R',
>
> > 's', 'S', 't', 'T', 'u', 'U', '?', '?', '?', '?', '?', '?', '?', '?', 'v',
>
> > 'V', 'w', 'W', 'x', 'X', 'y', 'Y', 'z', 'Z')
>
> >
>
> > How can I do this? The default sorted() does not give the desired result.
>
>
>
> I'm assuming that doesn't correspond to some standard locale's collating
>
> order, so we really do need to roll our own encoding (and that you have
>
> a good reason for wanting to do this).
It is for creating a Dutch dictionary. This sorting order is not to be found in
an existing locale.
> I'm also assuming that what I'm
>
> seeing as question marks are really accented characters in some encoding
>
> that my news reader just isn't dealing with (it seems to think your post
>
> was in ISO-2022-CN (Simplified Chinese).
>
>
>
> I'm further assuming that you're starting with a list of unicode
>
> strings, the contents of which are limited to the above alphabet.
Correct.
> I'm
>
> even further assuming that the volume of data you need to sort is small
>
> enough that efficiency is not a huge concern.
Well, it is for 200,000 - 450,000 words but the code is allowed be slow. It
will not be used for web application or something which requires a quick
response.
> Given all that, I would start by writing some code which turned your
>
> alphabet into a pair of dicts. One maps from the code point to a
>
> collating sequence number (i.e. ordinals), the other maps back.
>
> Something like (for python 2.7):
>
>
>
> alphabet = (' ', '.', '\'', '-', '0', '1', '2', '3', '4', '5',
>
> '6', '7', '8', '9', 'a', 'A', '?', '?', '?', '?',
>
> [...]
>
> 'v', 'V', 'w', 'W', 'x', 'X', 'y', 'Y', 'z', 'Z')
>
>
>
> map1 = {c: n for n, c in enumerate(alphabet)}
>
> map2 = {n: c for n, c in enumerate(alphabet)}
OK, similar to Thomas' proposal.
> Next, I would write some functions which encode your strings as lists of
>
> ordinals (and back again)
>
>
>
> def encode(s):
>
> "encode('foo') ==> [34, 19, 19]" # made-up ordinals
>
> return [map1[c] for c in s]
>
>
>
> def decode(l):
>
> "decode([34, 19, 19]) ==> 'foo'"
>
> return ''.join(map2[i] for i in l)
>
>
>
> Use these to convert your strings to lists of ints which will sort as
>
> per your specified collating order, and then back again:
>
>
>
> encoded_strings = [encode(s) for s in original_list]
>
> encoded_strings.sort()
>
> sorted_strings = [decode(l) for l in encoded_strings]
>
>
>
> That's just a rough sketch, and completely untested, but it should get
>
> you headed in the right direction. Or at least one plausible direction.
>
> Old-time perl hackers will recognize this as the Schwartzian Transform.
I will test it and let you know. :) Pander
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