Partial creative solution for Bird Language by veky - python coding challenges

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Partial solution in Creative category for Bird Language by veky

import re,functools
translate=functools.partial(re.sub,r"(\w)(\1\1|.)",r"\1")

June 25, 2014

Comments:

Cjkjvfnby on June 25, 2014, 11:12 p.m.
good solution :)

veky on June 25, 2014, 11:28 p.m.
Thanks. :-)

Sim0000 on June 25, 2014, 11:17 p.m.
Cool!

coells on June 26, 2014, 9:03 a.m.
veky ~ bottomless well full of surprises

veky on June 26, 2014, 9:20 a.m.
LOL. Strangest compliment I've received in a while. Thanks. :-D

gyahun_dash on June 28, 2014, 11:43 p.m.
Gooooooodo

drobotnik on May 5, 2015, 10:57 a.m.
Where do I go to understand what the hell is happening here? The Python docs for regex say that \w matches any alphanumeric character. Where do you define vowels???

veky on May 5, 2015, 1:17 p.m.
<blockquote> Where do I go to understand what the hell is happening here? </blockquote> You can always try asking me. Contrary to popular belief, I do like explaining my code. <blockquote> The Python docs for regex say that \w matches any alphanumeric character. </blockquote> And they are mostly correct (as long as you consider _ an alphanumeric, which is a historical accident:). <blockquote> Where do you define vowels??? </blockquote> Nowhere. I don't have to. Note that the task is not to translate something to the bird language, but from it. If I had to translate to bird language, I'd have to implement those two rules, and that would include knowing whether something is a vowel. But, since I have to translate from bird language (and I have an assurance that the input is a valid bird language string), I just have to apply the rules in reverse. And they are: <ul> <li>every vowel repeated three times should be replaced by just one instance</li> <li>every vowel preceded by a consonant should be deleted</li> </ul> If we write those with just lookahead (not lookbehind as in second rule), since consonants are surely different from vowels, we have <ul> <li>every letter repeated two more times should be echoed (just once)</li> <li>every letter not repeated (followed by something else) should be echoed (without that "something else")</li> </ul> See, wovels are recognized by the fact that they (and only they, of alphanumerics) are tripled. Nothing else is required. We can even say that in one rule: <ul> <li>every letter, followed by (either two more instances of same letter, or some other letter), should be echoed, without the parenthesized part</li> </ul> And that is what's written in the regex.

drobotnik on May 5, 2015, 1:20 p.m.
Thanks! Need to process this a bit :) Just out of interest - do you find Regexes natural and clear to work with? As a beginner they seem a bit esoteric... Although I felt a bit like that about list comprehensions not too long ago!

veky on May 6, 2015, 7:02 a.m.
<blockquote> do you find Regexes natural and clear to work with? </blockquote> No. They are Perl heritage: a microlanguage of its own, and not really aligned with Python values. (Comprehensions are just hard to get, since bindings are syntactically reversed, but other than that, they are aligned with Python values.:) As you can probably see, I very rarely use regexen on CiO, even when it's "obvious" (to everyone but me) they should be used. :-) But sometimes, like here, they really are the simplest thing that expresses the intent.

StefanPochmann on May 5, 2015, 8:31 p.m.
That regex is too long, you can make it four chars shorter.

veky on May 6, 2015, 6:57 a.m.
Now it's my turn to be confused. :-O Yes, I see some modifications (though not by 4 chars, leaving the rest of program unchanged), but they don't work for all precondition-satisfying inputs. If you still think you're right, write a solution, and let's bet two thumbs on it. :-D

StefanPochmann on May 6, 2015, 7:14 a.m.
<a href="http://www.checkio.org/mission/bird-language/publications/StefanPochmann/python-3/vek/">It's here</a>

veky on May 6, 2015, 7:38 a.m.
I like those bets when I'm happy whether I win or lose. Either my ego or my knowledge increases. :-D

bsquare on Sept. 18, 2019, 10:20 a.m.
Very quick and clear solution! Just for my information: here you find regexp useful, and the right tool to use?

veky on Sept. 18, 2019, 10:54 a.m.
Yup. It's practically made for it. Find a pattern and substitute.

Spiritusvini on Dec. 4, 2020, 9:50 p.m.
The logic is brilliant! But I can't figure out how partial works here.

veky on Dec. 4, 2020, 11:01 p.m.
<pre class='brush: python'>translate = partial(sub, pattern, replacement) </pre> means that <pre class='brush: python'>translate(word) = sub(pattern, replacement, word) </pre> Anything else?

Spiritusvini on Dec. 4, 2020, 11:46 p.m.
I see! Sorry for the dumb question, I didn't notice that there is no agument. Thanks a lot for answer!

veky on Dec. 5, 2020, 5:57 a.m.
There are plenty of arguments, only not the unknown one. :-)

veky on Dec. 4, 2020, 11:02 p.m.
<pre class='brush: python'>translate = partial(sub, pattern, replacement) ~~~> translate(word) = sub( pattern, replacement, word) </pre> Anything else? :-)

andromedk on Feb. 13, 2021, 1:21 p.m.
I love regular expressions, I will definitely study them further. Thanks!

veky on Feb. 14, 2021, 11:16 a.m.
They are not really Pythonic (they are mostly a legacy from Perl), but they are there if you need them. :-)

sawako.oono on May 26, 2021, 7:09 a.m.
Learned about "partial" for the first time here.\ Also, I was not aware that \n expression can be reused across arguments:\ \1 in the second argument in re.sub() here referring to \w in the first argument.

veky on May 26, 2021, 7:13 a.m.
I write my solutions so you learn something from them. :-) "Reusing" \1 and its cousins can be said to be the main reason for their existence: you extract some information from string with a regex, and then reassemble it into another string with sub. :-]

sawako.oono on May 26, 2021, 9:10 a.m.
Thanks, now I get how it's really useful. I tried using what you described here in several patterns and realized one thing. str method such as str.lower() or str.upper() does not work for \n ? <hr /> import re\ text = "ABcccBB" print(re.sub(r"([A-Z])([A-Z])",r"\1"+":"+r"\2",text)) <blockquote> A:BcccB:B </blockquote> print(re.sub(r"([A-Z])([A-Z])",r"\1".lower()+r"\2",text)) <blockquote> ABcccBB </blockquote> <hr />

veky on May 26, 2021, 7:26 p.m.
Of course it works. r'\1'.lower() == r'\1'. :-P I think you should learn the order in which Python does things. f(g(x)) first calls g. But you can solve your problem with callbacks. <pre class='brush: python'>>>> re.sub(r"([A-Z])" * 2, lambda m: m[1].lower() + m[2], text) </pre>

sawako.oono on May 27, 2021, 10:59 a.m.
Now I see. Thank you so much for your kind explanation.

alterGNU on July 2, 2021, 9:34 a.m.
Wow, that's a really nice way to solve it^^ (dayoum I really need to use REGEX more often in python XD )

veky on July 2, 2021, 10:20 a.m.
That's probably the wrong lesson to take out of this. I wrote about half a thousand solutions here, and only a handful of them use regexen. They aren't really Pythonic; see the <a href="https://py.checkio.org/mission/bird-language/publications/veky/python-3/partial/share/7296b239c1685e8f544f6787b4cec8f8/#comment-30133">above comment</a>.

spitfire_ch on Aug. 9, 2021, 7:58 p.m.
Excellent solution, cool! 👍

mr-quant on Aug. 26, 2021, 4:54 p.m.
Good solution to demondtrate basic regexp methods.

Alex_4444D on Oct. 20, 2021, 9:52 p.m.
Nice very good :)

marcopunteri on Nov. 15, 2021, 5:27 a.m.
i like the partial part, it's a way i always forget :)

lxf42 on March 22, 2022, 11:16 a.m.
good reason for me to look into partial and learn a new function :)

viktor.chyrkin on July 11, 2022, 5:47 p.m.
This is big brain code, and I don't understand a single part of it.

veky on July 11, 2022, 5:49 p.m.
You need a bird brain version? :-P Have you looked at the comments?

viktor.chyrkin on July 11, 2022, 5:50 p.m.
This is big brain code because i have NO idea of how to use functools.

veky on July 11, 2022, 5:55 p.m.
functools is a toolshed, nobody knows how to use all of it. For this you just need to know partial, and that's explained in <a href="https://py.checkio.org/mission/bird-language/publications/veky/python-3/partial/share/7296b239c1685e8f544f6787b4cec8f8/#comment-94167">this comment</a> (which is why I asked you if you read the comments).

viktor.chyrkin on July 11, 2022, 5:55 p.m.
I am a 11 year old dude working at my dads computer so not much to say :-P

Kverde on Aug. 11, 2022, 4:47 p.m.
It's very interesting solution!

veky on Aug. 12, 2022, 4:26 p.m.
Yes, but there are more interesting ones. Look at the comments. ;-)

Max0526 on May 5, 2023, 2:04 p.m.
Genius! Technically, no need to use functools, this lambda function will do: <pre class='brush: python'>translate = lambda text, s = __import__('re').sub: s(r"(\w)(\1\1|.)", r"\1", text) </pre>

veky on May 5, 2023, 2:30 p.m.
Sure, but every partial can be a lambda, there's no understanding to gain here. And you really don't need a separate argument `s`.

Max0526 on May 5, 2023, 3:08 p.m.
Got it, thank you! <pre class='brush: python'>translate = lambda text: __import__('re').sub(r"(\w)(\1\1|.)", r"\1", text) </pre>

rinatv99 on Aug. 12, 2024, 3:34 a.m.
Восхищает меня использование functools, надеюсь я научусь им пользоваться и писать так чтобы не надо было комментариев.

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