vek creative solution for Bird Language by StefanPochmann - python coding challenges

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vek solution in Creative category for Bird Language by StefanPochmann

import re

def translation(phrase):
    return re.sub(r'(\w)\1?.', r'\1', phrase)

May 5, 2015

Comments:

StefanPochmann on May 5, 2015, 8:30 p.m.
Unfortunately I looked at veky's solution before thinking much, so I can't claim this as my own. I made the regex shorter, though.

veky on May 6, 2015, 7:28 a.m.
Argh - there is a number between 1 and 3 after all! :-D And that is the number of thumbs you get. :-)

drobotnik on May 6, 2015, 9:34 a.m.
veky is stingy with his thumbs :)

veky on May 6, 2015, 9:42 a.m.
It was a bet, on a fixed amount of thumbs. But yes, I should probably give additional thumbs for the beauty of this solution, surpassing just the bet amount. There, squared it. :-D

nickie on Feb. 16, 2016, 11:07 p.m.
Just brilliant.

Dr__ON on March 21, 2019, 4:13 a.m.
Please, explain your solution by the group of symbols, cause I can't understand it... I read documentation, but it wasn't more clear for me(

Negamax on March 30, 2019, 10:59 p.m.
<pre class='brush: python'>re.sub(r'(\w)\1?.', r'\1', phrase) </pre> The search pattern `r'(\w)\1?.'` means: <ol> <li>`r'string'`: "raw string" - backslashes are treated as literal characters.</li> <li>`(\w)` - searches one alphanumeric character (<a href="https://docs.python.org/2/howto/regex.html#matching-characters">Matching Characters</a>) and captures it into the `group 1` (<a href="https://docs.python.org/2/howto/regex.html#grouping">Grouping</a>).</li> <li>`\1?` - checks if the next character matches a character captured in the `group 1`, using a backreference `\1`. When a character is followed by `?` in a regular expression it means to match zero or one instance of the character. So `\1?` matches the `group 1` content if there is one in the string, but otherwise matches nothing.</li> <li>`.` - "It matches anything except a newline character, and there’s an alternate mode (`re.DOTALL`) where it will match even a newline."</li> </ol> The substitution string `r'\1'` means: substitute all pattern matched characters to the captured `group 1` content. <a href="https://docs.python.org/2/howto/regex.html#grouping">From documentation:</a> <blockquote> Backreferences in a pattern allow you to specify that the contents of an earlier capturing group must also be found at the current location in the string. For example, `\1` will succeed if the exact contents of `group 1` can be found at the current position, and fails otherwise. Remember that Python’s string literals also use a backslash followed by numbers to allow including arbitrary characters in a string, so be sure to use a raw string when incorporating backreferences in a RE. </blockquote> <h2> </h2> <h2>Example:</h2> Search string = `hieeelalaooo`, result string = `''` First search: `(\w)` matches `h` character and put it into the `group 1`. `\1?` - is the next character matches the `group 1` one? No it is `i`. But the `?` quantifier allows this unmatching, so go to the next regex character - `.`. `.` take any next character - it is `i` in this case (it wasn't consumed in the previous step, due to unmatching). Matching ends, substitution happens: `hi` substring replaced to the `h` (the `group 1` content). Second search: Search string = `eeelalaooo`, result string = `h` `(\w)` matches `e` character and put it into the `group 1`. `\1?` - is the next character matches the `group 1` one? Yes, it is `e`. `.` take any next character - it is `e` in this case. Matching ends, substitution happens: `eee` substring replaced to the `e` (the `group 1` content). Third search: Search string = `lalaooo`, result string = `he` Etc.

Dr__ON on April 1, 2019, 3:21 a.m.
Thanks, now it is more clear for me. But I can't still understand how 'i', 'a' etc. replacing by nothing...

Negamax on April 1, 2019, 10:01 a.m.
Because the `(\w)\1?.` pattern matches to strings `hi`, `la`; as well as `eee`, `ooo` or `eeo` or `aae`. It says literally - the string should have: `\w` - one alphanumeric character - it is necessary condition, else pattern shouldn't match. `\1?` - it can have the previous character repeating (`group 1`). But can have not. `.` - and one more character. Any. So, if a string has "the previous character repeating" - good, pattern matches: `eee`, `ooo`, `eeo` or `aae`. Also, if a string doesn't have "the previous character repeating" - good, pattern matches: `hi`, `la`. It just has `\w` and `.`. `\1?` is not necessary for matching due the `?` sign. Example: `a?a?a?b` matches to `b`, `ab`, `aab`, `aaab`. This pattern says literally - find string, that has: `a?` - one or zero `a` character. `a?` - one or zero `a` character. `a?` - one or zero `a` character. `b` - one `b` character. Only this is necessary for matching, others are optional.

Dr__ON on April 1, 2019, 10:20 a.m.
Thanks,bro! I understood indeed)

imtiaz.rahi on Aug. 27, 2019, 9:12 a.m.
Many thanks for the detailed explanation. Helped me to understand what's going on with the regex and replacements.

Maks05 on Aug. 5, 2022, 6:27 p.m.
Thanks a lot for the explanation!

andromedk on Feb. 12, 2021, 7:31 a.m.
Great!

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