regexp clear solution for Three Words by knezi - python coding challenges

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regexp solution in Clear category for Three Words by knezi

import re
def checkio(words):
    return bool(re.compile("([a-zA-Z]+ ){2}[a-zA-Z]+").search(words))

#These "asserts" using only for self-checking and not necessary for auto-testing
if __name__ == '__main__':
    assert checkio("Hello World hello") == True, "Hello"
    assert checkio("He is 123 man") == False, "123 man"
    assert checkio("1 2 3 4") == False, "Digits"
    assert checkio("bla bla bla bla") == True, "Bla Bla"
    assert checkio("Hi") == False, "Hi"

April 2, 2014

Comments:

hyl3rid on Oct. 14, 2014, 12:22 a.m.
From what i understand this mean the code searchs for alphabetics only and return True else false. A few questions: How does it count to 3? What does the {2} mean in the middle? Why is [a-zA-Z]+ twice in there?...

c555 on Oct. 28, 2014, 5:01 p.m.
([a-zA-Z]+ ){2} means this ([a-zA-Z]+ ) twice. '+' is equals {1,} that means it needs to have more than one. than '([a-zA-Z]+ ){2}[a-zA-Z]+' is same as '[a-zA-Z]{1,} [a-zA-Z]{1,} [a-zA-Z]{1,}'.it's means it will only match when have a string with a sequence of 3 letter separated by space

tarjeii on Oct. 23, 2015, 7:08 p.m.
What does the '+' do? What would "[a-zA-Z] [a-zA-Z] [a-zA-Z]" mean?

koborC on Dec. 14, 2020, 2:56 a.m.
[a-zA-Z] matches an alphabet : <pre class='brush: python'>match : "a", "b", ..., "z", "A", "B", ..., "Z" miss : "word", "1", " ", "_" </pre> + means "one or more repetitions of what it follows". So "[a-zA-Z]+" means all of below: + [a-zA-Z] : an alphabet. Ex.) "I", "O" + [a-zA-Z][a-zA-Z] : two character alphabetical word. Ex.) "Hi", "oh" + [a-zA-Z][a-zA-Z][a-zA-Z] : three character alphabetical word. Ex.) "You", "you", "our" + so on... Then "[a-zA-Z]+" matches alphabetical word consist of 1+ character(s). <pre class='brush: python'>match : "word", "w", "hello" miss : "w2rd", "", "23", " " </pre> Also note that you can learn regular expressions easily using online tutorials..: <a href="https://regexone.com">https://regexone.com</a>

Rexze001 on May 8, 2020, 11:46 a.m.
Why don't write like this "([a-zA-Z]+ ){3}"?

Evelyn_He_Xinyi on Sept. 25, 2020, 7:51 a.m.
"([a-zA-Z]+ ){3}" gives you an extra trailing space at the end that you shouldn't have when matching for 3 successive words.

Alexandr_Svirbutovich on Dec. 2, 2014, 3:15 a.m.
import re def checkio(w): <pre class='brush: python'>re.sub(r'\D+ \D+ \D+','',w)!=w </pre>

brianfgonzalez on Oct. 24, 2020, 2:48 a.m.
i get the \D+ part but can you explain the rest?

koborC on Dec. 14, 2020, 3:03 a.m.
re.sub(pattern, replaced, string) --> search all part that match "pattern" in "string" and replace it with "replaced" then return the processed string. !=w tests w is processed (that mean re.sub can find three continuous alphabetical words) or not.

studioj on Jan. 14, 2015, 9:58 p.m.
regex is the way :-)

mreinhardt on May 29, 2015, 8:35 a.m.
Nice, using regex is the clear move here. There's no point in compiling a regex inside the function though since you'll have to recompile it every time the function is called. You might as well just use the raw string if you're going to do that. Instead, create the compiled regex as an external constant and then it's only compiled once: <pre class='brush: python'>THREE_WORDS_RE = re.compile("([a-zA-Z]+ ){2}[a-zA-Z]+") def checkio(words): return bool(THREE_WORDS_RE.search(words)) </pre>

Marcin_Coles on Oct. 31, 2017, 8:37 a.m.
I don't believe you even need to compile the regex in this case. From the standard library docs (3.6.3 at least) <a href="https://docs.python.org/3/library/re.html">https://docs.python.org/3/library/re.html</a> <blockquote> Note The compiled versions of the most recent patterns passed to re.compile() and the module-level matching functions are cached, so programs that use only a few regular expressions at a time needn’t worry about compiling regular expressions. </blockquote>

SteveHill on Aug. 18, 2016, 1:36 p.m.
I love regular expressions

Alexandr_Zhytenko on March 7, 2017, 9:13 p.m.
I like this code. Great!

hugoboss on Aug. 7, 2018, 3:13 p.m.
We mustn't forget regexes.

ArchTauruS on Nov. 9, 2018, 1:09 a.m.
your regx has bug, it will match 'hahaha 123abc def ghi'

siaimes on Dec. 5, 2019, 8:57 a.m.
Precondition: The input contains words and/or numbers. There are no mixed words (letters and digits combined).

leggewie on May 22, 2021, 6:18 p.m.
Indeed, thank you for pointing that out.

UlanDT on July 20, 2020, 7:18 a.m.
how fair is that to use modules?

mr.4c3 on Dec. 24, 2020, 7:01 p.m.
([a-zA-Z]+){3} also works if i am not mistaken

kdim on Jan. 10, 2021, 9:55 a.m.
Good work! Also you can use flag "re.IGNORECASE" <pre class='brush: python'>import re def checkio(words): return bool(re.search(r"[a-z]+ [a-z]+ [a-z]+", words, re.IGNORECASE)) </pre>

multicast on March 4, 2021, 8:22 p.m.
This is clever

GrumpyEconomist on March 28, 2021, 12:59 p.m.
Nice :)

leggewie on May 22, 2021, 6:17 p.m.
The code is elegant but faulty. It will incorrectly match "123abc def ghi".

Basinga on May 25, 2021, 3:58 p.m.
big brain move. I enjoy this one

2dfsdfl on June 8, 2021, 9:43 a.m.
match : "a", "b", ..., "z", "A", "B", ..., "Z" miss : "word", "1", " ", "_"

Aleksandr_Komarov on Sept. 16, 2022, 12:38 p.m.
Pretty cleare using of RE features

Aleksandr_Komarov on Sept. 16, 2022, 12:38 p.m.
Pretty clear using of RE features

viktor.chyrkin on Dec. 19, 2022, 1:44 a.m.
I like your use of bool and how short your solution is.

jv43 on Feb. 22, 2023, 4:22 a.m.
This one is good

nahandove on Jan. 31, 2024, 9:10 p.m.
Like this one. As clear as it gets and very efficient.

eecbmt on May 14, 2024, 7:38 a.m.
What is the advantage of using re.compile() instead of re.sub()?

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