Intersection clear solution for Common Words by bryukh - python coding challenges

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Intersection solution in Clear category for Common Words by bryukh

def checkio(first, second):
    """
    set data type has useful methods.
    """
    first_set, second_set = set(first.split(",")), set(second.split(","))
    common = first_set.intersection(second_set)
    return ",".join(sorted(common))

Feb. 20, 2014

Comments:

Cjkjvfnby on March 2, 2014, 1:30 p.m.
No need to set(second.split(",")), intersects accept any iterable

bryukh on March 2, 2014, 8:20 p.m.
EDIT: Got it (looked it in your solution). Thanks, great idea. Did not get it. Could you explain your idea?

Cjkjvfnby on March 2, 2014, 8:41 p.m.
Using two sets you can common = first_set & second_set but I prefer to use intersection.

lukewrites on Aug. 13, 2014, 5:58 a.m.
Thanks for this answer, I didn't figure out that I could use set() in this way. Why do you prefer intersection() over common()? Is it b/c common() will show repeats?

Cjkjvfnby on Aug. 13, 2014, 7:14 a.m.
I dont get common(), can you link to documentation.

lukewrites on Aug. 13, 2014, 7:16 a.m.
Sorry, typed without thinking…that's something I apparently made up (>.<) Will delete original comment.

ArchTauruS on June 22, 2014, 3:46 p.m.
set() can get rid of word repeatation

Cjkjvfnby on Aug. 13, 2014, 7:11 a.m.
set(...) & set(...) has same effect as set(...).intersection([...]) converting list to set is additional O(n).

inspace_3f5f9a599b0249d5a91220 on Aug. 17, 2014, 6:30 p.m.
Hi. I solved this in a similar way. I have a question: Why did it work only with the 'set()' constructor and not with the set literal notation '{, , }' for set creation? I tried creating the sets with {first.split(",")} but that wasn't working and I had to replace the braces with set(). Any idea? thanks

bryukh on Aug. 17, 2014, 8:56 p.m.
I think because {} is a dictionary.

inspace_3f5f9a599b0249d5a91220 on Aug. 17, 2014, 9:45 p.m.
I see. That would be an empty dict literal, and there's no empty set literal. I was assuming that because a={1,2,3,'foo','bar'} is a set literal, {} would be an empty one, but I was forgetting the dictionaries. Thanks

bryukh on Aug. 17, 2014, 9:48 p.m.
Honestly, when I work with sets too much, I do this error every time too :-)

koborC on Sept. 2, 2021, 9:26 a.m.
No, we can use {...} for <a href="https://docs.python.org/3/library/stdtypes.html#set">creating set</a> too. <blockquote> type( {1, 2, 3, 4} ) # => set type( {1:2, 3:4} ) # => dict </blockquote> What a high-level language...

zelen.danny on May 26, 2018, 6:45 p.m.
Hm, because you should unpack list created by .split method. Another way, you try to make set from one element (list) You can use * before object to unpack iterable object to function parameters. Try this: {*first.split(",")}

donBit on Nov. 11, 2014, 2:05 a.m.
I learn something new with this code, thanks. I wanna ask which is the difference between first_set.intersection(second_set) and use first_set & second_set ?

bryukh on Nov. 11, 2014, 9:39 a.m.
<a href="https://docs.python.org/2/library/sets.html">https://docs.python.org/2/library/sets.html</a>#set-objects Equivalents.

viktor.pecheniuk on Nov. 20, 2014, 2:05 p.m.
the same :) return ','.join(sorted(list(set(first.split(',')).intersection(second.split(',')))))

greenit on Dec. 11, 2014, 8:23 a.m.
TBH I don't think this line is pretty readable or 'clear'. I prefer OPs solution as it is a bit more readable. Just in case someone else wants to read the code. ;-)

viktor.pecheniuk on Dec. 11, 2014, 1:35 p.m.
cool, well done, man!

Cjkjvfnby on Dec. 12, 2014, 11:49 a.m.
<blockquote> I prefer OPs solution as it is a bit more readable. </blockquote> Can you give a link?

Happiness on Feb. 9, 2015, 3:43 p.m.
Is it more readable? :) <pre class='brush: python'>def checkio(words1, words2): return ','.join(sorted(set( words1.split(',')).intersection( set(words2.split(','))))) </pre>

RedheadA3_8469cf456c5840cf9112 on Dec. 9, 2014, 4:41 p.m.
This makes way more sense than what I was doing. I had no idea that intersection existed! Thanks :)

giolefpaceid on Feb. 20, 2015, 10:15 p.m.
Nice solution! Almost the same with mine! :)

Lucy2015 on June 3, 2015, 5:23 a.m.
return ",".join(sorted(common)) Clear and simple! I learn it.

saulwy on June 23, 2015, 8:25 a.m.
common‘s type is set. but set is disorderly, are you sure function sorted() can be used in set type variable.

saulwy on June 23, 2015, 8:26 a.m.
common‘s type is set. but set is disorderly, are you sure function sorted() can be used in set type variable.

fh2game86 on July 29, 2015, 10:27 p.m.
my solution , <a href="http://www.checkio.org/mission/common-words/publications/fh2game86/python-27/first/">http://www.checkio.org/mission/common-words/publications/fh2game86/python-27/first/</a> return ",".join(sorted(item for item in second.split(",") if item in first))

shuai on Oct. 3, 2015, 2:21 a.m.
you don't need two sets to perform intersection

tarjeii on Nov. 1, 2015, 6:32 p.m.
Why write: common = firstset.intersection(secondset) when one can write: common = firstset & secondset ? Is there any difference? Or is it just a matter of taste?

bryukh on Nov. 3, 2015, 8:26 a.m.
<blockquote> Or is it just a matter of taste? </blockquote> Yep.

ljy95135 on May 27, 2016, 2:19 a.m.
sorted(common) Learn it, I use list function for a set which is unnecessary.

Thenbacker on Oct. 2, 2016, 7:07 p.m.
Cool, I didn't know join would return an empty string with an empty set as input, otherwise I would have shortened my code by two lines.

wendychen on July 6, 2017, 1:54 a.m.
Ach so. Use split with set. I came up with split() and set() but didn't use them together... :(

a_l_e_x_e_y_b on July 11, 2017, 5:53 a.m.
Really interesting function

tigerhu3180 on June 25, 2018, 1:10 p.m.
set.intersection(any iterable)...so convenient

chchun on Oct. 1, 2018, 4:07 a.m.
good set operation and intersection

IsaacYang on Oct. 25, 2019, 12:52 a.m.
Mine is similar,but why do you use set instead of list? def checkio(first, second): <pre class='brush: python'>list1=first.split(',') list2=second.split(',') list3=[] for item in list1: if item in list2: list3.append(item) return ','.join(sorted(list3)) </pre> <h1>These "asserts" using only for self-checking and not necessary for auto-testing</h1> if name == 'main': print(checkio("hello,world", "hello,earth")) <pre class='brush: python'>print(checkio("one,two,three", "four,five,six")) print(checkio("one,two,three", "four,five,one,two,six,three")) </pre>

igharok on March 10, 2020, 4:52 p.m.
Cool avoiding of looping lists!

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