blocks|key|442541|text|如果你坚持使用一行代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|442542|[(x,+concat+$+["fizz"+%7C+mod+x+3+==+0]+%2B%2B+["buzz"+%7C+mod+x+5+==+0])+%7C+x+<-+[1..100]]|code-block|syntax|javascript|442543|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you insist on a one-liner:

<pre><code>[(x, concat $ ["fizz" | mod x 3 == 0] ++ ["buzz" | mod x 5 == 0]) | x &lt;- [1..100]]
</code></pre>

blocks|key|419377|text|我们不需要恶臭的mod..。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|419378|zip+[1..100]+$+zipWith+(%2B%2B)+(cycle+["","","fizz"])+(cycle+["","","","","buzz"])|code-block|syntax|javascript|419379|或者稍微短一点|419380|import+Data.Function(on)

zip+[1..100]+$+(zipWith+(%2B%2B)+`on`+cycle)+["","","fizz"]+["","","","","buzz"]|419381|或者使用暴力的方式：|419382|zip+[1..100]+$+cycle+["","","fizz","","buzz","fizz","","","fizz","buzz","","fizz","","","fizzbuzz"]|419383|entityMap^0|8|3|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|10|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|T|$]]

We need no stinkin' <code>mod</code>...

<pre><code>zip [1..100] $ zipWith (++) (cycle ["","","fizz"]) (cycle ["","","","","buzz"])
</code></pre>

or slightly shorter

<pre><code>import Data.Function(on)

zip [1..100] $ (zipWith (++) `on` cycle) ["","","fizz"] ["","","","","buzz"]
</code></pre>

Or the brute force way:

<pre><code>zip [1..100] $ cycle ["","","fizz","","buzz","fizz","","","fizz","buzz","","fizz","","","fizzbuzz"]
</code></pre>

blocks|key|441462|text|不如..。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|441463|fizzBuzz++=++[(x,+fizz+x+%2B%2B+buzz+x)+%7C+x+<-+[1..100]]
++where+fizz+n+%7C+n+`mod`+3+==+0++=++"fizz"
+++++++++++++++%7C+otherwise+++++++=++""
++++++++buzz+n+%7C+n+`mod`+5+==+0++=++"buzz"
+++++++++++++++%7C+otherwise+++++++=++""|code-block|syntax|javascript|441464|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

blocks|key|441512|text|我情不自禁地转向另一个方向，让事情变得更加复杂。听着没有mod..。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|441513|merge+as@(a@(ia,sa):as')+bs@(b@(ib,sb):bs')+=
++case+compare+ia+ib+of
++++LT+->+a+:+merge+as'+bs
++++GT+->+b+:+merge+as++bs'
++++EQ+->+(ia,+sa%2B%2Bsb)+:+merge+as'+bs'
merge+as+bs+=+as+%2B%2B+bs

zz+(n,s)+=+[(i,+s)+%7C+i+<-+[n,2*n..]]
fizzBuzz+=+foldr+merge+[]+$+map+zz+[(1,""),+(3,"fizz"),+(5,"buzz")]|code-block|syntax|javascript|441514|entityMap^0|S|3|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Couldn't resist going in the other direction and making it more complicated. Look, no <code>mod</code>...

<pre><code>merge as@(a@(ia,sa):as') bs@(b@(ib,sb):bs') =
 case compare ia ib of
 LT -&gt; a : merge as' bs
 GT -&gt; b : merge as bs'
 EQ -&gt; (ia, sa++sb) : merge as' bs'
merge as bs = as ++ bs

zz (n,s) = [(i, s) | i &lt;- [n,2*n..]]
fizzBuzz = foldr merge [] $ map zz [(1,""), (3,"fizz"), (5,"buzz")]
</code></pre>

blocks|key|4457415|text|与拉尔斯曼的答案相同：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4457416|fizzBuzz+=+[(x,+f+3+"fizz"+x+%2B%2B+f+5+"buzz"+x)+%7C+x+<-+[1..100]]
++where+f+k+s+n+%7C+n+`mod`+k+==+0+=+s
++++++++++++++++%7C+otherwise++++++=+""|code-block|syntax|javascript|4457417|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Along the same lines as larsmans' answer:

<pre><code>fizzBuzz = [(x, f 3 "fizz" x ++ f 5 "buzz" x) | x &lt;- [1..100]]
 where f k s n | n `mod` k == 0 = s
 | otherwise = ""
</code></pre>

blocks|key|1458178|text|我认为你感觉像是在对抗语法的原因是因为你混合了太多的类型。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1458179|而不是尝试打印：|1458180|[(1,+""),+(2,""),+(3,"Fizz")...]|code-block|syntax|javascript|1458181|只需考虑打印字符串：|1458182|["1","2","Fizz"...]|1458183|我的尝试是：|1458184|Prelude>+let+fizzBuzz+x+%7C+x+`mod`+15+==+0+=+"FizzBuzz"+%7C+x+`mod`+5+==+0+=+"Buzz"+%7C+x+`mod`+3+==+0+=+"Fizz"+%7C+otherwise+=+show+x
Prelude>+[fizzBuzz+x+%7C+x+<-[1..100]]

["1","2","Fizz","4","Buzz","Fizz","7","8","Fizz","Buzz","11","Fizz","13","14","FizzBuzz"...]|1458185|为了将Int转换为字符串，您可以使用：|1458186|show+x|1458187|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|11|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|T|5|F|7|14|8|@]|9|@]|A|$G|H]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

I think the reason why you feel like you are fighting the syntax is because you are mixing too many types.

Instead of trying to print: 

<pre><code>[(1, ""), (2,""), (3,"Fizz")...]
</code></pre>

Just think of printing strings:

<pre><code>["1","2","Fizz"...]
</code></pre>

My attempt:

<pre><code>Prelude&gt; let fizzBuzz x | x `mod` 15 == 0 = "FizzBuzz" | x `mod` 5 == 0 = "Buzz" | x `mod` 3 == 0 = "Fizz" | otherwise = show x
Prelude&gt; [fizzBuzz x | x &lt;-[1..100]]

["1","2","Fizz","4","Buzz","Fizz","7","8","Fizz","Buzz","11","Fizz","13","14","FizzBuzz"...]
</code></pre>

In order to convert an Int to String you use the:

<pre><code>show x
</code></pre>

blocks|key|1458251|text|只是为了学习|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1458252|zipWith+(\a+b+->+b+a)+(map+show+[1..100])+$+cycle+[id,id,const+"fizz",id,const+"buzz",const+"fizz",id,id,const+"fizz",const+"buzz",id,const+"fizz",id,id,const+"fizzbuzz"]|code-block|syntax|javascript|1458253|产生|1458254|["1","2","fizz","4","buzz","fizz","7","8","fizz","buzz","11","fizz","13","14","fizzbuzz","16","17","fizz","19","buzz","fizz","22","23","fizz","buzz","26","fizz","28","29","fizzbuzz","31","32","fizz","34","buzz","fizz","37","38","fizz","buzz","41","fizz","43","44","fizzbuzz","46","47","fizz","49","buzz","fizz","52","53","fizz","buzz","56","fizz","58","59","fizzbuzz","61","62","fizz","64","buzz","fizz","67","68","fizz","buzz","71","fizz","73","74","fizzbuzz","76","77","fizz","79","buzz","fizz","82","83","fizz","buzz","86","fizz","88","89","fizzbuzz","91","92","fizz","94","buzz","fizz","97","98","fizz","buzz"]|1458255|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Just for studying

<pre><code>zipWith (\a b -&gt; b a) (map show [1..100]) $ cycle [id,id,const "fizz",id,const "buzz",const "fizz",id,id,const "fizz",const "buzz",id,const "fizz",id,id,const "fizzbuzz"]
</code></pre>

produces

<pre><code>["1","2","fizz","4","buzz","fizz","7","8","fizz","buzz","11","fizz","13","14","fizzbuzz","16","17","fizz","19","buzz","fizz","22","23","fizz","buzz","26","fizz","28","29","fizzbuzz","31","32","fizz","34","buzz","fizz","37","38","fizz","buzz","41","fizz","43","44","fizzbuzz","46","47","fizz","49","buzz","fizz","52","53","fizz","buzz","56","fizz","58","59","fizzbuzz","61","62","fizz","64","buzz","fizz","67","68","fizz","buzz","71","fizz","73","74","fizzbuzz","76","77","fizz","79","buzz","fizz","82","83","fizz","buzz","86","fizz","88","89","fizzbuzz","91","92","fizz","94","buzz","fizz","97","98","fizz","buzz"]
</code></pre>

blocks|key|4457641|text|编写器monad可能看起来不错(如果你不喜欢concat)：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4457642|fizzBuzz+=+[(x,+execWriter+$+when+(x+`mod`+3+==+0)+(tell+"fizz")+>>+when+(x+`mod`+5+==+0)++(tell+"buzz"))+%7C+x+<-+[1..100]]|code-block|syntax|javascript|4457643|不过，它并不是特别简洁。|4457644|entityMap^0|M|6|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

Writer monad may look nice (if you don't like <code>concat</code>):

<pre><code>fizzBuzz = [(x, execWriter $ when (x `mod` 3 == 0) (tell "fizz") &gt;&gt; when (x `mod` 5 == 0) (tell "buzz")) | x &lt;- [1..100]]
</code></pre>

It's not particularly succinct though.

I'm still learning Haskell, and I was wondering if there is a less verbose way to express the below statement using 1 line of code:

<pre><code>map (\x -&gt; (x, (if mod x 3 == 0 then "fizz" else "") ++ 
 if mod x 5 == 0 then "buzz" else "")) [1..100]
</code></pre>

Produces:
<code>[(1,""),(2,""),(3,"fizz"),(4,""),(5,"buzz"),(6,"fizz"),(7,""),(8,""),(9,"fizz"),(10,"buzz"),(11,""),(12,"fizz"),(13,""),(14,""),(15,"fizzbuzz"),(16,""),(17,""),(18,"fizz"),(19,""),(20,"buzz"),(21,"fizz"),(22,""),(23,""),(24,"fizz"),(25,"buzz"),(26,""),(27,"fizz"),(28,""),(29,""),(30,"fizzbuzz")</code>, etc

It just feels like I'm fighting the syntax more than I should. I've seen other questions for this in Haskell, but I'm looking for the most optimal way to express this in a single statement (trying to understand how to work the syntax better).

FizzBuzz cleanup

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我还在学习Haskell，我想知道是否有一种不那么冗长的方式来用一行代码来表达下面的语句：map (\x -> (x, (if mod x 3 == 0 then "fizz" else "") ++  if mod x 5 == 0 then "buzz" else "")) [1..100]产品：[(1,""),(2,""),(3,"fizz"),(4,""),(5,"buzz"),(6,"f

问FizzBuzz清理
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