在简化代码时最有优势的是 Python 吗？

如打印1-100中可被2和3整除的總和
最一般的情況可能这样写

<span class="n">TheSum</span><span class="o">=</span><span class="mi">0</span>
<span class="k">for</span> <span class="n">x</span> <span class="ow">in</span> <span class="nb">xrange</span><span class="p">(</span><span class="mi">101</span><span class="p">):</span>
    <span class="k">if</span> <span class="n">x</span> <span class="o">%</span><span class="mi">2</span><span class="o">==</span><span class="mi">0</span> <span class="ow">and</span> <span class="n">x</span><span class="o">%</span><span class="mi">3</span><span class="o">==</span><span class="mi">0</span><span class="p">:</span>
        <span class="n">TheSum</span><span class="o">+=</span><span class="n">x</span>
<span class="k">print</span> <span class="n">TheSum</span>

回复内容：

不是

J

+/ (#~ (0=2&|) *. (0=3&|)) 1+i.100

Scala:

6 to 100 by 6 sum

不妨考虑一下可读性吧 sum(range(6,101,6))

1 to(100) asList select(%6==0) sum

Haskell ：sum [x | x

<span class="nf">sum</span> <span class="p">[</span><span class="n">x</span> <span class="o">|</span> <span class="n">x</span> <span class="ow"><-</span> <span class="p">[</span><span class="mi">1</span><span class="o">..</span><span class="mi">101</span><span class="p">],</span> <span class="n">x</span> <span class="p">`</span><span class="n">mod</span><span class="p">`</span> <span class="mi">2</span> <span class="o">==</span> <span class="mi">0</span> <span class="o">&&</span> <span class="n">x</span> <span class="p">`</span><span class="n">mod</span><span class="p">`</span> <span class="mi">3</span> <span class="o">==</span> <span class="mi">0</span><span class="p">]</span>

可以被2和3整除，不就是被6整除嘛
sum [6,12..101] Ruby 也好简单

(1..100).select{|x| x%6 == 0 }.inject(:+)

这种东西就是比语法糖和标准库，比出来也是意义不大。。。 sum [6,12..100]


旁边那些好意思说自己写的是Haskell么…好歹也写成下面这样吧…

sum . filter ((==0).('mod' 6)) $ [1..100]


还有那些把[1..100]写成[0..101]的那些泥们垢了！有审过题么，就答？知乎都这样了还怎么玩耍...