Python:为什么下面这段程序只删除1个0？
s=[0,0,5,3]
for x in s:
if x==0
s.remove(x)
结果为什么是[0,5,3]而不是[5,3]呢？
回复内容：
初始s[0] = 0, s[1] = 0, s[2] = 5, s[3] = 3for执行第一次的时候,
x = s[0], 也就是 x = 0 然后将这个元素剔除,
s变成了[0, 5, 3], 也就是 s[0] = 0, s[1] = 5, s[2] = 3
for 执行第二次的时候,
x = s[1], 也就是 x = 5
因为x不是0,s不变
造成出现你这个结果的原因是列表s在运算过程中发生了变化.
合理的方法是将不是0的的元素重新放到一个新列表里面
<span class="k">print</span> <span class="p">[</span><span class="n">x</span> <span class="k">for</span> <span class="n">x</span> <span class="ow">in</span> <span class="n">s</span> <span class="k">if</span> <span class="n">x</span> <span class="o">!=</span> <span class="mi">0</span><span class="p">]</span>
在for x in s:中的s是你迭代的列表, 而在迭代中, 你又使用了s.remove(x), 这一行代码修改了原来的列表s, 这样, 第二次执行"for x in s:"的时候, 这里的s和第一次执行的已经不是同一个列表了, 第一次s = [0, 0, 5, 3], 第二次s = [0, 5, 3], 但是, 他们使用的下标却又是同一组的下标, 即第一次下标为0, 第二次下标为1, 那么, 对应两次的列表s, 可知, 第一次s[0]为0, 第二次s[1]为2(对应第二个s)
所以, 这个很容易犯的错误, 就是因为迭代的列表发生了变化, 因此在python的迭代中, 一定不要修改列表(这里提一个你以后会犯的错误--列表的可变性和多引用, 导致修改其他列表时会改变原列表的问题)
那么, 除了洛克的列表推倒式(python中最好用, 最具特色, 最X炸天的技巧, 你必须要会哦), 其他解法:
1.)新建一个列表result = [], 保存符合条件的值result.append(x), 这样就没有修改原列表了
2.)使用a = s, 然后迭代a, for x in a:, 再使用s.remove(x)修改列表s, 那恭喜你, 你错了, 就是上面提到的多引用问题, 这个自己去好好看文档弄明白吧
最后如果看不太明白, 那分享一句话"书读百遍, 其义自见" @洛克 的解释很正确，但是方法可以改进。
这其实更适合通过 Python 的 filter 来解决：
<span class="c">#!/usr/bin/env python3</span>
<span class="k">def</span> <span class="nf">check_number</span><span class="p">(</span><span class="n">x</span><span class="p">):</span>
<span class="k">if</span> <span class="n">x</span> <span class="o">==</span> <span class="mi">0</span><span class="p">:</span>
<span class="k">return</span> <span class="bp">False</span>
<span class="k">else</span><span class="p">:</span>
<span class="k">return</span> <span class="bp">True</span>
<span class="n">s</span> <span class="o">=</span> <span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">5</span><span class="p">,</span><span class="mi">3</span><span class="p">]</span>
<span class="k">print</span><span class="p">(</span><span class="nb">filter</span><span class="p">(</span><span class="n">check_number</span><span class="p">,</span><span class="n">s</span><span class="p">))</span>
<span class="n">s</span> <span class="o">=</span> <span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">5</span><span class="p">,</span><span class="mi">3</span><span class="p">]</span> <span class="k">print</span><span class="p">(</span><span class="nb">filter</span><span class="p">(</span><span class="k">lambda</span> <span class="n">x</span><span class="p">:</span> <span class="n">x</span> <span class="o">!=</span> <span class="mi">0</span><span class="p">,</span> <span class="n">s</span><span class="p">))</span>
官方文档里有很好的例子，就在2.7版本的list章节。
循环时不实用原列单a[]，而是使用a[:]
这个方法最简单，也不需要增加额外的代码。
另外，@黑溜儿 的问题很好，@洛克 的解释是对的。因为for循环的时候不是先获得列单长度，这样效率太低了。for是通过iter迭代器实现的，所以是动态的按下标迭代直到结尾。
Life is short, I use python. s已经不是原来的s，x确实原来的x 这样解释的话，循环体会执行四次，那第四次应该是数组越界了，为毛不报错？python是怎么个检查机制？
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