python 用list of lists表示矩阵的问题？

题主刷leetCode的时候发现的一个小情况>_
python 初始化 a list of lists of integer，就用了如下方式（假设是4 x 4的方阵）：
n = 4
matrix = [[0]*n]*n
print matrix
[[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
然后诡异的是，假如我只想给第二行的中间两个元素赋值：
matrix[1][1:3] = [1, 2]
结果会是：
print matrix
[[0, 1, 2, 0], [0, 1, 2, 0], [0, 1, 2, 0], [0, 1, 2, 0]]

但是如果用如下方式初始化matrix:
matrix = [[0 for col in range(n)] for row in range(n)]
matrix[1][1:3] = [1, 2]
print matrix
[[0, 0, 0, 0], [0, 1, 2, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
会是我想要的答案。

但是不知道第一种初始化方式错在哪里...

回复内容：

<code class="language-text">matrix = [[0]*n]*n
</code>

<code class="language-python"><span class="c">#!/usr/bin/python</span>
<span class="c">#encoding=utf-8</span>

<span class="c"># 内容来自python cookbook 第二版 第4.5章节</span>
<span class="c"># 在无须共享引用的条件下创建列表的列表</span>
<span class="c"># 避免隐式的引用共享</span>

<span class="n">multi</span> <span class="o">=</span> <span class="p">[</span> <span class="p">[</span><span class="mi">0</span><span class="p">]</span> <span class="o">*</span> <span class="mi">5</span> <span class="p">]</span> <span class="o">*</span> <span class="mi">3</span> 
<span class="k">print</span> <span class="n">multi</span> 
<span class="n">multi</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="o">=</span> <span class="s">'oops'</span>
<span class="k">print</span> <span class="n">multi</span>
<span class="c"># [ [ 'oops', 0, 0, 0, 0 ],[ 'oops', 0, 0, 0, 0],[ 'oops', 0, 0, 0, 0 ] ] </span>

<span class="c"># 等价方式</span>
<span class="n">row</span> <span class="o">=</span> <span class="p">[</span><span class="mi">0</span><span class="p">]</span> <span class="o">*</span> <span class="mi">5</span> <span class="c"># row 列表中的5个子项都是引用0</span>
<span class="n">multi</span> <span class="o">=</span> <span class="n">row</span> <span class="o">*</span> <span class="mi">3</span> <span class="c"># multi 列表中的3个子项都是引用row</span>
<span class="c"># 解释: 在row创建中， 有无引用被复制完全不重要， 因为被引用的</span>
<span class="c"># 是数字, 而数字不可改变，换句话说，如果对象是不可改变的，则</span>
<span class="c"># 对象和对对象的引用实际没有区别。</span>
<span class="c"># multi创建中，包含了3个对[row] 内容引用，而其内容则是对一个</span>
<span class="c"># 列表的引用。因此修改时候其他3个引用也改变了，甚至row也改变</span>

<span class="c"># 解决方法:</span>
<span class="n">multilist_method1</span> <span class="o">=</span> <span class="p">[</span> <span class="p">[</span> <span class="mi">0</span> <span class="k">for</span> <span class="n">col</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">5</span><span class="p">)</span> <span class="p">]</span> <span class="k">for</span> <span class="n">row</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">3</span><span class="p">)</span> <span class="p">]</span>
<span class="n">multilist_method2</span> <span class="o">=</span> <span class="p">[</span> <span class="p">[</span><span class="mi">0</span><span class="p">]</span> <span class="o">*</span> <span class="mi">5</span> <span class="k">for</span> <span class="n">row</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">3</span><span class="p">)</span> <span class="p">]</span>
<span class="n">multilist_method1</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="o">=</span> <span class="s">'abc'</span>
<span class="n">multilist_method2</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="o">=</span> <span class="s">'edf'</span>
<span class="k">print</span> <span class="n">multilist_method1</span>
<span class="k">print</span> <span class="n">multilist_method2</span>
</code>

赞同楼上，另外要处理矩阵相关的话用 NumPy 第一种是浅拷贝 这个问题，在《Python cookbook》第二版p148页提到，即隐式的引用共享问题。
解决的方案是使用列表推导。
详细你可以看看书。（不好意思手机码字懒得写太多了，希望能解决你的问题。） 刷leetcode用python坑多啊，昨天刚遇到一个python的除法和余数机制和c不一样啊