blocks|key|3155797|text|[[]]*2是对相同列表的两个引用的列表。您正在附加它，然后看到它两次。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3155798|entityMap^0|0|6|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

<code>[[]]*2</code> is a list of two references to the same list. You are appending to it and then seeing it twice.

blocks|key|258075|text|因为实际上只有一个列表。请考虑以下内容：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|258076|>>>+l+=+[[]]
>>>+l2+=+l*2
>>>+l2[0]+is+l[0]
True
>>>+l2[1]+is+l[0]
True|code-block|syntax|javascript|258077|在列表上执行的*2不会复制列表，但会返回一个长度为2的列表，其中填充了相同的引用。|offset|length|style|CODE|258078|你可能想要的是：|258079|>>>+l+=+[[]+for+_+in+xrange(2)]|258080|正如@Asterisk在评论中提到的那样，所有常见的集合都会公开相同的行为。因此，根据经验，最好只在具有值语义的不可变类型上使用乘法。|258081|entityMap^0|0|0|7|2|P|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@$I|X|J|Y|K|L]|$I|Z|J|10|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|11|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|12|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|T|$]]

Because there is really only one list. Consider this:

<pre><code>&gt;&gt;&gt; l = [[]]
&gt;&gt;&gt; l2 = l*2
&gt;&gt;&gt; l2[0] is l[0]
True
&gt;&gt;&gt; l2[1] is l[0]
True
</code></pre>

<code>*2</code> performed on a list does not copy the list but return a list of length <code>2</code> filled with the same reference.

What you probably wanted was this:

<pre><code>&gt;&gt;&gt; l = [[] for _ in xrange(2)]
</code></pre>

As @Asterisk mentions in a comment, the same behaviour is exposed by all common collections. As a rule of thumb it is therefore best to only use multiplication on immutable types with value-semantics.

blocks|key|254198|text|下面是我初始化列表列表的方法。行的变化最慢。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|254199|nrows+=+3;+ncols+=+5

l_of_ls+=+[[0]*ncols+for+i+in+range(nrows+)]

for+rix,+r+in+enumerate(l_of_ls):
++++for+cix,+c+in+enumerate(r):
++++++++print+rix,+cix,+'val+=+',c|code-block|syntax|javascript|254200|结果|offset|length|style|BOLD|254201|0+0+val+=++0
0+1+val+=++0
0+2+val+=++0
0+3+val+=++0
0+4+val+=++0
1+0+val+=++0
1+1+val+=++0
1+2+val+=++0
1+3+val+=++0
1+4+val+=++0
2+0+val+=++0
2+1+val+=++0
2+2+val+=++0
2+3+val+=++0
2+4+val+=++0+|254202|出于索引目的，也值得注意|254203|for+rix+in+range(nrows):
++++for+cix+in+range(ncols):
++++++++print+l_of_ls[rix][cix],
++++print|254204|254205|0+0+0+0+0
0+0+0+0+0
0+0+0+0+0++|254206|entityMap^0|0|0|0|2|0|0|7|5|0|0|0|2|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Z|8|@$I|10|J|11|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|12|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|13|8|@$I|14|J|15|K|L]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|16|8|@]|9|@]|A|$E|F]]|$1|S|3|H|5|6|7|17|8|@$I|18|J|19|K|L]]|9|@]|A|$]]|$1|T|3|U|5|D|7|1A|8|@]|9|@]|A|$E|F]]|$1|V|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|W|$]]

Here's how i initialize a list of lists. Rows vary slowest.

<pre><code>nrows = 3; ncols = 5

l_of_ls = [[0]*ncols for i in range(nrows )]

for rix, r in enumerate(l_of_ls):
 for cix, c in enumerate(r):
 print rix, cix, 'val = ',c
</code></pre>

RESULT

<pre><code>0 0 val = 0
0 1 val = 0
0 2 val = 0
0 3 val = 0
0 4 val = 0
1 0 val = 0
1 1 val = 0
1 2 val = 0
1 3 val = 0
1 4 val = 0
2 0 val = 0
2 1 val = 0
2 2 val = 0
2 3 val = 0
2 4 val = 0 
</code></pre>

Also Worth Noting for Indexing Purposes

<pre><code>for rix in range(nrows):
 for cix in range(ncols):
 print l_of_ls[rix][cix],
 print
</code></pre>

RESULT

<pre><code>0 0 0 0 0
0 0 0 0 0
0 0 0 0 0 
</code></pre>

blocks|key|256013|text|展示内存布局差异的：|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|256014|listOfLists+=+[[]]+*+3
listOfListsRange+=+[[]+for+i+in+range(0,+3)]|code-block|syntax|javascript|256015|​|256016|📷|atomic|256017|256018|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/2433ca4216056b2fb60f57be2763988f.jpeg|imageAlt^0|9|1|0|0|0|0|1|0|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|13|8|@]|D|@]|E|$]]|$1|M|3|N|5|O|7|14|8|@]|D|@$9|15|A|16|1|17]]|E|$]]|$1|P|3|L|5|6|7|18|8|@]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|19|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X|Y|-4]]]]

Showcasing the difference with memory layout:

<pre><code>listOfLists = [[]] * 3
listOfListsRange = [[] for i in range(0, 3)]
</code></pre>

<a href="https://i.stack.imgur.com/hVozf.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hVozf.jpg" alt="enter image description here"></a>

Don't really know how to formulate the question...

Suppose I do the following:

<pre><code> &gt;&gt;&gt; l = [[]]*2
 &gt;&gt;&gt; l
 [[], []]
 &gt;&gt;&gt; l[0].append(1)
 &gt;&gt;&gt; l
 [[1], [1]]
</code></pre>

Why does 1 gets appended to both lists?

Python list append behavior

我真的不知道如何表达这个问题。假设我执行以下操作：    >>> l = [[]]*2    >>> l    [[], []]    >>> l[0].append(1)    >>> l    [[1], [1]]为什么1会同时附加到两个列表？

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