python建立二维数组问题

淡淡的烟草味﹌ 2022-06-17 05:24 213阅读 0赞

# [在python中定义二维数组][python] #

# 发表于 http://liamchzh.0fees.net/?p=234&i=1 #

一次偶然的机会，发现python中list非常有意思。

先看一段代码

\[py\]  
array = \[0, 0, 0\]  
matrix = \[array\*3\]  
print matrix  
\#\# \[\[0,0,0,0,0,0,0,0,0\]\]\[/py\]

这段代码其实没有新建一个二维数组

再看一段代码

\[py\]  
array = \[0, 0, 0\]  
matrix = \[array\] \* 3  
print matrix  
\#\# \[\[0, 0, 0\], \[0, 0, 0\], \[0, 0, 0\]\]\[/py\]

咋一看这段代码应该创建一个二维数组了

测试一下

\[py\]  
matrix\[0\]\[1\] = 1  
print matrix  
\#\# \[\[0, 1, 0\], \[0, 1, 0\], \[0, 1, 0\]\]\[/py\]

照理matrix\[0\]\[1\]修改的应该只是二维数组中的一个元素，但是测试结果表明，修改的是每个List的第二个元素。  
有问题看文档，然后我找到了[The Python Standard Library][]  
其中[5.6. Sequence Types][]是这样描述的：

Note also that **the copies are shallow; nested structures are not copied**. This often haunts new Python programmers; consider:  
>>> lists = \[\[\]\] \* 3  
>>> lists  
\[\[\], \[\], \[\]\]  
>>> lists\[0\].append(3)  
>>> lists  
\[\[3\], \[3\], \[3\]\]

What has happened is that \[\[\]\] is a one-element list containing an empty list, so all three elements of \[\[\]\] \* 3 are (pointers to) this single empty list. Modifying any of the elements of lists modifies this single list. You can create a list of different lists this way:

>>>  
>>> lists = \[\[\] for i in range(3)\]  
>>> lists\[0\].append(3)  
>>> lists\[1\].append(5)  
>>> lists\[2\].append(7)  
>>> lists  
\[\[3\], \[5\], \[7\]\]  
也就是说matrix = \[array\] \* 3操作中，只是创建3个指向array的引用，所以一旦array改变，matrix中3个list也会随之改变。

那如何才能在python中创建一个二维数组呢？  
例如创建一个3\*3的数组  
方法1 直接定义

\[py\]matrix = \[\[0, 0, 0\], \[0, 0, 0\], \[0, 0, 0\]\]\[/py\]

方法2 间接定义

matrix = \[\[0 for i in range(3)\] for i in range(3)\]

[python]: http://www.cnblogs.com/woshare/p/5823303.html
[The Python Standard Library]: http://docs.python.org/2/library/index.html
[5.6. Sequence Types]: http://docs.python.org/2/library/stdtypes.html#sequence-types-str-unicode-list-tuple-bytearray-buffer-xrange