在python中定义二维数组
- 2018-02-03 13:32:00
- 六月
- 来源:
- https://www.cnblogs.com/woshare/p/5823303.html
- 转贴 1016
一次偶然的机会,发现python中list非常有意思。
先看一段代码
array = [0, 0, 0] matrix = [array*3] print matrix ## [[0,0,0,0,0,0,0,0,0]]
这段代码其实没有新建一个二维数组
再看一段代码
array = [0, 0, 0] matrix = [array] * 3 print matrix ## [[0, 0, 0], [0, 0, 0], [0, 0, 0]]
咋一看这段代码应该创建一个二维数组了
测试一下
matrix[0][1] = 1 print matrix ## [[0, 1, 0], [0, 1, 0], [0, 1, 0]
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
matrix = [[0, 0, 0], [0, 0, 0], [0, 0, 0]]
方法2 间接定义
matrix = [[0 for i in range(3)] for i in range(3)]