Question

我有一个函数应该用“3D”数组表示一个立方体，但是一个图层中间有一个正方形。 I.E.

def cube(n):
    list1 = []
    list2 = []
    list3 = []


    for a in range(n+2):
        list3.append(str('c'))
    for b in range(n+2):
        list2.append(list(list3))
    for c in range(n+2):
        list1.append(list(list2))


    for d in range(1,n+1):
        for e in range(1,n+1):
            list1[0][d][e]='h'
    return list1

pprint.pprint(cube(2))

>>>

[[['c', 'c', 'c', 'c'],
['c', 'h', 'h', 'c'],
['c', 'h', 'h', 'c'],
['c', 'c', 'c', 'c']],

[['c', 'c', 'c', 'c'],
['c', 'h', 'h', 'c'],
['c', 'h', 'h', 'c'],
['c', 'c', 'c', 'c']],

[['c', 'c', 'c', 'c'],
['c', 'h', 'h', 'c'],
['c', 'h', 'h', 'c'],
['c', 'c', 'c', 'c']],

[['c', 'c', 'c', 'c'],
['c', 'h', 'h', 'c'],
['c', 'h', 'h', 'c'],
['c', 'c', 'c', 'c']]]

但我想这样：

>>>

[[['c', 'c', 'c', 'c'],
['c', 'h', 'h', 'c'],
['c', 'h', 'h', 'c'],
['c', 'c', 'c', 'c']],

[['c', 'c', 'c', 'c'],
['c', 'c', 'c', 'c'],
['c', 'c', 'c', 'c'],
['c', 'c', 'c', 'c']],

[['c', 'c', 'c', 'c'],
['c', 'c', 'c', 'c'],
['c', 'c', 'c', 'c'],
['c', 'c', 'c', 'c']],

[['c', 'c', 'c', 'c'],
['c', 'c', 'c', 'c'],
['c', 'c', 'c', 'c'],
['c', 'c', 'c', 'c']]]

仅限第一层中的h。为什么python会这样做？

Answer 1

这是获得您想要的最小变化。首先，添加

from copy import deepcopy

然后替换：

list1.append(list(list2))

使用：

list1.append(deepcopy(list2))

为什么python复制列表在这里？

1 个答案: