Python 二维查询..

发布于 2024-11-26 22:48:40 字数 364 浏览 0 评论 0原文

我在 python 中看到了一个非常不寻常的行为。请让我知道我做错了什么！

bc = [[0]*(n+1)]*(n+1)

for i in range(n+1):
    bc[i][i] = 1

print (bc)

输出

[[1, 1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1]]

我试图将二维数组的对角线元素初始化为 1，但它用 1 初始化所有元素。我认为我在访问二维数组时做错了。

另外，请让我知道如何使用两个循环来访问二维数组的所有元素..我的下一步..

谢谢。

原文

I am seeing a very unusual behavior in python.. Kindly let me know what am i doing wrong!!

bc = [[0]*(n+1)]*(n+1)

for i in range(n+1):
    bc[i][i] = 1

print (bc)

Output

[[1, 1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1]]

I am trying to initialize the diagonal elements of two dimensional array to 1, but it is initializing all the elements with 1. I think I am doing something wrong with accessing two dimensional Array..

Also, kindly let me know how can I use two loops to access all the elements of two dimensional array.. my next step..

Thanks.

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葬心 2024-12-03 22:48:40

您的数组初始化不正确。初始化二维数组的正确方法是这样的：

bc = [[0 for i in xrange(n + 1)] for i in xrange(n + 1)]

这是一个常见的错误，但是 * 运算符将指针复制到列表而不是复制列表，因此虽然看起来有一个二维列表，但实际上有一个一维列表指向同一个列表的指针。

Your array is initialized incorrectly. The correct way to initialize a 2d array is this:

bc = [[0 for i in xrange(n + 1)] for i in xrange(n + 1)]

It's a common mistake, but the * operator copies the pointer to an list rather than copying the list, so while it looks like you have a 2d list, you actually have a 1d list of pointers to the same list.

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じее 2024-12-03 22:48:40

问题是数组中的每个数组都是内存中的同一个数组。你每次都需要一个新的数组。例如，[[0]]*6 将在一个数组中创建 6 个相同的数组，编辑其中一个将更新其他数组。

例如

>>> x=[1]
>>> y=x
>>> x.append(3)
>>> x
[1, 3]
>>> y
[1, 3]
>>> z=[x]*3
>>> x.append(6)
>>> z
[[1, 3, 4, 6], [1, 3, 4, 6], [1, 3, 4, 6]]

，这里通过简单地将 bc 编辑为 n+1 个不同的数组来修复：

n=4
bc = [[0]*(n+1) for i in range(n+1)]

for i in range(n+1):
    bc[i][i] = 1

print (bc)

[[1, 0, 0, 0, 0], [0, 1, 0, 0, 0], [0, 0, 1, 0, 0], [0, 0, 0, 1, 0], [0, 0, 0, 0, 1]]

the problem is that each array in your array is the same array in memory. you need a new array each time. the [[0]]*6 will for example make 6 of the same arrays in an array, editing one of them will update the other ones.

e.g.

>>> x=[1]
>>> y=x
>>> x.append(3)
>>> x
[1, 3]
>>> y
[1, 3]
>>> z=[x]*3
>>> x.append(6)
>>> z
[[1, 3, 4, 6], [1, 3, 4, 6], [1, 3, 4, 6]]

here is a fix by simply editing bc to be n+1 different arrays:

n=4
bc = [[0]*(n+1) for i in range(n+1)]

for i in range(n+1):
    bc[i][i] = 1

print (bc)

[[1, 0, 0, 0, 0], [0, 1, 0, 0, 0], [0, 0, 1, 0, 0], [0, 0, 0, 1, 0], [0, 0, 0, 0, 1]]

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寂寞陪衬 2024-12-03 22:48:40

试试这个：

bc = [[0 for i in range(n+1)] for j in range(n+1)]

在您的示例中，您只有一个 (!) [0] 实例，该实例被多次引用。因此，如果您更改该实例，所有引用都会更改。

Try this one:

bc = [[0 for i in range(n+1)] for j in range(n+1)]

In your example you have only one (!) instance of [0] which is referenced multiple times. So if you change that instance, all references are changed.

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