在numpy中生成对称矩阵的语义 [英] semantics of generating symmetric matrices in numpy

问题描述

我试图制作一个随机对称矩阵来测试我的程序.只要对称，我就根本不在乎数据(完全没有随机性).

I tried to make a random symmetric matrix to test my program. I don't care about the data at all as long as it is symmetric (sufficient randomness is no concern at all).

我的第一次尝试是

x=np.random.random((100,100))
x+=x.T

但是，np.all(x==x.T)返回False. print x==x.T产量

However, np.all(x==x.T) returns False. print x==x.T yields

array([[ True,  True,  True, ..., False, False, False],
   [ True,  True,  True, ..., False, False, False],
   [ True,  True,  True, ..., False, False, False],
   ..., 
   [False, False, False, ...,  True,  True,  True],
   [False, False, False, ...,  True,  True,  True],
   [False, False, False, ...,  True,  True,  True]], dtype=bool)

我尝试运行一个n = 10的小型测试示例，以了解发生了什么，但是该示例可以正常运行.

I tried to run a small test example with n=10 to see what was going on, but that example works just as you would expect it to.

如果我改为这样做:

x=np.random.random((100,100))
x=x+x.T

然后就可以了.

这是怎么回事?这些语句在语义上不是等效的吗?有什么区别?

What's going on here? Aren't these statements semantically equivalent? What's the difference?

推荐答案

这些语句在语义上并不等效. x.T返回原始数组的 view .在+=情况下，实际上是在遍历x时更改了x的值(这会更改x.T的值).

Those statements aren't semantically equivalent. x.T returns a view of the original array. in the += case, you're actually changing the values of x as you iterate over it (which changes the values of x.T).

这样想吧...当您的算法索引到(3,4)时，它在伪代码中看起来像这样:

Think of it this way ... When your algorithm gets to index (3,4), it looks something like this in pseudocode:

x[3,4] = x[3,4] + x[4,3]

现在，当您迭代到(4,3)时，您便会

now, when your iteration gets to (4,3), you do

x[4,3] = x[4,3] + x[3,4]

但是，x[3,4]不是您开始迭代时的样子.

but, x[3,4] is not what it was when you started iterating.

在第二种情况下，您实际上创建了一个新的(空)数组并更改了空数组中的元素(切勿写入x).因此，伪代码如下所示:

In the second case, you actually create a new (empty) array and change the elements in the empty array (never writing to x). So the pseudocode looks something like:

y[3,4] = x[3,4] + x[4,3]

和

y[4,3] = x[4,3] + x[3,4]

这显然会给您一个对称矩阵(y.

which obviously will give you a symmetric matrix (y.

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