numpy：按列点积

发布于 2024-11-13 09:48:54 字数 420 浏览 1 评论 0原文

给定一个 2D numpy 数组，我需要计算每列与其自身的点积，并将结果存储在 1D 数组中。以下工作原理：

In [45]: A = np.array([[1,2,3,4],[5,6,7,8]])

In [46]: np.array([np.dot(A[:,i], A[:,i]) for i in xrange(A.shape[1])])
Out[46]: array([26, 40, 58, 80])

有没有一种简单的方法来避免 Python 循环？上面的内容并不是世界末日，但如果有一个 numpy 原语可以实现这一点，我想使用它。

编辑实际上，矩阵有很多行和相对较少的列。因此，我不太热衷于创建大于 O(A.shape[1]) 的临时数组。我也无法就地修改 A 。

原文

Given a 2D numpy array, I need to compute the dot product of every column with itself, and store the result in a 1D array. The following works:

In [45]: A = np.array([[1,2,3,4],[5,6,7,8]])

In [46]: np.array([np.dot(A[:,i], A[:,i]) for i in xrange(A.shape[1])])
Out[46]: array([26, 40, 58, 80])

Is there a simple way to avoid the Python loop? The above is hardly the end of the world, but if there's a numpy primitive for this, I'd like to use it.

edit In practice the matrix has many rows and relatively few columns. I am therefore not overly keen on creating temporary arrays larger than O(A.shape[1]). I also can't modify A in place.

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糖粟与秋泊 2024-11-20 09:48:54

怎么样：

>>> A = np.array([[1,2,3,4],[5,6,7,8]])
>>> (A*A).sum(axis=0)
array([26, 40, 58, 80])

编辑：嗯，好吧，你不需要中间的大对象。也许：

>>> from numpy.core.umath_tests import inner1d
>>> A = np.array([[1,2,3,4],[5,6,7,8]])
>>> inner1d(A.T, A.T)
array([26, 40, 58, 80])

无论如何，这似乎有点快。这应该在幕后执行您想要的操作，因为 AT 是一个视图（它不会创建自己的副本，IIUC），并且inner1d似乎按照它需要的方式循环。

非常迟来的更新：另一种选择是使用 np.einsum ：

>>> A = np.array([[1,2,3,4],[5,6,7,8]])
>>> np.einsum('ij,ij->j', A, A)
array([26, 40, 58, 80])
>>> timeit np.einsum('ij,ij->j', A, A)
100000 loops, best of 3: 3.65 us per loop
>>> timeit inner1d(A.T, A.T)
100000 loops, best of 3: 5.02 us per loop
>>> A = np.random.randint(0, 100, (2, 100000))
>>> timeit np.einsum('ij,ij->j', A, A)
1000 loops, best of 3: 363 us per loop
>>> timeit inner1d(A.T, A.T)
1000 loops, best of 3: 848 us per loop
>>> (np.einsum('ij,ij->j', A, A) == inner1d(A.T, A.T)).all()
True

How about:

>>> A = np.array([[1,2,3,4],[5,6,7,8]])
>>> (A*A).sum(axis=0)
array([26, 40, 58, 80])

EDIT: Hmm, okay, you don't want intermediate large objects. Maybe:

>>> from numpy.core.umath_tests import inner1d
>>> A = np.array([[1,2,3,4],[5,6,7,8]])
>>> inner1d(A.T, A.T)
array([26, 40, 58, 80])

which seems a little faster anyway. This should do what you want behind the scenes, as A.T is a view (which doesn't make its own copy, IIUC), and inner1d seems to loop the way it needs to.

VERY BELATED UPDATE: Another alternative would be to use np.einsum:

>>> A = np.array([[1,2,3,4],[5,6,7,8]])
>>> np.einsum('ij,ij->j', A, A)
array([26, 40, 58, 80])
>>> timeit np.einsum('ij,ij->j', A, A)
100000 loops, best of 3: 3.65 us per loop
>>> timeit inner1d(A.T, A.T)
100000 loops, best of 3: 5.02 us per loop
>>> A = np.random.randint(0, 100, (2, 100000))
>>> timeit np.einsum('ij,ij->j', A, A)
1000 loops, best of 3: 363 us per loop
>>> timeit inner1d(A.T, A.T)
1000 loops, best of 3: 848 us per loop
>>> (np.einsum('ij,ij->j', A, A) == inner1d(A.T, A.T)).all()
True

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原谅我要高飞 2024-11-20 09:48:54

您可以计算所有元素的平方并使用逐列求和

np.sum(np.square(A),0);

（我不完全确定 sum 函数的第二个参数，它标识求和所沿的轴，并且我当前没有安装 numpy，也许你必须尝试一下:) ...)

编辑

看看DSM的帖子，似乎你应该使用轴=0。使用 square 函数可能比使用 A*A 的性能更高一些。

You can compute the square of all elements and sum up column-wise using

np.sum(np.square(A),0);

(I'm not entirely sure about the second parameter of the sum function, which identifies the axis along which to take the sum, and I have no numpy currently installed. Maybe you'll have to experiment :) ...)

EDIT

Looking at DSM's post, it seems that you should use axis=0. Using the square function might be a little more performant than using A*A.

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自由如风 2024-11-20 09:48:54

根据线性代数，第 i 行与第 j 行的点积是 AA^T 的第 i,j 个条目。类似地，第 i 列与第 j 列的点积是 (A^T)A 的第 i,j 个条目。

因此，如果您想要 A 的每个列向量与其自身的点积，可以使用 ColDot = np.dot(np.transpose(A), A).diagonal()。另一方面，如果您想要每行与其自身的点积，则可以使用 RowDot = np.dot(A, np.transpose(A)).diagonal()。

这两行都返回一个数组。

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