Python / Numpy - 矩阵乘以2D阵列和另一个2D阵列的每一行

What is the best way to do this?

做这个的最好方式是什么？

a = 3x3 array
b = 20x3 array
c = 20x3 array = some_dot_function(a, b) where:
c[0] = np.dot(a, b[0])
c[1] = np.dot(a, b[1])
c[2] = np.dot(a, b[2])
...etc...

I know this can be done with a simple python loop or using numpy's apply_along_axis, but I'm wondering if there is any good way to do this entirely within the underlying C code of numpy. I looked at tensordot and some other functions, but didn't have any luck. I also tried the following:

我知道这可以通过一个简单的python循环或使用numpy的apply_along_axis来完成，但我想知道是否有任何好方法可以在numpy的底层C代码中完成。我查看了tensordot和其他一些功能，但没有任何运气。我也尝试了以下内容：

c = np.dot(a, b[:, :, np.newaxis]
#c.shape = (3, 59, 1)

This actually ran and gave results that looked approximately right, except that the resulting array is not 20x3. I may be able to find a way to reshape it into the array I want, but I figured that there must be an easier/cleaner/clearer built-in method that I'm missing?

这实际上运行并给出了看起来大致正确的结果，除了结果数组不是20x3。我或许可以找到一种方法将其重塑为我想要的数组，但我认为必须有一个更容易/更清晰/更清晰的内置方法，我缺少？

2 个解决方案

#1

This gives (what looks to me like) the correct result:

这给了（看起来像我的样子）正确的结果：

numpy.dot(b, a.T)

Here's some example output:

这是一些示例输出：

>>> a = numpy.arange(9).reshape(3, 3)
>>> b = numpy.arange(60).reshape(20, 3)
>>> numpy.dot(b, a.T)
array([[   5,   14,   23],
       [  14,   50,   86],
       [  23,   86,  149],
       [  32,  122,  212],
       ....

#2

import numpy
a = numpy.arange(9).reshape(3,3)
b = numpy.arange(60).reshape(20,3)
c1 = numpy.dot(b, a.T) # as in the answer of senderle
c2 = numpy.einsum('ji,ki->kj',a,b)

and the resulting c1 and c2 are both the same as what you wish (verified with your c[i] = np.dot(a, b[i]) )

并且得到的c1和c2都与你想要的一样（用c [i] = np.dot（a，b [i]）验证）

the advantage of numpy.einsum is that this trick 'ji,ki->kj' telling what has to be done on what dimension also works for larger dimensions.

numpy.einsum的优势在于，这个技巧'ji，ki-> kj'告诉必须在什么尺寸上做什么也适用于更大尺寸。

more explanation on einsum

关于einsum的更多解释

for example, if you want to do the following operation:

例如，如果要执行以下操作：

a = numpy.arange(60.).reshape(3,4,5)
b = numpy.arange(24.).reshape(4,3,2)
d1 = numpy.zeros((5,2))

for i in range(5):
    for j in range(2):
        for k in range(3):
            for n in range(4):
                d1[i,j] += a[k,n,i] * b[n,k,j]

you can do the same thing much faster by doing:

你可以通过以下方式更快地做同样的事情：

d2 = numpy.einsum('kni,nkj->ij', a, b) 
# the 'kni,nkj->ij' is what you otherwise do with the indices as in 
# d1[i,j] += a[k,n,i] * b[n,k,j]

or if you do not like this way of specifying what has to happen, you can also use numpy.tensordot instead of numpy.einsum, and specify the axes as follows:

或者如果你不喜欢这种指定必须发生的方式，你也可以使用numpy.tensordot而不是numpy.einsum，并指定轴如下：

d3 = numpy.tensordot(a,b, axes=([1,0],[0,1]))

so this einsum method is very general and can be used to shortcut for-loops (that are otherwise slow, if you do them in python), and are very interesting for funky tensor-stuff

所以这个einsum方法非常通用，可用于快捷方式for循环（如果你在python中执行它们会很慢），并且对于时髦的张量东西非常有趣

for more info, see http://docs.scipy.org/doc/numpy/reference/generated/numpy.tensordot.html and http://docs.scipy.org/doc/numpy/reference/generated/numpy.einsum.html

有关详细信息，请参阅http://docs.scipy.org/doc/numpy/reference/generated/numpy.tensordot.html和http://docs.scipy.org/doc/numpy/reference/generated/numpy.einsum html的

#1