Python numpy 矩阵向量乘法

numpy matrix vector multiplication

提问by user3272574

When I multiply two numpy arrays of sizes (n x n)*(n x 1), I get a matrix of size (n x n). Following normal matrix multiplication rules, a (n x 1) vector is expected, but I simply cannot find any information about how this is done in Python's Numpy module.

当我将两个大小为 (nxn)*(nx 1) 的 numpy 数组相乘时，我得到一个大小为 (nxn) 的矩阵。按照正常的矩阵乘法规则，需要一个 (nx 1) 向量，但我在 Python 的 Numpy 模块中找不到任何有关如何完成此操作的信息。

The thing is that I don't want to implement it manually to preserve the speed of the program.

问题是我不想手动实现它以保持程序的速度。

Example code is shown below:

示例代码如下所示：

a = np.array([[ 5, 1 ,3], [ 1, 1 ,1], [ 1, 2 ,1]])
b = np.array([1, 2, 3])

print a*b
   >>
   [[5 2 9]
   [1 2 3]
   [1 4 3]]

What i want is:

我想要的是：

print a*b
   >>
   [16 6 8]

采纳答案by wflynny

Simplest solution

最简单的解决方案

Use numpy.dotor a.dot(b). See the documentation here.

使用numpy.dot或a.dot(b)。请参阅此处的文档。

>>> a = np.array([[ 5, 1 ,3], 
                  [ 1, 1 ,1], 
                  [ 1, 2 ,1]])
>>> b = np.array([1, 2, 3])
>>> print a.dot(b)
array([16, 6, 8])

This occurs because numpy arrays are not matrices, and the standard operations *, +, -, /work element-wise on arrays. Instead, you could try using numpy.matrix, and *will be treated like matrix multiplication.

发生这种情况是因为 numpy 数组不是矩阵，并且标准操作*, +, -, /在数组上按元素工作。相反，您可以尝试使用numpy.matrix,*并将被视为矩阵乘法。

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Other Solutions

其他解决方案

Also know there are other options:

还知道还有其他选择：

• As noted below, if using python3.5+ the @operator works as you'd expect:

  >>> print(a @ b)
  array([16, 6, 8])
  

• If you want overkill, you can use numpy.einsum. The documentation will give you a flavor for how it works, but honestly, I didn't fully understand how to use it until reading this answerand just playing around with it on my own.

  >>> np.einsum('ji,i->j', a, b)
  array([16, 6, 8])
  

• As of mid 2016 (numpy 1.10.1), you can try the experimental numpy.matmul, which works like numpy.dotwith two major exceptions: no scalar multiplication but it works with stacks of matrices.

  >>> np.matmul(a, b)
  array([16, 6, 8])
  

• numpy.innerfunctions the same way as numpy.dotfor matrix-vector multiplication but behaves differentlyfor matrix-matrix and tensor multiplication (see Wikipedia regarding the differences between the inner product and dot productin general or see this SO answerregarding numpy's implementations).

  >>> np.inner(a, b)
  array([16, 6, 8])
  
  # Beware using for matrix-matrix multiplication though!
  >>> b = a.T
  >>> np.dot(a, b)
  array([[35,  9, 10],
         [ 9,  3,  4],
         [10,  4,  6]])
  >>> np.inner(a, b) 
  array([[29, 12, 19],
         [ 7,  4,  5],
         [ 8,  5,  6]])
  

• 如下所述，如果使用 python3.5+，该@运算符将按您的预期工作：

  >>> print(a @ b)
  array([16, 6, 8])
  

• 如果你想要矫枉过正，你可以使用numpy.einsum. 该文档将让您了解它的工作原理，但老实说，直到阅读此答案并自己玩弄它之前，我才完全了解如何使用它。

  >>> np.einsum('ji,i->j', a, b)
  array([16, 6, 8])
  

• 从 2016 年年中（numpy 1.10.1）开始，您可以尝试实验性的numpy.matmul，它的工作原理numpy.dot与两个主要例外类似：没有标量乘法，但它适用于矩阵堆栈。

  >>> np.matmul(a, b)
  array([16, 6, 8])
  

• numpy.inner以与numpy.dot矩阵向量乘法相同的方式运行，但对于矩阵矩阵和张量乘法的行为不同（请参阅维基百科关于内积和点积之间的一般差异或查看有关 numpy 实现的SO 答案）。

  >>> np.inner(a, b)
  array([16, 6, 8])
  
  # Beware using for matrix-matrix multiplication though!
  >>> b = a.T
  >>> np.dot(a, b)
  array([[35,  9, 10],
         [ 9,  3,  4],
         [10,  4,  6]])
  >>> np.inner(a, b) 
  array([[29, 12, 19],
         [ 7,  4,  5],
         [ 8,  5,  6]])
  

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Rarer options for edge cases

边缘情况的罕见选项

• If you have tensors (arrays of dimension greater than or equal to one), you can use numpy.tensordotwith the optional argument axes=1:

  >>> np.tensordot(a, b, axes=1)
  array([16,  6,  8])
  

• Don't use numpy.vdotif you have a matrix of complex numbers, as the matrix will be flattened to a 1D array, then it will try to find the complex conjugate dot product between your flattened matrix and vector (which will fail due to a size mismatch n*mvs n).

• 如果您有张量（维度大于或等于 1 的数组），则可以使用numpy.tensordot可选参数axes=1：

  >>> np.tensordot(a, b, axes=1)
  array([16,  6,  8])
  

• numpy.vdot如果您有一个复数矩阵，请不要使用，因为该矩阵将被展平为一维数组，然后它会尝试找到展平后的矩阵和向量之间的复共轭点积（由于大小不匹配而失败）n*m对n）。