似乎我在做一些可能很愚蠢的事情。我有一个n维的numpy数组，我想沿着某个维度（可以改变）将它与向量（1d数组）相乘。
举个例子，假设我想将一个2d数组沿着第0维与一个1d数组相乘，我可以这样做：
a=np.arange(20).reshape((5,4))
b=np.ones(5)
c=a*b[:,np.newaxis]
容易，但我想将这个想法扩展到n维（其中a是n维的，而b始终为1d）并且应用于任意轴。换句话说，我想知道如何在正确的位置生成带有np.newaxis的切片。假设a是3d，我想沿axis=1进行乘法，我想生成能正确给出以下结果的切片：
c=a*b[np.newaxis,:,np.newaxis]
即：给定一个数组的维数（比如说3），以及想要进行乘法的轴（比如说axis=1），我该如何生成并传递切片：
np.newaxis,:,np.newaxis
谢谢。
解决方案代码 -
import numpy as np
# Given axis along which elementwise multiplication with broadcasting
# is to be performed
given_axis = 1
# Create an array which would be used to reshape 1D array, b to have
# singleton dimensions except for the given axis where we would put -1
# signifying to use the entire length of elements along that axis
dim_array = np.ones((1,a.ndim),int).ravel()
dim_array[given_axis] = -1
# Reshape b with dim_array and perform elementwise multiplication with
# broadcasting along the singleton dimensions for the final output
b_reshaped = b.reshape(dim_array)
mult_out = a*b_reshaped
演示步骤的示例运行 -
In [149]: import numpy as np
In [150]: a = np.random.randint(0,9,(4,2,3))
In [151]: b = np.random.randint(0,9,(2,1)).ravel()
In [152]: whos
Variable Type Data/Info
-------------------------------
a ndarray 4x2x3: 24 elems, type `int32`, 96 bytes
b ndarray 2: 2 elems, type `int32`, 8 bytes
In [153]: given_axis = 1
现在，我们想要沿着给定的轴 axis = 1 执行元素乘法。让我们创建 dim_array：
In [154]: dim_array = np.ones((1,a.ndim),int).ravel()
...: dim_array[given_axis] = -1
...:
In [155]: dim_array
Out[155]: array([ 1, -1, 1])
最后，重塑b并执行逐元素乘法：
In [156]: b_reshaped = b.reshape(dim_array)
...: mult_out = a*b_reshaped
...:
再次查看whos信息，并特别注意b_reshaped和mult_out：
In [157]: whos
Variable Type Data/Info
---------------------------------
a ndarray 4x2x3: 24 elems, type `int32`, 96 bytes
b ndarray 2: 2 elems, type `int32`, 8 bytes
b_reshaped ndarray 1x2x1: 2 elems, type `int32`, 8 bytes
dim_array ndarray 3: 3 elems, type `int32`, 12 bytes
given_axis int 1
mult_out ndarray 4x2x3: 24 elems, type `int32`, 96 bytes
shape = [ -1 if ax == given_axis else 1 for ax in range(a.ndim)]，然后进行计算 a*b.reshape(shape)。 - Puco4避免复制数据和浪费资源！
使用类型转换和视图，而不是将数据 N 次复制到具有适当形状的新数组中（如现有答案所做），这样更加高效地利用内存。以下是一种基于 @ShuxuanXU 代码的方法：
def mult_along_axis(A, B, axis):
# ensure we're working with Numpy arrays
A = np.array(A)
B = np.array(B)
# shape check
if axis >= A.ndim:
raise AxisError(axis, A.ndim)
if A.shape[axis] != B.size:
raise ValueError(
"Length of 'A' along the given axis must be the same as B.size"
)
# np.broadcast_to puts the new axis as the last axis, so
# we swap the given axis with the last one, to determine the
# corresponding array shape. np.swapaxes only returns a view
# of the supplied array, so no data is copied unnecessarily.
shape = np.swapaxes(A, A.ndim-1, axis).shape
# Broadcast to an array with the shape as above. Again,
# no data is copied, we only get a new look at the existing data.
B_brc = np.broadcast_to(B, shape)
# Swap back the axes. As before, this only changes our "point of view".
B_brc = np.swapaxes(B_brc, A.ndim-1, axis)
return A * B_brc
import numpy as np
a = np.arange(18).reshape((3,2,3))
b = np.array([1,3])
ss = [None] * a.ndim
ss[1] = slice(None) # set the dimension along which to broadcast
print ss # [None, slice(None, None, None), None]
c = a*b[tuple(ss)] # convert to tuple to avoid FutureWarning from newer versions of Python
None] * a.ndim 等同于 [None for i in range(a.ndim)] 但看起来更简单。 - onestop:）、省略号（...）、numpy.newaxis（None）和整数或布尔数组是有效的索引。 - klmnss = tuple((None if i != axis else slice(None) for i in range(a.ndim))) - TheIdealisdef multiply_along_axis(A, B, axis):
return np.swapaxes(np.swapaxes(A, axis, -1) * B, -1, axis)
这个例子还避免了复制和浪费内存。通过将A中所需的轴交换到最后一个位置，执行乘法，然后将轴交换回原始位置来避免显式广播。额外的好处是numpy会处理错误处理和类型转换。
import numpy as np
from numpy.core._internal import AxisError
def multiply_along_axis(A, B, axis):
A = np.array(A)
B = np.array(B)
# shape check
if axis >= A.ndim:
raise AxisError(axis, A.ndim)
if A.shape[axis] != B.size:
raise ValueError("'A' and 'B' must have the same length along the given axis")
# Expand the 'B' according to 'axis':
# 1. Swap the given axis with axis=0 (just need the swapped 'shape' tuple here)
swapped_shape = A.swapaxes(0, axis).shape
# 2. Repeat:
# loop through the number of A's dimensions, at each step:
# a) repeat 'B':
# The number of repetition = the length of 'A' along the
# current looping step;
# The axis along which the values are repeated. This is always axis=0,
# because 'B' initially has just 1 dimension
# b) reshape 'B':
# 'B' is then reshaped as the shape of 'A'. But this 'shape' only
# contains the dimensions that have been counted by the loop
for dim_step in range(A.ndim-1):
B = B.repeat(swapped_shape[dim_step+1], axis=0)\
.reshape(swapped_shape[:dim_step+2])
# 3. Swap the axis back to ensure the returned 'B' has exactly the
# same shape of 'A'
B = B.swapaxes(0, axis)
return A * B
以下是一个示例：
In [33]: A = np.random.rand(3,5)*10; A = A.astype(int); A
Out[33]:
array([[7, 1, 4, 3, 1],
[1, 8, 8, 2, 4],
[7, 4, 8, 0, 2]])
In [34]: B = np.linspace(3,7,5); B
Out[34]: array([3., 4., 5., 6., 7.])
In [35]: multiply_along_axis(A, B, axis=1)
Out[34]:
array([[21., 4., 20., 18., 7.],
[ 3., 32., 40., 12., 28.],
[21., 16., 40., 0., 14.]])
你也可以使用一个简单的矩阵技巧
c = np.matmul(a,diag(b))