看来我迷失在潜在的愚蠢的事情中。 我有一个n维数阵列，我想沿某个维度（可以更改！）将其用向量（1D数组）乘以。例如，说我想将2D数组乘以第一个数组的轴0乘以1D数组，我可以做这样的事情：
a=np.arange(20).reshape((5,4))
b=np.ones(5)
c=a*b[:,np.newaxis]
简单，但我想将此想法扩展到n维（对于A，而B始终为1D）和任何轴。换句话说，我想知道如何在正确的位置使用np.newaxis生成切片。说一个是3D，我想沿轴= 1乘，我想生成正确给出的切片：
c=a*b[np.newaxis,:,np.newaxis]
即。鉴于A（例如3）的尺寸和我要乘的轴（例如轴= 1），我如何生成和传递切片：
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
示例运行以进行步骤的演示 -
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，我们想沿given axis = 1执行ElementWise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＆amp;执行元素乘法：
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＆amp;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
避免复制数据和废物资源！
利用铸造和视图，而不是实际将数据n n次复制到具有适当形状的新数组（如现有答案所做的那样）是更有效的。这是这样的方法（基于 @shuxuanxu的代码）：
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
您可以构建一个切片对象，然后在其中选择所需的维度：
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
我在进行一些数值计算时也有类似的需求。
假设我们有两个数组（A和B）和一个用户指定的"轴"。A是多维阵列。B是1-D阵列。
基本思想是扩展B，以使A和B具有相同的形状。这是解决方案代码
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
这是一个示例
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.]])
简化 @neinstein的解决方案，我到达
def multiply_along_axis(A, B, axis):
return np.swapaxes(np.swapaxes(A, axis, -1) * B, -1, axis)
此示例还避免复制和浪费内存。通过将所需轴交换到最后一个位置，执行乘法并将轴交换回原始位置，可以避免显式广播。另一个优势是Numpy负责处理错误处理和键入转换。
您也可以使用简单的矩阵技巧
c = np.matmul(a,diag(b))
基本上只是在a和对角的矩阵之间进行矩阵乘法，其对角线是b的元素。也许不是那么高效，但这是一个不错的单行解决方案