问题:如何在numpy中获得按元素矩阵乘法(Hadamard积)?

我有两个矩阵

a = np.matrix([[1,2], [3,4]])
b = np.matrix([[5,6], [7,8]])

我想得到元素乘积[[1*5,2*6], [3*7,4*8]],等于

[[5,12], [21,32]]

我努力了

print(np.dot(a,b)) 

print(a*b)

但两者都给出结果

[[19 22], [43 50]]

这是矩阵乘积,而不是元素乘积。如何使用内置函数获取按元素分类的产品(又名Hadamard产品)?

I have two matrices

a = np.matrix([[1,2], [3,4]])
b = np.matrix([[5,6], [7,8]])

and I want to get the element-wise product, [[1*5,2*6], [3*7,4*8]], equaling

[[5,12], [21,32]]

I have tried

print(np.dot(a,b)) 

and

print(a*b)

but both give the result

[[19 22], [43 50]]

which is the matrix product, not the element-wise product. How can I get the the element-wise product (aka Hadamard product) using built-in functions?


回答 0

对于matrix对象的元素乘法,可以使用numpy.multiply

import numpy as np
a = np.array([[1,2],[3,4]])
b = np.array([[5,6],[7,8]])
np.multiply(a,b)

结果

array([[ 5, 12],
       [21, 32]])

但是,您应该真正使用array而不是matrixmatrix对象与常规ndarray具有各种可怕的不兼容性。使用ndarrays时,您可以仅使用*元素级乘法:

a * b

如果您使用的是Python 3.5+,则您甚至都不会失去使用运算符执行矩阵乘法的能力,因为@矩阵乘法现在可以

a @ b  # matrix multiplication

For elementwise multiplication of matrix objects, you can use numpy.multiply:

import numpy as np
a = np.array([[1,2],[3,4]])
b = np.array([[5,6],[7,8]])
np.multiply(a,b)

Result

array([[ 5, 12],
       [21, 32]])

However, you should really use array instead of matrix. matrix objects have all sorts of horrible incompatibilities with regular ndarrays. With ndarrays, you can just use * for elementwise multiplication:

a * b

If you’re on Python 3.5+, you don’t even lose the ability to perform matrix multiplication with an operator, because @ does matrix multiplication now:

a @ b  # matrix multiplication

回答 1

只是这样做:

import numpy as np

a = np.array([[1,2],[3,4]])
b = np.array([[5,6],[7,8]])

a * b

just do this:

import numpy as np

a = np.array([[1,2],[3,4]])
b = np.array([[5,6],[7,8]])

a * b

回答 2

import numpy as np
x = np.array([[1,2,3], [4,5,6]])
y = np.array([[-1, 2, 0], [-2, 5, 1]])

x*y
Out: 
array([[-1,  4,  0],
       [-8, 25,  6]])

%timeit x*y
1000000 loops, best of 3: 421 ns per loop

np.multiply(x,y)
Out: 
array([[-1,  4,  0],
       [-8, 25,  6]])

%timeit np.multiply(x, y)
1000000 loops, best of 3: 457 ns per loop

两者np.multiply*都会产生元素明智的乘法,称为Hadamard积

%timeit 是ipython的魔力

import numpy as np
x = np.array([[1,2,3], [4,5,6]])
y = np.array([[-1, 2, 0], [-2, 5, 1]])

x*y
Out: 
array([[-1,  4,  0],
       [-8, 25,  6]])

%timeit x*y
1000000 loops, best of 3: 421 ns per loop

np.multiply(x,y)
Out: 
array([[-1,  4,  0],
       [-8, 25,  6]])

%timeit np.multiply(x, y)
1000000 loops, best of 3: 457 ns per loop

Both np.multiply and * would yield element wise multiplication known as the Hadamard Product

%timeit is ipython magic


回答 3

试试这个:

a = np.matrix([[1,2], [3,4]])
b = np.matrix([[5,6], [7,8]])

#This would result a 'numpy.ndarray'
result = np.array(a) * np.array(b)

在此,np.array(a)返回类型为2D的2D数组,ndarray并且ndarray将导致元素相乘。因此结果将是:

result = [[5, 12], [21, 32]]

如果您想获取矩阵,请执行以下操作:

result = np.mat(result)

Try this:

a = np.matrix([[1,2], [3,4]])
b = np.matrix([[5,6], [7,8]])

#This would result a 'numpy.ndarray'
result = np.array(a) * np.array(b)

Here, np.array(a) returns a 2D array of type ndarray and multiplication of two ndarray would result element wise multiplication. So the result would be:

result = [[5, 12], [21, 32]]

If you wanna get a matrix, the do it with this:

result = np.mat(result)

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