是否有可以完成第三个“等式”的python(numpy)功能?
将其用作返回的lambda函数
1。向量*标量
import numpy as np
a = np.array([3,4])
b = 2
print(a*b)
>>[6,8]
或作为lambda函数:
import numpy as np
def multiply():
return lambda a,b: a*b
a = np.array([3,4])
b = 2
j = multiply()
print(j(a,b))
>>[6,8]
2。矩阵*向量
import numpy as np
a = np.array([[3,4],[2,5]])
b = np.array([2,4])
print(a*b)
print()
print(np.multiply(a,b))
print()
print(a.dot(b))
print()
print(b.dot(a))
>>[[ 6 16]
>>[ 4 20]]
>>
>>[[ 6 16]
>>[ 4 20]]
>>
>>[22 24]
>>
>>[14 28]
或作为lambda函数:
import numpy as np
def multiply():
return lambda a,b: a.dot(b)
a = np.array([[3,4],[2,5]])
b = np.array([2,4])
j = multiply()
print(j(a,b))
>>[22 24]
3。矩阵(解释为许多(2,1)-向量)*向量(解释为许多标量)或:每行向量*标量
import numpy as np
a = np.array([[3,4],[2,5]])
b = np.array([2,4])
?
import numpy as np
a = np.array([[3,2], [4, 5]])
b = np.array([2, 4])
c = np.vstack((b, b)).T
d = np.multiply(a,c)
print(d)
array([[ 6, 8],
[8, 20]])
如果需要功能
def elementwisemult(a, b): b = np.vstack((b, b)).T d = np.multiply(a,b) return d