print("\ndot product of two ndarray objects") print(np.dot(a, b)) print("\ndot product of two matrix objects") print(np.dot(c, d)) 当使用*操作符将两个ndarray对象相乘时,结果是逐元素相乘。另一方面,当使用*操作符将两个矩阵对象相乘时,结果是点(矩阵)乘积,相当于前面的np.dot()。 import...
c=np.matrix([[1,2],[3,4]])d=np.matrix([[5,6,7],[8,9,10]])print("\nc",type(c))print(c)print("\nd",type(d))print(d)print("\ndot product of two ndarray objects")print(np.dot(a,b))print("\ndot product of two matrix objects")print(np.dot(c,d)) 当使用*操作符将...
print("\ndot product of two ndarray objects") print(np.dot(a, b)) print("\ndot product of two matrix objects") print(np.dot(c, d)) 当使用*操作符将两个ndarray对象相乘时,结果是逐元素相乘。另一方面,当使用*操作符将两个矩阵对象相乘时,结果是点(矩阵)乘积,相当于前面的np.dot()。 import...
另一种方法是使用numpy矩阵类。 ndarray和matrix对象的点积都可以使用np.dot()得到。 importnumpyasnp#Matricesasndarrayobjectsa=np.array([[1,2], [3,4]])b=np.array([[5,6,7], [8,9,10]])print("a",type(a))print(a)print("\nb",type(b))print(b)#Matricesasmatrixobjectsc=np.matrix([[...
矩阵乘法可以使用 np.dot() 函数或者 @ 操作符: # 方法1:使用 np.dot() 函数 E = np.dot(A, B) print("Matrix multiplication using np.dot():\n", E) # 方法2:使用 @ 操作符 F = A @ B print("Matrix multiplication using @:\n", F) 元素级乘法 如果你想进行元素级的乘法(即Hadamard积...
矩阵乘法是线性代数中的重要运算。我们可以使用dot函数或@运算符进行矩阵乘法: python 复制代码 # 矩阵乘法 matrix_product = np.dot(matrix_a, matrix_b) print("\nMatrix A * Matrix B (using np.dot):") print(matrix_product) # 使用 @ 运算符进行矩阵乘法 ...
print("\nDeterminant of A:",np.linalg.det(A)) # 矩阵 A 的逆 print("\nInverse of A:\n",np.linalg.inv(A)) print("\nMatrix A raised to power 3:\n", np.linalg.matrix_power(A,3)) 1. 2. 3. 4. 5. 6. 7. 8. 9. ...
dot(a,b) print("c as inner product of vectors:", c) # c = 14 # 情况2:If both a and b are 2-D arrays, it is matrix multiplication, but using matmul or a @ b is preferred. a = np.arange(1,5).reshape((2,2)) b = np.arange(5,9).reshape((2,2)) c = np.dot(a,b...
Dot product of two arrays. Specifically, If bothaandbare 1-D arrays, it is inner product of vectors (without complex conjugation). If bothaandbare 2-D arrays, it is matrix multiplication, but usingmatmulora@bis preferred. 为了保持一致性,都使用了转置操作。如下图所示,矩阵乘法按线性代数定义,...
np.dot(矩阵1,矩阵2) a = np.array([[1,2,3],[4,8,16]])a:1 2 34 8 16b = np.array([5,6,11]).reshape(-1,1)b:5611np.dot(a,b) produces38160Just like any dot product of a matrix with a column vector would produce. ...