Last Updated : 08 Mar, 2024
Prerequisite - numpy.dot() in Pythonnumpy.vdot(vector_a, vector_b)
returns the dot product of vectors a and b. If first argument is complex the complex conjugate of the first argument(this is where
vdot()
differs working of
dot()
method) is used for the calculation of the dot product. It can handle multi-dimensional arrays but working on it as a flattened array.
Parameters -dot Product of vectors a and b.
Code 1 : Python3
# Python Program illustrating
# numpy.vdot() method
import numpy as geek
# 1D array
vector_a = 2 + 3j
vector_b = 4 + 5j
product = geek.vdot(vector_a, vector_b)
print("Dot Product : ", product)
Output :
Dot Product : (23-2j)How Code1 works ?
vector_a = 2 + 3j vector_b = 4 + 5j As per method, take conjugate of vector_a i.e. 2 - 3j now dot product = 2(4 - 5j) + 3j(4 - 5j) = 8 - 10j + 12j + 15 = 23 - 2j
Code 2 : Python3
# Python Program illustrating
# numpy.vdot() method
import numpy as geek
# 1D array
vector_a = geek.array([[1, 4], [5, 6]])
vector_b = geek.array([[2, 4], [5, 2]])
product = geek.vdot(vector_a, vector_b)
print("Dot Product : ", product)
product = geek.vdot(vector_b, vector_a)
print("\nDot Product : ", product)
"""
How Code 2 works :
array is being flattened
1 * 2 + 4 * 4 + 5 * 5 + 6 * 2 = 55
"""
Output :
Dot Product : 55 Dot Product : 55
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