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numpy.vdot() in Python
  • Last Updated : 05 Oct, 2017

Prerequisite – numpy.dot() in Python
numpy.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 –

  1. vector_a : [array_like] if a is complex its complex conjugate is used for the calculation of the dot product.
  2. vector_b : [array_like] if b is complex its complex conjugate is used for the calculation of the dot product.

Return – dot Product of vectors a and b.

Code 1 :




# 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 :




# 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

References :
https://docs.scipy.org/doc/numpy-dev/reference/generated/numpy.vdot.html#numpy.vdot

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