numpy.dot() in Python
Last Updated :
18 Nov, 2022
numpy.dot(vector_a, vector_b, out = None) returns the dot product of vectors a and b. It can handle 2D arrays but considers them as matrix and will perform matrix multiplication. For N dimensions it is a sum-product over the last axis of a and the second-to-last of b :
dot(a, b)[i,j,k,m] = sum(a[i,j,:] * b[k,:,m])
Parameters
- vector_a : [array_like] if a is complex its complex conjugate is used for the calculation of the dot product.
- vector_b : [array_like] if b is complex its complex conjugate is used for the calculation of the dot product.
- out : [array, optional] output argument must be C-contiguous, and its dtype must be the dtype that would be returned for dot(a,b).
Dot Product of vectors a and b. if vector_a and vector_b are 1D, then scalar is returned
Code 1:
Python
import numpy as geek
product = geek.dot( 5 , 4 )
print ( "Dot Product of scalar values : " , product)
vector_a = 2 + 3j
vector_b = 4 + 5j
product = geek.dot(vector_a, vector_b)
print ( "Dot Product : " , product)
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Output:
Dot Product of scalar values : 20
Dot Product : (-7+22j)
How Code1 works ?
vector_a = 2 + 3j
vector_b = 4 + 5j
now dot product
= 2(4 + 5j) + 3j(4 +5j)
= 8 + 10j + 12j - 15
= -7 + 22j
Code 2:
Python
import numpy as geek
vector_a = geek.array([[ 1 , 4 ], [ 5 , 6 ]])
vector_b = geek.array([[ 2 , 4 ], [ 5 , 2 ]])
product = geek.dot(vector_a, vector_b)
print ( "Dot Product : \n" , product)
product = geek.dot(vector_b, vector_a)
print ( "\nDot Product : \n" , product)
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Output:
Dot Product :
[[22 12]
[40 32]]
Dot Product :
[[22 32]
[15 32]]
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