What is the pythonic way to calculate dot product?

问题

I have two lists, one is named as A, another is named as B. Each element in A is a triple, and each element in B is just an number. I would like to calculate the result defined as :

result = A[0][0] * B[0] + A[1][0] * B[1] + ... + A[n-1][0] * B[n-1]

I know the logic is easy but how to write in pythonic way?

Thanks!

回答1:

import numpy
result = numpy.dot( numpy.array(A)[:,0], B)

http://docs.scipy.org/doc/numpy/reference/

If you want to do it without numpy, try

sum( [a[i][0]*b[i] for i in range(len(b))] )

回答2:

Python 3.5 has an explicit operator @ for the dot product, so you can write

a = A @ B

instead of

a = numpy.dot(A,B)

回答3:

My favorite Pythonic dot product is:

sum([i*j for (i, j) in zip(list1, list2)])

So for your case we could do:

sum([i*j for (i, j) in zip([K[0] for K in A], B)])

回答4:

from operator import mul

sum(map(mul, A, B))

回答5:

Using the operator and the itertools modules:

from operator import mul
from itertools import imap

sum(imap(mul, A, B))

回答6:

Probably the most Pythonic way for this kind of thing is to use numpy. ;-)

回答7:

>>> X = [2,3,5,7,11]
>>> Y = [13,17,19,23,29]
>>> dot = lambda X, Y: sum(map(lambda x, y: x * y, X, Y))
>>> dot(X, Y)
652

And that's it.

回答8:

This might be repeated solution, however:

>>> u = [(1, 2, 3), (4, 5, 6)]
>>> v = [3, 7]

In plain Python:

>>> sum([x*y for (x, *x2), y in zip(u,v)])
31

Or using numpy (as described in user57368's answer) :

import numpy as np
>>> np.dot(np.array(u)[:,0], v)
31

回答9:

All above answers are correct, but in my opinion the most pythonic way to calculate dot product is:

>>> a=[1,2,3]
>>> b=[4,5,6]
>>> sum(map(lambda pair:pair[0]*pair[1],zip(a,b)))
32

回答10:

Using more_itertools, a third-party library that implements the dotproduct itertools recipe:

import more_itertools as mit


a = [1, 2, 3]
b = [7, 8, 9]

mit.dotproduct(a, b)
# 50

来源：https://stackoverflow.com/questions/5919530/what-is-the-pythonic-way-to-calculate-dot-product

标签

python

dot-product