Python | Give list [1,2,3,4] -> return 1*2 2*3 3*4-CodePudding

So as i said in the title, i want to multipy each two neigbors in the list and sum them all - with a single code line.

I'm looking for the most elegant and efficient method, without using extra memory or such.

Here's what I do have now:

import numpy as np
G = lambda array: sum(np.multiply(array[:-1], array[1:])))

That works, but when i write array[:-1] and array[1:], doesn't it create two new arrays? If so, is there a way to do that with the same original array?

Or maybe you can come up with a nicer way to do that :)

CodePudding user response：

Try the following:

    lst = [1,2,3,4]
    func = lambda lst: sum([lst[i]*lst[i 1] for i in range(len(lst)-1)])
    func(lst)

CodePudding user response：

If you're not constrained by the lambda you can:

def sum_with_neighbours(arr: list[int]) -> int:
    sum_ = 0
    for i in range(len(arr) - 1):  # -1 because we want to go until second-last item
        sum_  = arr[i] * arr[i   1]
    return sum_

CodePudding user response：

For normal lists, using sum and zip should be fairly efficient:

f = lambda lst: sum(x * y for x, y in zip(lst[1:], lst))
f([1, 2, 3, 4])

Output:

CodePudding user response：

itertools.pairwise may be the most pythonic choice:

>>> from itertools import pairwise
>>> [*pairwise(range(5))]
[(0, 1), (1, 2), (2, 3), (3, 4)]
>>> sum(i * j for i, j in pairwise(range(5)))
20

CodePudding user response：

The actual numpy way

array = np.array([1, 2, 3, 4])

array[1:] @ array[:-1]

Out[]: 20

@ is the dot-product operator, which is what that multiply-sum operation is called normally.