Question

如何优雅地完成（Pythonic方式）

您有两个矩阵A和B，每个矩阵由一个列表列表。

您想计算A和B的元素乘积矩阵的总和。

示例代码：

sum = 0
nrows = len(A)
ncols = len(A[0])
for i in range(nrows):
    for j in range(ncols):
        sum += A[i][j]*B[i][j]

# interested finally in the value of sum
# return sum

例如

A = [[1,2,3]，[3,4,5]]

和

B = [[3,2,2]，[1,1,1]]

结果是：

1 * 3 + 2 * 2 + 3 * 2 + 3 * 1 + 4 * 1 + 5 * 1 = 25

Answer 1

如果矩阵总是2D，你可以做

sum(sum(a * b for a, b in zip(*rows)) for rows in zip(A, B))

IDEOne Link

Answer 2

您也可以尝试这样的事情（假设矩阵是2D）：

import itertools

A=[[1,2,3],[3,4,5]]
B=[[3,2,2],[1,1,1]]

def flatten(lst):
    return list(itertools.chain.from_iterable(lst))

>>> print(sum(x * y for x, y in zip(flatten(A), flatten(B))))
25

甚至：

>>> print(sum(x * y for x, y in zip(sum(A, []), sum(B, []))))
25

Answer 3

这是@ MadPhysicist的生成器expr的平面版本。

sum(a*b for ab in zip(A, B) for a, b in zip(*ab))
# 25

Answer 4

import itertools

A = [[1,2,3],[3,4,5]]
B = [[3,2,2],[1,1,1]]

# flatten list A => [1,2,3,3,4,5]]
A_flat = itertools.chain.from_iterable(A)

# flatten list B = > [3,2,2,1,1,1]]
B_flat = itertools.chain.from_iterable(B)

# making sum of 1+3 , 2+2 , 3+2 result=> [4, 4, 5, 4, 5, 6]
pair_sum = [ sum(pair) for pair in zip(A_flat,B_flat)]

# total
print('Total : ', sum(pair_sum))


############

import itertools

A = [[1,2,3],[3,4,5]]
B = [[3,2,2],[1,1,1]]
A_flat = itertools.chain.from_iterable(A)
B_flat = itertools.chain.from_iterable(B)
total = sum(sum(pair) for pair in zip(A_flat,B_flat))
print(total)

两个列表的点积和的Pythonic方式

4 个答案: