我有两个包含数字列表的列表(m1和m2)。我正在尝试按元素方式进行乘法和求和,包括交叉乘积以获得最终列表(结果)如下:
m1 = [[1, 2, 3], [4, 5, 6]]
m2 = [[7, 9, 2], [8, 1, 3]]
[[1*7+2*9+3*2,1*8+2*1+3*3],[4*7+5*9+6*2,4*8+5*1+6*3]]
result = [[31,19],[85,55]]
答案 0 :(得分:5)
您可以使用python built-in functions和嵌套列表理解:
>>> [[sum(t*k for t,k in zip(i,j)) for j in m2] for i in m1]
[[31, 19], [85, 55]]
您还可以使用itertools.product
查找子列表之间的产品:
>>> from itertools import product
>>> [sum(t*k for t,k in zip(i,j)) for i,j in product(m1,m2)]
[31, 19, 85, 55]
答案 1 :(得分:3)
让我们把问题分解成更小的部分。在最低级别,我们有两个小列表:[1, 2, 3]
和[7, 9, 2]
,并希望逐个项目相乘:
item1 = [1, 2, 3]
item2 = [7, 9, 2]
zip(item1, item2) # ==> [(1, 7), (2, 9), (3, 2)]
[x * y for x, y in zip(item1, item2)] # ==> [7, 18, 6]
sum(x * y for x, y in zip(item1, item2)) # ==> 31
现在,我们可以在双循环中使用它:
[[sum(x * y for x, y in zip(item1, item2)) for item2 in m2] for item1 in m1]
# ==> [[31, 19], [85, 55]]
答案 2 :(得分:0)
如果你想要它而不导入任何模块,你可以这样做:
>>> m1 = [[1, 2, 3], [4, 5, 6]]
>>> m2 = [[7, 9, 2], [8, 1, 3]]
>>> [[sum(map(lambda (s,t):s*t, zip(x,y))) for y in m2] for x in m1]
[[31, 19], [85, 55]]