我正在尝试多个两个矩阵A和B,其中B的列比A使用python和numpy更多。
示例:
A = numpy.matrix([[2,3,15],[5,8,12],[1,13,4]], dtype=numpy.object)
B = numpy.matrix([[2,15,6,15,8,14],[17,19,17,7,18,14],[24,14,0,24,2,11]], dtype=numpy.object)
( A*B ) = [[415,297,63,411,100,235],[434,395,166,419,208,314], [319,318,227,202,250,240]]
我发现了一些例子,如果它们是数组,但如果它们是矩阵则没有。有人可以帮助我吗?
答案 0 :(得分:3)
你真的试过这个吗?它在这里工作正常:
import numpy as np
def main():
A = np.matrix([[2,3,15],[5,8,12],[1,13,4]], dtype=np.object)
B = np.matrix([[2,15,6,15,8,14],[17,19,17,7,18,14],[24,14,0,24,2,11]], dtype=np.object)
C = ( A*B ) % 26 # = [[25,11,11,21,22,1],[18,5,10,3,0,2], [7,6,19,20,16,6]]
print(C)
return 0
if __name__ == '__main__':
main()
打印:
[[25 11 11 21 22 1]
[18 5 10 3 0 2]
[7 6 19 20 16 6]]
答案 1 :(得分:2)
您也可以使用点积。
theory ToyList
imports Main
begin
datatype 'a list = Nil | Cons 'a "'a list"
fun app :: "'a list ⇒ 'a list ⇒ 'a list" where
"app Nil ys = ys" |
"app (Cons x xs) ys = Cons x (app xs ys)"
它会输出您想要的相同答案。