Question

我的代码如下：

X = [[12,7,3],
    [4 ,5,6],
    [7 ,8,9]]


result = [[0,0,0],
         [0,0,0],
         [0,0,0]]

Final = [[0,0,0],
         [0,0,0],
         [0,0,0]]



for i in range(len(X)):
    for j in range(len(X)):
        for x in range(4):
                    result[i][j] = X[i][j] + X[i][j]
                    Final[i][j]=result[i][j]+X[i][j]


for r in Final: 
    print(r)

我想将X矩阵与其自身相加4次，并将答案作为最终求和矩阵返回。任何人都可以帮我吗？

Answer 1

yourlist = [5,6,7,8]
for i in range(len(yourlist)-1):
    print "Source: " + str(yourlist[i])
    print "Destination: " + str(yourlist[i+1]) + "\n"

输出：

Source = 5  
Destination = 6

Source = 6  
Destination = 7

Source = 7  
Destination = 8

Answer 2

一种方法是使用zip()函数从列表中生成对，然后迭代这些对以打印您的消息：

l = [5, 6, 7, 8]
for source, dest in zip(l, l[1:]):
    print('Source = {}'.format(source))
    print('Destination = {}'.format(dest))

输出：

Source = 5
Destination = 6
Source = 6
Destination = 7
Source = 7
Destination = 8

这里的关键是zip()函数，它通过将每个项目与其后续列表组合在一起工作：

>>> l = [1, 2, 3, 4, 5, 6]
>>> l[1:]
[2, 3, 4, 5, 6]
>>> zip(l, l[1:])
[(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)]

如果列表的长度不均匀，则最后一项将被删除：

>>> l = [1, 2, 3]
>>> zip(l, l[1:])
[(1, 2), (2, 3)]

如何以迭代方式将矩阵添加到自身？

2 个答案: