打印2x2矩阵

时间:2018-03-03 02:59:41

标签: python matrix

我编写了一个程序来打印 2x2矩阵,其中每个矩阵中表示的数字是通过用户输入给出的。

例如:

userin = 1 2     #two digit input with a spacing in between
userin2 = 3 4    #userin and userin2 is first matrix
userin3 = 5 6
userin4 = 7 8    #userin3 and userin4 is second matrix

The program would then print out: [[1,2],[3,4]],   #first matrix
                                  [[5,6],[7,8]]    #second matrix

我让程序按照我的意愿工作,但我觉得我编写的代码效率很低,以至于我创建了太多空列表来满足用户输入。

userin = input("Enter first two digit for first 2x2 matrix: ").split(' ')
userin2 = input("Enter last two digit for first 2x2 matrix: ").split(' ')
userin3 = input("Enter first two digit for second 2x2 matrix: ").split(' ')
userin4 = input("Enter last two digit for second 2x2 matrix: ").split(' ')
lst1 = []
lst2 = []
lst3 = []
lst4 = []
matrix = []
matrix2 = []

for x in userin:
    x = int(x)
    lst1.append(x)

for j in userin2:
    j = int(j)
    lst2.append(j)

for x in userin3:
    x = int(x)
    lst3.append(x)

for j in userin4:
    j = int(j)
    lst4.append(j)

matrix.append(lst1)
matrix.append(lst2)
matrix2.append(lst3)
matrix2.append(lst4)
print(matrix)
print(matrix2)

程序以我想要的格式输出 [[1,2],[3,4]],[[5,6],[7,8]] 但我和# 39;我正在寻找一种更有效的方法。

请注意,我不建议使用numpy和其他内置库。

感谢您的帮助。

3 个答案:

答案 0 :(得分:0)

以下是其他方式的几个例子。我不确定第一个例子是你在寻找什么,但我把它扔进去玩!

>>> userin = '1 2'.split()
... userin2 = '3 4'.split()
... userin3 = '5 6'.split()
... userin4 = '7 8'.split()
>>> print('[[{}, {}], [{}, {}]]'.format(*userin, *userin2))
... print('[[{}, {}], [{}, {}]]'.format(*userin3, *userin4))
[[1, 2], [3, 4]]
[[5, 6], [7, 8]]
>>> print([[int(a) for a in pair] for pair in (userin, userin2)])
... print([[int(a) for a in pair] for pair in (userin3, userin4)])
[[1, 2], [3, 4]]
[[5, 6], [7, 8]]

答案 1 :(得分:0)

我就是这样做的:

Enter first two digit for first 2x2 matrix: 1 2
Enter last two digit for first 2x2 matrix: 3 4
Enter first two digit for second 2x2 matrix: 5 6
Enter last two digit for second 2x2 matrix: 7 8
[[1, 2], [3, 4]],
[[5, 6], [7, 8]]

如果它总是只是那4个输入语句,我认为这很好。如果输入是可变的,那么使函数更有意义。

输出:

@Value("#{${validators}}")
private Map<String,String> validators;

答案 2 :(得分:0)

为什么不放入while循环来获取任意数量的矩阵,为什么不一次接收所有4个元素呢?

类似的东西:

user_in = ['0']
matrices = []
while (user_in[0].lower() != 'q'):
   user_in = input("Input 2x2 matrix(space separated), enter 'q' to quit: ").split(' ')
   if len(user_in) == 4:
      a, b, c, d = user_in
      m2x2 = [[int(a), int(b)], [int(c), int(d)]]
      matrices.append(m2x2)

for matrix in matrices:
    print(matrix)