2017-06-19 63 views
-2
raw = input("Enret Number of Raws : ") 
cols = input("Enter Number of Cols : ") 

obj = [] 
mat = [[[] for j in range(raw)] for i in range(cols)] 
for k in range(2): 
    for i in range(cols): 
     for j in range(raw): 
      number = int(input("Please Enter Elements of Matrix : ")) 
      mat[i][j] = number 
    obj.append(mat) 
    print obj 

输出:动态矩阵输入在Python

 
Enret Number of Raws : 2 
Enter Number of Cols : 2 
Please Enter Elements of Matrix 1:9 
Please Enter Elements of Matrix 1:3 
Please Enter Elements of Matrix 1:7 
Please Enter Elements of Matrix 1:1 
[[[9, 3], [7, 1]]] 
Please Enter Elements of Matrix 2:8 
Please Enter Elements of Matrix 2:2 
Please Enter Elements of Matrix 2:4 
Please Enter Elements of Matrix 2:6 
[[[8, 2], [4, 6]], [[8, 2], [4, 6]]] 

在列表追加第二矩阵后,第一次拿到找零的价值。

+1

你已添加相同的'对mat''obj'多次。如果你想'mat'成为一个不同的列表,每次都要通过迭代,在你的外层循环中移动'mat = [']'行。 – khelwood

回答

0

原始=输入( “原糖的Enret数量: ”) COLS =输入(“ 列数的输入号码:”)

OBJ = []

对于k在范围(2):

垫= [[[]用于范围Ĵ(原始)用于i的范围(COLS)]

for i in range(cols): 
    for j in range(raw): 
     number = int(input("Please Enter Elements of Matrix : ")) 
     mat[i][j] = number 
obj.append(mat) 
print obj 
+0

试试这个家伙。 它肯定会工作 –