2017-06-19 66 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 

輸出:動態矩陣輸入在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]]] 

在列表追加第二矩陣後,第一次拿到找零的價值。

+1

你已添加相同的'對mat''obj'多次。如果你想'mat'成爲一個不同的列表,每次都要通過迭代,在你的外層循環中移動'mat = [']'行。 – khelwood

回答

0

原始=輸入( 「原糖的Enret數量: 」) COLS =輸入(「 列數的輸入號碼:」)

OBJ = []

對於k在範圍(2):

墊= [[[]用於範圍Ĵ(原始)用於i的範圍(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

試試這個傢伙。 它肯定會工作 –