A = [[1,0,0], [0,1,0], [0,0,1]]
B = []
m,n = 3,3
scalar = 5
for i in range(n):
B.append([])
for j in range(n):
result = A[i][j] * scalar
B[i].append(result)
#B = [[5, 0, 0], [0, 5, 0], [0, 0, 5]]
for i in B:
print(i)
然后你的输出是：
[5, 0, 0]
[0, 5, 0]
[0, 0, 5]
You were very close! Here is my solution
A = [[1,0,0], [0,1,0], [0,0,1]]
B = []
m,n = 3,3
scalar = 5
for i in range(n):
B.append([])
for j in range(n):
result = A[i][j] * scalar
B[i].append(result)
#B = [[5, 0, 0], [0, 5, 0], [0, 0, 5]]
for i in B:
print(i)
m, n = 3, 3
scalar = 5
A = [ [1, 0, 0],
[0, 1, 0],
[0, 0, 1] ]
B = []
for row_index,row in enumerate(A):
for column_index,column in enumerate(row):
A[row_index][column_index] *= scalar
for i in A:
print(i)
输出：
[5, 0, 0]
[0, 5, 0]
[0, 0, 5]
注意：我根本没有使用B。
My solution:
m, n = 3, 3
scalar = 5
A = [ [1, 0, 0],
[0, 1, 0],
[0, 0, 1] ]
B = []
for row_index,row in enumerate(A):
for column_index,column in enumerate(row):
A[row_index][column_index] *= scalar
for i in A:
print(i)
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评论（4）
你们非常接近！这是我的解决方案
然后你的输出是：
You were very close! Here is my solution
And, then your output is:
你能使用 numpy 吗？
它的工作原理如下：
Can you use
numpy?It works like this:
在不依赖其他模块/库的情况下，最好的选择是在循环之外处理打印。这是一种方法：
Without relying on another module/library, your best bet is to handle the printing outside the loop. Here's one way to do it:
我的解决方案：
输出：
注意：我根本没有使用
B。My solution:
Output:
Note: I didn´t use
Bat all.