[已解决]嵌套列表相减

JoeJoy

a=[[6, 2, 0, 0, 0, 0], [2, 1, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0]]
b=[[4, 0, 0, 0, 0, 0], [2, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0]]

如何得出 [[2, 0, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0]]?

最佳答案

月排行榜 / 总排行榜

Twilight6

2020-7-22 22:05:57

1. a=[[6, 2, 0, 0, 0, 0], [2, 1, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0]]
   
2. b=[[4, 0, 0, 0, 0, 0], [2, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0]]
   
3. new_list = []
   
4. for i in range(len(a)):
   
5.     temp = []
   
6.     for j in range(len(a[i])):
   
7.         temp.append(a[i][j]-b[i][j])
   
8.     new_list.append(temp)
   
9. print(new_list)

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Twilight6

1. a=[[6, 2, 0, 0, 0, 0], [2, 1, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0]]
   
2. b=[[4, 0, 0, 0, 0, 0], [2, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0]]
   
3. new_list = []
   
4. for i in range(len(a)):
   
5.     temp = []
   
6.     for j in range(len(a[i])):
   
7.         temp.append(a[i][j]-b[i][j])
   
8.     new_list.append(temp)
   
9. print(new_list)

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JoeJoy

Twilight6 发表于 2020-7-22 22:05

感谢

zltzlt

用列表推导式：

1. >>> a = [[6, 2, 0, 0, 0, 0], [2, 1, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0]]
   
2. >>> b = [[4, 0, 0, 0, 0, 0], [2, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0]]
   
3. >>> [[c - d for c, d in zip(a, b)] for a, b in zip(a, b)]
   
4. [[2, 2, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0]]

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Twilight6

JoeJoy 发表于 2020-7-22 22:07
感谢

如果列表中的元素个数都是相同的，还可以这样用列表推导式：

1. a=[[6, 2, 0, 0, 0, 0], [2, 1, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0]]
   
2. b=[[4, 0, 0, 0, 0, 0], [2, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0]]
   
3. 
   
4. new_list = [ [a[i][j]-b[i][j] for i in range(len(a))] for j in range(len(a[0]))]
   
5. print(new_list)

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Twilight6

zltzlt 发表于 2020-7-22 22:08
用列表推导式：

  厉害，我都忘记可以用 zip 了 思维定式了