[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

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]]
new_list = []
for i in range(len(a)):
    temp = []
    for j in range(len(a[i])):
        temp.append(a[i][j]-b[i][j])
    new_list.append(temp)
print(new_list)

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Twilight6

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]]
new_list = []
for i in range(len(a)):
    temp = []
    for j in range(len(a[i])):
        temp.append(a[i][j]-b[i][j])
    new_list.append(temp)
print(new_list)

JoeJoy

Twilight6 发表于 2020-7-22 22:05

感谢

zltzlt

用列表推导式：

>>> 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]]
>>> [[c - d for c, d in zip(a, b)] for a, b in zip(a, b)]
[[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]]

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：

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]]

new_list = [ [a[i][j]-b[i][j] for i in range(len(a))] for j in range(len(a[0]))]
print(new_list)

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