Python-List返回[[…],6][重复]

Question

data = [[1,2],[3,4],[5,6]]

for x in data:
    print(x[0])

for x[0] in data:
    print(x)

1
3
5
[[1, 2], 6]
[[3, 4], 6]
[[...], 6]

>>> print(y)
[[Ellipsis], 6]

>>> print(y[0])
[0]

>>> print(x)
[[...], 6]

>>> print(x[0])
[[...], 6]

>>> print(x[0][0])
[[...], 6]

>>> print(x[0][0][0])
[[...], 6]

>>> print(x[1])
6

>>> print(x[0][1])
6

Answer 1

让我们给您的子列表命名：

a = [1, 2]
b = [3, 4]
c = [5, 6]
data = [a, b, c]

第一个循环将A、B和C依次绑定到X。当循环终止时，您实际上已经设置了x=c。

第二个循环现在将A、B和C依次绑定到X[0]。这对于a和B很好，但是对于C，实际上是在执行C[0]=C，创建循环引用。由于列表能够捕捉到此信息，因此它不会尝试打印