Question

I have a list of tuples `data`:

data =[(array([[2, 1, 3]]), array([1])),
 (array([[2, 1, 2]]), array([1])),
 (array([[4, 4, 4]]), array([0])),
 (array([[4, 1, 1]]), array([0])),
 (array([[4, 4, 3]]), array([0]))]

为简单起见，此列表中只有5个元组。

当我运行以下代码时，似乎我能够解压缩每个元组每次迭代：

for x,y in data2:
    print(x,y)


output:

[[2 1 3]] [1]
[[2 1 2]] [1]
[[4 4 4]] [0]
[[4 1 1]] [0]
[[4 4 3]] [0]


This also works:

for x,y in data2[:2]:
    print(x,y)


output:

[[2 1 3]] [1]
[[2 1 2]] [1]


However, when I take only a single tuple from the list:

for x,y in data2[0]:
    print(x,y)


output: 
ValueError                                Traceback (most recent call last)
<ipython-input-185-1eed1fccdb3a> in <module>()
----> 1 for x,y in data2[0]:
      2     print(x,y)

ValueError: not enough values to unpack (expected 2, got 1)

我对在早期案例中如何解包元组感到困惑防止最后一个案例也成功解压缩元组。

谢谢。

Answer 1

在前两种情况下，您正在循环浏览list，在您访问的最后一个tuple

不确定要实现的目标，但data[0]代替data[:1]，而不是{{1}}。

Answer 2

如果您的数据如下所示：

data =[([[2, 1, 3]], [1]),
([[2, 1, 2]], [1]),
([[4, 4, 4]]), [0]),
([[4, 1, 1]], [0]),
([[4, 4, 3]], [0])]

for [a], b in data:
    print a, b

输出：

[2, 1, 3] [1]
[2, 1, 2] [1]
[4, 4, 4] [0]
[4, 1, 1] [0]
[4, 4, 3] [0]

元组解包的混乱行为 - 用于元组列表上的循环

2 个答案: