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Asked:2023-08-23 17:39:36 +0000 UTC2023-08-23 17:39:36 +0000 UTC

如何循环遍历嵌套结构？

  • 772

有一个变量，它存储以下结构：

a = [
    {
        "key": [
            0,
            1,
            2,
            ...
        ]
    },
    {...},
    ...
]

尝试这样：

for i in a:
    for k, v in i:
        for j in k:
            print (j)

出现错误：

ValueError: too many values to unpack (expected 2)
python-3.x

2 个回答

  1. Best Answer
    a = [{"key": [0, 1, 2] },
    {"key1": [0, 1, 2, 3, 4,{"subkey":[1,2]}]}
    ]

def p(a):
    if isinstance(a,dict):
        for k,z in a.items():
            p(k)
            p(z)
    elif isinstance(a,list):
        for l in a:
            p(l)
    else:
        print(a)
        return
p(a)

    key
0
1
2
key1
0
1
2
3
4
subkey
1
2
    • 4

  2. 当然不是一个循环，而是作为一种选择。

    该解决方案的最大优点是yaml您可以从格式重新组装对象。

    pip install PyYaml

    import yaml

a = [
    {"key": [0, 1, 2]},
    {"key1": [0, 1, 2, 3, 4, {"subkey": [1, 2]}]},
    {"key25": [0, 4, 2, 10, 4, {"subkey": [1, 6, 9]}]},
    {"key48": [0, {"subkey": [11, 62, 80]}, 2, 10, 4, {"subkey": [1, 6, 9]}]},
]

with open('result.yaml', 'w', encoding='utf-8') as f:
    yaml.dump(
        a,
        f,
        sort_keys=False
    )

    result.yaml

    - key:
  - 0
  - 1
  - 2
- key1:
  - 0
  - 1
  - 2
  - 3
  - 4
  - subkey:
    - 1
    - 2
- key25:
  - 0
  - 4
  - 2
  - 10
  - 4
  - subkey:
    - 1
    - 6
    - 9
- key48:
  - 0
  - subkey:
    - 11
    - 62
    - 80
  - 2
  - 10
  - 4
  - subkey:
    - 1
    - 6
    - 9

    如果像这样打印：

    import yaml
import io

buf = io.StringIO()

yaml.dump(
    a,
    buf,
    sort_keys=False
)

print(
    buf.getvalue()
)
    • 0

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