將列表字符串轉換為值列表

Question

我有一個列表字符串，我試圖將其轉換為列表，但是ast.literal_eval返回錯誤：

y = "[
        {
           'ake_ii': 88888, 
           'azefmlkh_amrlgba_dd': datetime.datetime(2022, 1, 31, 16, 52), 
           'sklmfs_qujdf_': datetime.datetime(2022, 1, 31, 23, 4)
        }
     ]"
ast.literal_eval(y)

返回錯誤

ValueError: malformed node or string: <ast.Call object at 0x7fbaec22da60>

我想要一個返回的列表而不是一個字符串。

Answer 1

這是改編自我的其他相關答案的答案：

import ast
import datetime

data = """
[
        {
           'ake_ii': 88888, 
           'azefmlkh_amrlgba_dd': datetime.datetime(2022, 1, 31, 16, 52), 
           'sklmfs_qujdf_': datetime.datetime(2022, 1, 31, 23, 4)
        }
]
""".strip()


def literal_eval_with_datetime(source):
    # Adapted from `ast.literal_eval`
    def _convert(node):
        if isinstance(node, list):
            return [_convert(arg) for arg in node]
        if isinstance(node, ast.Constant):
            return node.value
        if isinstance(node, ast.Tuple):
            return tuple(map(_convert, node.elts))
        if isinstance(node, ast.List):
            return list(map(_convert, node.elts))
        if isinstance(node, ast.Set):
            return set(map(_convert, node.elts))
        if isinstance(node, ast.Call) and isinstance(node.func, ast.Name) and node.func.id == 'set' and node.args == node.keywords == []:
            return set()
        if isinstance(node, ast.Dict):
            return dict(zip(map(_convert, node.keys), map(_convert, node.values)))
        if isinstance(node, ast.Expression):
            return _convert(node.body)
        if isinstance(node, ast.Call) and ast.get_source_segment(source, node.func) == 'datetime.datetime':
            return datetime.datetime(*_convert(node.args))
        return {
            f'${node.__class__.__name__}': ast.get_source_segment(source, node),
        }

    return _convert(ast.parse(source, mode='eval'))


print(literal_eval_with_datetime(data))

output 是（當然是真實列表）：

[
    {
        "ake_ii": 88888,
        "azefmlkh_amrlgba_dd": datetime.datetime(2022, 1, 31, 16, 52),
        "sklmfs_qujdf_": datetime.datetime(2022, 1, 31, 23, 4),
    }
]