如何将由列表组成的字符串转换为真实列表[关闭]

Question

现在我有了 unicode 对象

u'[
    [
        [
            ["Hello","w_0","w_0,w_10",1],
            [".","w_1","",0," "],
            ["This","w_2","",0," "],
        ]
        [
            ["\u4f60\u597d","w_10","w_0,w_10",1],
            ["\uff0c","w_11","",0],
            ["\u6211","w_12","",0],
        ],
        "http://www.kekenet.com/menu/201206/187574.shtml",154191
    ]
]'

如何将其转换为列表对象，例如

[
    [
        [
            ["Hello","w_0","w_0,w_10",1],
            [".","w_1","",0," "],
            ["This","w_2","",0," "],
        ],
        [
            ["\u4f60\u597d","w_10","w_0,w_10",1],
            ["\uff0c","w_11","",0],
            ["\u6211","w_12","",0],
        ],
        "http://www.kekenet.com/menu/201206/187574.shtml",154191
    ],
]

Answer 1

您可以使用 ast.literal_eval 安全地评估包含 Python 文字的字符串。

from ast import literal_eval

a = '["Hello", "World!", 2]'
b = literal_eval(a)
# ["Hello", "World!", 2]

请注意，字符串只能包含以下内容：字符串、字节、数字、元组、列表、字典、集合、布尔值和 None（取自文档 here）

Answer 2

在这种情况下你很幸运，因为你的字符串似乎是有效的 JSON，所以你可以这样做：

from simplejson import json

value = u'[
    [
        [
            ["Hello","w_0","w_0,w_10",1],
            [".","w_1","",0," "],
            ["This","w_2","",0," "],
        ]
        [
            ["\u4f60\u597d","w_10","w_0,w_10",1],
            ["\uff0c","w_11","",0],
            ["\u6211","w_12","",0],
        ],
        "http://www.kekenet.com/menu/201206/187574.shtml",154191
    ]
]'

array_value = json.loads(value)

为此，它依赖于使用双引号字符串的字符串（根据 json 规范），即 ["Hello","w_0","w_0,w_10",1] 而不是 ['Hello','w_0','w_0,w_10',1]

与使用eval 的解决方案相比，它的优势在于更安全……如果此值来自用户输入或其他不受信任的代码，则使用eval 非常危险，而json 是安全的。

编辑：
正如已经指出的那样，发布的字符串不是有效的 json ......虽然我想知道这是否是一个错字？

问题是：

value = u'[
    [
        [
            ["Hello","w_0","w_0,w_10",1],
            [".","w_1","",0," "],
            ["This","w_2","",0," "],
        ] <-- should be a comma here
        [
            ["\u4f60\u597d","w_10","w_0,w_10",1],