Question

我有一个元组列表，并希望从中创建JSON

[(1, 'a', 'A'), (2, 'b', 'B'), (3, 'c', 'C'), (4, 'd', 'D'), (5, 'e', 'E'), (6, 'f', 'F'), (7, 'g', 'G'), (8, 'h', 'H'), (9, 'i', 'I')]

预期的JSON

{
    "List": [
                {
                    "name": "1",
                    "description": "a",
                    "type": "A"
                },
                {
                    "name": "2",
                    "description": "b",
                    "type": "B"
                },
                {
                    "name": "3",
                    "description": "c",
                    "type": "C"
                },
               and so on....
            ]
}

我该怎么做？

重复的已确定答案无效并发出此错误

ValueError: dictionary update sequence element #0 has length 3; 2 is required

Answer 1

zip键和每个元组在结果上调用dict然后 json.dumps ：

import json
keys = ["name", "description","type"]

l=[(1, 'a', 'A'), (2, 'b', 'B'), (3, 'c', 'C'), (4, 'd', 'D'), (5, 'e', 'E'), (6, 'f', 'F'), (7, 'g', 'G'), (8, 'h', 'H'), (9, 'i', 'I')]


js =  json.dumps({"List":[dict(zip(keys, tup)) for tup in l]})

这给了你：

'{"List": [{"type": "A", "name": 1, "description": "a"}, {"type": "B", "name": 2, "description": "b"}, {"type": "C", "name": 3, "description": "c"}, {"type": "D", "name": 4, "description": "d"}, {"type": "E", "name": 5, "description": "e"}, {"type": "F", "name": 6, "description": "f"}, {"type": "G", "name": 7, "description": "g"}, {"type": "H", "name": 8, "description": "h"}, {"type": "I", "name": 9, "description": "i"}]}'

从元组的值列表和单独的键列表中创建JSON dict

1 个答案: