代码之家 › 专栏 › 技术社区 › Akash

Python-从字典列表创建动态嵌套字典

recursive-datastructures python-3.x python-2.7 python

Akash · 技术社区 · 7 年前

result = [
    {
        "standard": "119",
        "score": "0",
        "type": "assignment",
        "student": "4"
    },
    {
        "standard": "119",
        "score": "0",
        "type": "assignment",
        "student": "5"
    },
    {
        "standard": "118",
        "score": "0",
        "type": "assessment",
        "student": "4"
    }
]

我想创建一个函数conv\u to\u nested\u dict(*args,data),它将所有键列表动态转换为dictional。

{
    "118": {
        "4": [{
            "score": "0",
            "type": "assessment"
        }]
    },
    "119": {
        "4": [{
            "score": "0",
            "type": "assignment"
        }],
        "5": [{
            "score": "0",
            "type": "assignment"
        }]
    }

}

conv\u to\u nested\u dict(['standard','type',result)

{
    "118": {
        "assessment": [{
            "score": 0,
            "student": "4"
        }]
    },
    "119": {
        "assignment": [{
            "score": 0,
            "student": "4"
        },{
            "score": 0,
            "student": "5"
        }]
    }

}

1 回复 | 直到 7 年前

wroniasty 7 年前

这是一个总体思路。

def conf_to_nested_dict(keys, result):
    R = {}
    for record in result: 
        node = R
        for key in keys[:-1]:
            kv = record[key]
            next_node = node.get(kv, {})
            node[kv] = next_node
            node = next_node
        last_node = node.get(record[keys[-1]], [])
        last_node.append(record)
        node[record[keys[-1]]] = last_node


    return R

#R is your structure

result 是源阵列, keys 是要根据其对结果进行分组的键。迭代结果,为每个记录创建基于键值的树结构(记录[key])。对于最后一个键,创建一个列表并将记录附加到其中。