下面是示例列表数据,我想将其转换为动态字典。
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_to_nested_dict(*args,data),它将所有键列表动态转换为字典。
例如: conv_to_nested_dict(['standard','student'],result) 应该给出 op :
{
"118": {
"4": [{
"score": "0",
"type": "assessment"
}]
},
"119": {
"4": [{
"score": "0",
"type": "assignment"
}],
"5": [{
"score": "0",
"type": "assignment"
}]
}
}
conv_to_nested_dict(['标准','类型'],结果)
{
"118": {
"assessment": [{
"score": 0,
"student": "4"
}]
},
"119": {
"assignment": [{
"score": 0,
"student": "4"
},{
"score": 0,
"student": "5"
}]
}
}
最佳答案
这是一个总体想法。
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
是您的源数组,keys
是您想要对结果进行分组的键。对于每条记录,迭代结果 - 基于键值 ( record[key] ) 创建树结构。对于最后一个键 - 创建一个列表并将记录附加到其中。
关于Python-从字典列表创建动态嵌套字典,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/45673242/