将pandas数据帧转换为嵌套的dictionary/json格式

Question

我正在尝试将以下pandas数据帧(python(转换为嵌套字典格式。输入数据熊猫数据帧：

statetraffic   |state    | act       | traffic| reward | header      | time |   id
stateacttraf |     1   |    1      | 12      |  22     |   str1    |   1572340221000 | 34022100
stateacttraf |     1   |    2      | 87      |  30     |   str1    |   1572340221000 | 34022100
stateacttraf |     1   |    3      | 1       |  48     |   str1    |   1572340221000 | 34022100
stateacttraf |     2   |    1      | 10      |  13     |   str1    |   1572340221000 | 34022100
stateacttraf |     2   |    2      | 80      |  27     |   str1    |   1572340221000 | 34022100
stateacttraf |     2   |    3      | 10      |  60     |   str1    |   1572340221000 | 34022100

有尝试过以下代码，但不起作用：

1)final_op = input_df.to_dict(orient='records') -> does not provide the answer       
2)from jsonmerge import merge; 
message = {'statetraffic': 'stateacttraf'}; 
message1 = {'time': time.time()}; 
result = merge(final_op, message, message2) -> Neither does this provide the answer either

需要某种形式的嵌套字典

期望dictionary/json输出如下：

{
{  "statetraffic":"stateacttraf",
"time":1572340221000,
"str1":{ 
"id":34022100,
"state":1,
"act":1,
"trafficSplit":12,
"reward":22
}
{ 
"statetraffic":"stateacttraf",
"time":1572340221000,
"str1":{ 
"id":34022100,
"state":1,
"act":2,
"trafficSplit":87,
"reward":30
}
{ 
"statetraffic":"stateacttraf",
"time":1572340221000,
"str1":{ 
"id":34022100,
"state":1,
"act":3,
"trafficSplit":1,
"reward":48
}
{  "statetraffic":"stateacttraf",
"time":1572340221000,
"str1":{ 
"id":34022100,
"state":2,
"act":1,
"trafficSplit":10,
"reward":13
}
}

迫切需要这种格式的输出。因此，任何帮助都将不胜感激。

Answer 1

试试这个，假设你的数据帧是df

main_dict = df.to_dict()
uprow= ["statetraffic","time","header"]
drow = ["id","state" ,"act" ,"traffic","reward"]
datalist = []
for c in range(df.shape[0]):
subd = {}
for k,v in main_dict.items():
subd[k] = v[c]
subd_ = subd.copy()
tmp = subd.get("header")
subd[tmp] = 0
for i in uprow: del subd_[i]
subd[tmp]=subd_
for i in drow: del subd[i] 
del subd["header"]
datalist.append(subd)
print(datalist)