在Python中重新测试Json输出

Question

我在循环中有一个json输出，在循环结束时，我需要为json分配一个唯一的键。如下所示：

out = {"34267":{"Annual Expenditure":250,"Annual Frequency":1,"Avg days":null,"First visit":1456012800000,"Frequency":1,"Last visit":1456012800000,"Merchants":32,"Monetory":250,"Recency":621,"Visit_Ids":[116644],"RFMClass":"244","AvgLTV":13305.7692307692},"101":{"Annual Expenditure":250,"Annual Frequency":1,"Avg days":null,"First visit":1423958400000,"Frequency":1,"Last visit":1423958400000,"Merchants":32,"Monetory":250,"Recency":992,"Visit_Ids":[1883],"RFMClass":"444","AvgLTV":13305.7692307692},"8877":{"Annual Expenditure":250,"Annual Frequency":1,"Avg days":null,"First visit":1459900800000,"Frequency":1,"Last visit":1459900800000,"Merchants":32,"Monetory":250,"Recency":576,"Visit_Ids":[141943],"RFMClass":"144","AvgLTV":13305.7692307692}}

但是分配一些类似于out_final = {32: out}， 使内部Json成为如下所示的字符串：

{32: '{"34267":{"Annual Expenditure":250,"Annual Frequency":1,"Avg days":null,"First visit":1456012800000,"Frequency":1,"Last visit":1456012800000,"Merchants":32,"Monetory":250,"Recency":621,"Visit_Ids":[116644],"RFMClass":"244","AvgLTV":13305.7692307692},"101":{"Annual Expenditure":250,"Annual Frequency":1,"Avg days":null,"First visit":1423958400000,"Frequency":1,"Last visit":1423958400000,"Merchants":32,"Monetory":250,"Recency":992,"Visit_Ids":[1883],"RFMClass":"444","AvgLTV":13305.7692307692},"8877":{"Annual Expenditure":250,"Annual Frequency":1,"Avg days":null,"First visit":1459900800000,"Frequency":1,"Last visit":1459900800000,"Merchants":32,"Monetory":250,"Recency":576,"Visit_Ids":[141943],"RFMClass":"144","AvgLTV":13305.7692307692}}'}

这使得无法搜索特定值。有没有一种方法可以在不丢失内部结构的情况下分配唯一的密钥。你知道吗

如下所示：

{32: {"34267":{"Annual Expenditure":250,"Annual Frequency":1,"Avg days":null,"First visit":1456012800000,"Frequency":1,"Last visit":1456012800000,"Merchants":32,"Monetory":250,"Recency":621,"Visit_Ids":[116644],"RFMClass":"244","AvgLTV":13305.7692307692},"101":{"Annual Expenditure":250,"Annual Frequency":1,"Avg days":null,"First visit":1423958400000,"Frequency":1,"Last visit":1423958400000,"Merchants":32,"Monetory":250,"Recency":992,"Visit_Ids":[1883],"RFMClass":"444","AvgLTV":13305.7692307692},"8877":{"Annual Expenditure":250,"Annual Frequency":1,"Avg days":null,"First visit":1459900800000,"Frequency":1,"Last visit":1459900800000,"Merchants":32,"Monetory":250,"Recency":576,"Visit_Ids":[141943],"RFMClass":"144","AvgLTV":13305.7692307692}}}

Answer 1

一种可能的解决方案是创建嵌套字典，然后通过this转换为json：

d = df.set_index('Guest').to_dict(orient='index')
d_final = {32: j}

import json
with open('result.json', 'w') as fp:
    json.dump(d, fp)

Answer 2

您可以用{{简单地转义{，如下所示：

out_final = '{{32:{}}}'.format(out)