Python 如何将pandas dataframe转换为结构独特的嵌套json
我有一个DF,其结构如下:Python 如何将pandas dataframe转换为结构独特的嵌套json,python,json,pandas,dataframe,Python,Json,Pandas,Dataframe,我有一个DF,其结构如下: traffic_group app_id key category factors 0 desktop app1 CI html 16.618628 1 desktop app1 CI xhr 35.497082 2 desktop app1 IP html 18.294468 3 desktop ap
traffic_group app_id key category factors
0 desktop app1 CI html 16.618628
1 desktop app1 CI xhr 35.497082
2 desktop app1 IP html 18.294468
3 desktop app1 IP xhr 30.422464
4 desktop app2 CI html 11.028240
5 desktop app2 CI json 33.548279
6 mobile app1 IP html 12.808367
7 mobile app1 IP image 14.410633
我需要将其输出到以下结构的json:
{ "desktop": {
app1: [ {
"key": "CI",
"threshold: 1,
"window": 60,
"factors: {
"html" : 16.618628
"xhr" : 35.497082
}
}, {
"key": "IP",
"threshold: 1,
"window": 60,
"factors: {
"html" : 18.294468
"xhr" : 30.422464
}
],
app2: [ {
"key": "CI",
"threshold: 1,
"window": 60,
"factors: {
"html" : 11.028240
"json" : 33.548279
}
}
},
"mobile": {
app1: [ {
"key": "IP",
"threshold: 1,
"window": 60,
"factors: {
"html" : 12.808367
"xhr" : 14.410633
}
]
}
}
这一结构无疑是错综复杂的
我考虑了以下先前的答案,并试图模仿它们的逻辑,但没有效果:
感谢您的帮助。请不要只是发布解决方案,还要解释你的逻辑。好吧,我已经用“老式”的方式解决了。为将来可能需要我的解决方案的任何人发布我的解决方案。尽管如此,如果有人能用熊猫做这件事,我很想看看
json_output = {}
for traffic_group in sorted_df.traffic_group.unique():
json_output[traffic_group] = {}
for app_id in sorted_df[sorted_df.traffic_group == traffic_group].app_id.unique():
json_output[traffic_group][app_id] = []
for key in sorted_df[(sorted_df.traffic_group == traffic_group) &
(sorted_df.app_id == app_id)].key.unique():
inner_dict = {"key" : key, "threshold" : 1, "window" : 60, "factors" : {}}
for category in sorted_df[(sorted_df.traffic_group == traffic_group) &
(sorted_df.app_id == app_id) &
(sorted_df.key == key)].category.unique():
value = sorted_df[(sorted_df.traffic_group == traffic_group) &
(sorted_df.app_id == app_id) &
(sorted_df.key == key) &
(sorted_df.category == category)].factors
inner_dict["factors"][category] = value.iloc[0]
json_output[traffic_group][app_id].append(inner_dict)
我在输入中没有看到嵌套字典的任何“阈值”和“窗口”键。假设它们有固定的值。根据您的输出,似乎每个三元组(流量组、应用程序id、密钥)都希望(通常)创建一个不同的嵌套字典。因此,我们需要使用这三个键进行初始groupby操作。我们为每个组创建嵌套字典:
def创建嵌套命令(df):
返回{'key':df['key'].unique()[0],'threshold':1,'window':60,'factors':dict(zip(df['category'],df['factors']))
df=df.groupby(['traffic\u group','app\u id','key'])。应用(创建嵌套目录)
下一步是将行组合到每个(流量组、应用程序id)doublet的列表中,并将它们作为dict返回:
df=df.groupby(['traffic\u group','app\u id']).apply(lambda-df:df.tolist())
最后一步是将df
转换为输出。做这件事有多种方法。一个简单的例子如下:
df=df.reset_index().groupby('traffic_group').apply(lambda df:df.values)
output=dict(zip(df.index,[{app_id:val for u,app_id,val in vals}for vals in df.values]))
使用以下方法:
In [208]: d = {}
In [209]: grouped = df.groupby(['traffic_group', 'app_id', 'key']).agg(pd.Series.to_dict).to_dict(orient='index')
In [210]: for t, v in grouped.items():
...: traff_gr, app_id, key = t
...: inner_d = {"key": key, "threshold": 1, "window": 60, 'factors': dict(zip(v['category'].values(), v['f
...: actors'].values()))}
...: d.setdefault(traff_gr, {}).setdefault(app_id, []).append(inner_d)
...:
In [211]: d
Out[211]:
{'desktop': {'app1': [{'key': 'CI',
'threshold': 1,
'window': 60,
'factors': {'html': 16.618628, 'xhr': 35.497082}},
{'key': 'IP',
'threshold': 1,
'window': 60,
'factors': {'html': 18.294468, 'xhr': 30.422464}}],
'app2': [{'key': 'CI',
'threshold': 1,
'window': 60,
'factors': {'html': 11.02824, 'json': 33.548279}}]},
'mobile': {'app1': [{'key': 'IP',
'threshold': 1,
'window': 60,
'factors': {'html': 12.808367, 'image': 14.410632999999999}}]}}
您的解决方案似乎有效,但@Georgios的答案更为“pythonic”,他/她的流程通过解释变得清晰。