如何从pandas DataFrame生成n级分层JSON? [英] How to generate n-level hierarchical JSON from pandas DataFrame?

问题描述

是否有一种有效的方法来创建层次化JSON(深度为n级)，其中父值是键而不是变量标签?即:

Is there an efficient way to create hierarchical JSON (n-levels deep) where the parent values are the keys and not the variable label? i.e:

{"2017-12-31":
    {"Junior":
        {"Electronics":
            {"A":
                {"sales": 0.440755
                }
            },
            {"B":
                {"sales": -3.230951
                }
            }
        }, ...etc...
    }, ...etc...
}, ...etc... 

---

1.我的测试DataFrame:

colIndex=pd.MultiIndex.from_product([['New York','Paris'],
                                     ['Electronics','Household'],
                                     ['A','B','C'],
                                     ['Junior','Senior']],
                               names=['City','Department','Team','Job Role'])

rowIndex=pd.date_range('25-12-2017',periods=12,freq='D')

df1=pd.DataFrame(np.random.randn(12, 24), index=rowIndex, columns=colIndex)
df1.index.name='Date'
df2=df1.resample('M').sum()
df3=df2.stack(level=0).groupby('Date').sum()

2.我正在进行的转换，因为它似乎是从以下位置构建JSON的最合乎逻辑的结构:

2. Transformation I'm making as it seems to be the most logical structure to build the JSON from:

df4=df3.stack(level=[0,1,2]).reset_index() \
    .set_index(['Date','Job Role','Department','Team']) \
    .sort_index()

3.我到目前为止的尝试

我遇到了这个非常有帮助的SO问题，该问题使用如下代码使用代码解决了一层嵌套的问题:

I came across this very helpful SO question which solves the problem for one level of nesting using code along the lines of:

j =(df.groupby(['ID','Location','Country','Latitude','Longitude'],as_index=False) \
    .apply(lambda x: x[['timestamp','tide']].to_dict('r'))\
    .reset_index()\
    .rename(columns={0:'Tide-Data'})\
    .to_json(orient='records'))

...但是我找不到找到嵌套.groupby()的方法:

...but I can't find a way to get nested .groupby()s working:

j=(df.groupby('date', as_index=True).apply(
    lambda x: x.groupby('Job Role', as_index=True).apply(
        lambda x: x.groupby('Department', as_index=True).apply(
            lambda x: x.groupby('Team', as_index=True).to_dict())))  \
                .reset_index().rename(columns={0:'sales'}).to_json(orient='records'))

推荐答案

您可以使用itertuples生成嵌套的dict，然后转储到json.为此，您需要将日期时间戳更改为string

You can use itertuples to generate a nested dict, and then dump to json. To do this, you need to change the date timestamp to string

df4=df3.stack(level=[0,1,2]).reset_index() 
df4['Date'] = df4['Date'].dt.strftime('%Y-%m-%d')
df4 = df4.set_index(['Date','Job Role','Department','Team']) \
    .sort_index()

创建嵌套字典

def nested_dict():
    return collections.defaultdict(nested_dict)
result = nested_dict()

使用itertuples填充

for row in df4.itertuples():
    result[row.Index[0]][row.Index[1]][row.Index[2]][row.Index[3]]['sales'] = row._1
    # print(row)

，然后使用json模块将其转储.

and then use the json module to dump it.

import json
json.dumps(result)

'{"2017-12-31":{初级":{电子":{"A":{销售":-0.3947134370101142}，"B":{销售":-0.9873530754403204}， "C":{销售":-1.1182598058984508}，家庭":{"A":{销售":-1.1211850078098677}，"B":{销售":2.0330914483907847}，"C":{销售":3.94762379718749}}}，高级":{电子":{"A":{销售":1.4528493451404196}，"B":{销售":-2.3277322345261005}，"C":{销售:-2.8040263791743922}}，"家庭:{" A:{"销售:3.0972591929279663}，" B:{"销售:9.884565742502392}，" C:{"销售:2.9359830722457576}}}} ，"2018-01-31":{初级":{电子":{"A":{销售":-1.3580300149125217}，"B":{销售":1.414665000013205}，"C":{ 销售":-1.432795129108244}}，家庭":{"A":{销售":2.7783259569115346}，"B":{销售":2.717700275321333}，"C":{销售":1.4358377416259644}} }，"Senior":{"Electronics":{"A":{"sales":2.8981726774941485}，"B":{"sales":12.022897003654117}，"C":{"sales":0.01776855733076088}}，家庭":{"A":{销售":-3.3421637766130 92}，"B":{"sales":-5.283208386572307}，"C":{"sales":2.942580121975619}}}}}''

'{"2017-12-31": {"Junior": {"Electronics": {"A": {"sales": -0.3947134370101142}, "B": {"sales": -0.9873530754403204}, "C": {"sales": -1.1182598058984508}}, "Household": {"A": {"sales": -1.1211850078098677}, "B": {"sales": 2.0330914483907847}, "C": {"sales": 3.94762379718749}}}, "Senior": {"Electronics": {"A": {"sales": 1.4528493451404196}, "B": {"sales": -2.3277322345261005}, "C": {"sales": -2.8040263791743922}}, "Household": {"A": {"sales": 3.0972591929279663}, "B": {"sales": 9.884565742502392}, "C": {"sales": 2.9359830722457576}}}}, "2018-01-31": {"Junior": {"Electronics": {"A": {"sales": -1.3580300149125217}, "B": {"sales": 1.414665000013205}, "C": {"sales": -1.432795129108244}}, "Household": {"A": {"sales": 2.7783259569115346}, "B": {"sales": 2.717700275321333}, "C": {"sales": 1.4358377416259644}}}, "Senior": {"Electronics": {"A": {"sales": 2.8981726774941485}, "B": {"sales": 12.022897003654117}, "C": {"sales": 0.01776855733076088}}, "Household": {"A": {"sales": -3.342163776613092}, "B": {"sales": -5.283208386572307}, "C": {"sales": 2.942580121975619}}}}}'

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