使用2级嵌套数组将数据帧转换为JSON [英] Convert dataframe to JSON with 2 level nested array

问题描述

我对Python编程有点陌生.我有一个小的要求，我需要以JSON格式列出给定两周内的所有客户及其金额.

I am a bit new to Python programming. I have a small requirement where in I need to list down all customers and their amounts for a given fortnight in a JSON format.

目前，我以这种方式拥有一个数据框:

Currently, I have a dataframe this way:

  FortNight      Amount     Customer    Parameter
  Apr-2FN-2018   339632.00    10992     CustomerSales
  Apr-2FN-2018   27282.00     10994     CustomerSales 
  Apr-2FN-2018   26353.00     10995     CustomerSales 
  Apr-2FN-2018   24797.00     11000     CustomerSales
  Apr-2FN-2018   21093.00     10990     CustomerSales

期望的JSON:

"CustomerSales" : [                                                                
    {"FortNight" : "Apr-2FN-2018",                                                                                      
         "Details" :[
             {"Customer":  "10992","Amount" : 339632.00},                                                                                                                                
             {"Customer":  "10994","Amount" : 27282.00},
             {"Customer":  "10995","Amount" : 26353.00},  
             {"Customer":  "11000","Amount" : 24797.00},
             {"Customer":  "10990","Amount" : 21093.00}
           ]
    }
]

我尝试过:

dict(df.set_index('Parameter').groupby(level=0).apply(lambda  x : ast.literal_eval(x.to_json(orient = 'records', date_format = "iso"))))

它检索此:

 [{'CustomerSales': 
[{'Customer': '10992', 'Amount': 339632.00, 'FortNight': 'Apr-2FN-2018'}, {'Customer': '10994', 'Amount': 27282.00, 'FortNight': 'Apr-2FN-2018'},{'Customer': '10995', 'Amount': 26353.00, 'FortNight': 'Apr-2FN-2018'},
{'Customer': '11000', 'Amount': 24797.00, 'FortNight': 'Apr-2FN-2018'},
{'Customer': '10990', 'Amount': 21093.00, 'FortNight': 'Apr-2FN-2018'}]}]

我也尝试了其他方法，但徒劳无功.欢迎任何帮助. 预先感谢！

I tried other ways too but in vain. Any help is welcome. Thanks in advance!

推荐答案

首先对参数和 FortNight列进行分组，然后在结果分组的行上使用.to_dict()来产生内部大多数词典:

Start by grouping on both the Parameter and FortNight columns, and using .to_dict() on the resulting grouped rows to produce the inner-most dictionaries:

details = df.groupby(['Parameter', 'FortNight']).apply(
    lambda r: r[['Customer', 'Amount']].to_dict(orient='records'))

这为您提供了一个在Parameter和FortNight上具有多个索引的系列，并且值都是正确格式的所有列表，每个条目都有一个包含Customer和Amount列的字典.如果需要转换值类型，请先对r[['Customer', 'Amount']]数据框结果进行转换，然后再对其调用to_dict().

This gives you a series with a multi-index over Parameter and FortNight, and the values are all the lists in the correct format, each entry a dictionary with Customer and Amount columns. If you need to convert the value types, do so on the r[['Customer', 'Amount']] dataframe result before calling to_dict() on it.

然后您可以取消堆叠系列化为数据框，为您提供嵌套的参数-> FortNight->详细信息结构；参数值变成列，每个列表由FortNight索引的客户/金额字典:

You can then unstack the series into a dataframe, giving you a nested Parameter -> FortNight -> details structure; the Parameter values become columns, each list of Customer / Amount dictionaries indexed by FortNight:

nested = details.unstack('Parameter')

如果将其变成字典，您将获得一本最正确的字典:

If you turn this into a dictionary, you'd get a dictionary that's mostly correct already:

>>> pprint(grouped.unstack('Parameter').to_dict())
{'CustomerSales': {'Apr-2FN-2018': [{'Amount': 339632.0, 'Customer': '10992'},
                                    {'Amount': 27282.0, 'Customer': '10994'},
                                    {'Amount': 26353.0, 'Customer': '10995'},
                                    {'Amount': 24797.0, 'Customer': '11000'},
                                    {'Amount': 21093.0, 'Customer': '10990'}]}}

但是对于您的格式，您需要将每一列中的值转换为{'FortNight': indexvalue, 'Details': value}映射的列表，然后然后将整个结构转换为字典:

but for your format, you'd convert the values in each column to a list of {'FortNight': indexvalue, 'Details': value} mappings, then converting the whole structure to a dictionary:

output = nested.apply(lambda s: [
    {s.index.name: idx, 'Details': value}
    for idx, value in s.items()
]).to_dict('records')

这将为您提供最终输出:

This gives you your final output:

>>> pprint(output)
[{'CustomerSales': {'Details': [{'Amount': 339632.0, 'Customer': '10992'},
                                {'Amount': 27282.0, 'Customer': '10994'},
                                {'Amount': 26353.0, 'Customer': '10995'},
                                {'Amount': 24797.0, 'Customer': '11000'},
                                {'Amount': 21093.0, 'Customer': '10990'}],
                    'FortNight': 'Apr-2FN-2018'}}]

如果需要JSON文档，请使用.to_json(orient='records')而不是.to_dict('records').

If you need a JSON document, use .to_json(orient='records') rather than .to_dict('records').

放在一起作为一个表达式:

Put together as one expression:

df.groupby(['Parameter', 'FortNight']).apply(
        lambda r: r[['Customer', 'Amount']].to_dict(orient='records')
    ).unstack('Parameter').apply(lambda s: [
        {s.index.name: idx, 'Details': value}
        for idx, value in s.items()]
    ).to_json(orient='records')

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