转换嵌套的DataJSON框架

Question

我的问题与此相反：

Create a Pandas DataFrame from deeply nested JSON

我想知道是否可以反过来。给一张桌子如下：

     Library  Level           School Major  2013 Total
200  MS_AVERY  UGRAD  GENERAL STUDIES  GEST        5079
201  MS_AVERY  UGRAD  GENERAL STUDIES  HIST           5
202  MS_AVERY  UGRAD  GENERAL STUDIES  MELC           2
203  MS_AVERY  UGRAD  GENERAL STUDIES  PHIL          10
204  MS_AVERY  UGRAD  GENERAL STUDIES  PHYS           1
205  MS_AVERY  UGRAD  GENERAL STUDIES  POLS          53

是否可以生成一个嵌套的dict（或JSON），如：

口述：

^{pr2}$

Answer 1

以下是我在处理this question时提出的解决方案：

def rollup_to_dict_core(x, values, columns, d_columns=None):
    if d_columns is None:
        d_columns = []

    if len(columns) == 1:
        if len(values) == 1:
            return x.set_index(columns)[values[0]].to_dict()
        else:
            return x.set_index(columns)[values].to_dict(orient='index')
    else:
        res = x.groupby([columns[0]] + d_columns).apply(lambda y: rollup_to_dict_core(y, values, columns[1:]))
        if len(d_columns) == 0:
            return res.to_dict()
        else:
            res.name = columns[1]
            res = res.reset_index(level=range(1, len(d_columns) + 1))
            return res.to_dict(orient='index')

def rollup_to_dict(x, values, d_columns=None):
    if d_columns is None:
        d_columns = []

    columns = [c for c in x.columns if c not in values and c not in d_columns]
    return rollup_to_dict_core(x, values, columns, d_columns)

>>> pprint(rollup_to_dict(df, ['2013 Total']))
{'MS_AVERY': {'UGRAD': {'GENERAL STUDIES': {'GEST': 5079,
                                            'HIST': 5,
                                            'MELC': 2,
                                            'PHIL': 10,
                                            'PHYS': 1,
                                            'POLS': 53}}}}

Answer 2

如果给定DataFrame对象，创建一个函数来构建递归字典似乎并不难：

def fdrec(df):
    drec = dict()
    ncols = df.values.shape[1]
    for line in df.values:
        d = drec
        for j, col in enumerate(line[:-1]):
            if not col in d.keys():
                if j != ncols-2:
                    d[col] = {}
                    d = d[col]
                else:
                    d[col] = line[-1]
            else:
                if j!= ncols-2:
                    d = d[col]
    return drec

将产生：

^{pr2}$