python - 如何将多索引 DataFrame 转换为嵌套字典结构？

Question

如何制作一个具有多索引的数据框并将其制作成一个漂亮的嵌套字典？

这是我到目前为止所尝试过的，它很接近，但是键是元组。希望将它们分解为更多的字典键。

我尝试过的:

that = {'Food':['Apple','Apple','Apple','Apple','Banana','Banana','Orange','Orange'],
    'Color':['Red','Green','Yellow','Red','Red','Green','Green','Yellow'],
    'Type':['100','4','7','101','100','100','4','7'],
    'time':[np.linspace(0,10,2) for i in range(8)]}

nn = pd.DataFrame(that)
nn = nn.set_index(['Food','Color','Type'])
vv = {}
for idx in nn.index:
   vv[idx] = nn.loc[idx]

vv
Out[1]: 
{('Apple', 'Red', '100'): time    [0.0, 10.0]
 Name: (Apple, Red, 100), dtype: object,
 ('Apple', 'Green', '4'): time    [0.0, 10.0]
 Name: (Apple, Green, 4), dtype: object,
 ('Apple', 'Yellow', '7'): time    [0.0, 10.0]
 Name: (Apple, Yellow, 7), dtype: object,
 ('Apple', 'Red', '101'): time    [0.0, 10.0]
 Name: (Apple, Red, 101), dtype: object,
 ('Banana', 'Red', '100'): time    [0.0, 10.0]
 Name: (Banana, Red, 100), dtype: object,
 ('Banana', 'Green', '100'): time    [0.0, 10.0]
 Name: (Banana, Green, 100), dtype: object,
 ('Orange', 'Green', '4'): time    [0.0, 10.0]
 Name: (Orange, Green, 4), dtype: object,
 ('Orange', 'Yellow', '7'): time    [0.0, 10.0]
 Name: (Orange, Yellow, 7), dtype: object}

我想要的输出是什么样的。

vv = {'Apple':{'Red':{'100':[0,10],'101':[0,10]},
               'Green':{'4':[0,10]},
               'Yellow':{'7':[0,10]}},
      'Banana':{'Red':{'100':[0,10]},
                'Green':{'100':[0,10]}}
      'Orange':{'Green':{'4':[0,10]},
                'Yellow':{'7':[0,10]}}}

编辑: 将范围改回 8...是一个拼写错误，并将 linspace 中的点数更改为 2 点，以便简单地反射(reflect)示例。

编辑2: 寻找一种通用的方法来做到这一点。特别是，我的一位同事在 pyqt 中编写了一个 treeView 模型，它接受树的嵌套字典。我只是希望能够将我创建的数据帧快速转换为所需的格式。

对于那些对一般如何执行此操作感到好奇的人，请看这里。我写的不错的小函数。更能满足我的需要。

that = {'Food':['Apple','Apple','Apple','Apple','Banana','Banana','Orange','Orange'],
        'Color':['Red','Green','Yellow','Red','Red','Green','Green','Yellow'],
        'Type':['100','4','7','101','100','100','4','7'],
        'time':[np.linspace(0,10,2) for i in range(8)]}

x = pd.DataFrame(that)

def NestedDict_fromDF(iDF,keyorder,values):
    if not isinstance(keyorder,list):
        keyorder = [keyorder]
    if not isinstance(values,list):
        values = [values]
    for i in reversed(range(len(keyorder))):
        if keyorder[i] not in iDF:
            keyorder.pop(i)
    for i in reversed(range(len(values))):
        if values[i] not in iDF:
            values.pop(i)
    rdict = {}
    if keyorder:
        ndf = iDF.set_index(keyorder)
        def makeDict(basedict,group):
            for k,g in group:
                basedict[k] = {}
                try:
                    makeDict(basedict[k], g.droplevel(0).groupby(level=0))
                except:
                    if values:
                        basedict[k] = g[values].reset_index(drop=True)
                    else:
                        basedict[k] = []
            return basedict
        rdict = makeDict({}, ndf.groupby(level=0))
    return rdict

yy = NestedDict_fromDF(x,['Food','Color','Type','Integer'],['time'])


{'Apple': {'Green': {'4':DataFrame},
           'Red': {'100':DataFrame,
                   '101':DataFrame},
           'Yellow': {'7':DataFrame}},
 'Banana': {'Green': {'100':DataFrame},
            'Red': {'100':DataFrame}},
 'Orange': {'Green': {'4':DataFrame},
            'Yellow': {'7':DataFrame}}}

Answer 1

它变得太复杂、太快:

from pprint import pprint
import pandas as pd

that = {'Food':['Apple','Apple','Apple','Apple','Banana','Banana','Orange','Orange'],
    'Color':['Red','Green','Yellow','Red','Red','Green','Green','Yellow'],
    'Type':['100','4','7','101','100','100','4','7'],
    'time':[np.linspace(0,10,2) for i in range(8)]}

nn = pd.DataFrame(that)
df = nn.groupby(['Food', 'Color', 'Type']).agg(list)
d = {}
new_df = df.groupby(level=[0,1]).apply(lambda df:df.xs(df.name).to_dict()).to_dict() #[1]
for (food, color), v in new_df.items():
    if not food in d:
        d[food] = {color: {Type: time[0].tolist() for Type, time in v['time'].items()}}
    else:
        d[food][color] = {Type: time[0].tolist() for Type, time in v['time'].items()}
pprint(d)

输出:

{'Apple': {'Green': {'4': [0.0, 10.0]},
           'Red': {'100': [0.0, 10.0], '101': [0.0, 10.0]},
           'Yellow': {'7': [0.0, 10.0]}},
 'Banana': {'Green': {'100': [0.0, 10.0]}, 'Red': {'100': [0.0, 10.0]}},
 'Orange': {'Green': {'4': [0.0, 10.0]}, 'Yellow': {'7': [0.0, 10.0]}}}

[1] 取自:DataFrame with MultiIndex to dict

哈!终于明白了!

that = {'Food':['Apple','Apple','Apple','Apple','Banana','Banana','Orange','Orange'],
    'Color':['Red','Green','Yellow','Red','Red','Green','Green','Yellow'],
    'Type':['100','4','7','101','100','100','4','7'],
    'time':[np.linspace(0,10,2) for i in range(8)]}

nn = pd.DataFrame(that)
nn = nn.set_index(['Food','Color','Type'])
group = nn.groupby(level=0)
d = {k: g.droplevel(0).groupby(level=0)
       .apply(lambda df:df.xs(df.name)['time']
       .apply(lambda x:x.tolist()).to_dict())
       .to_dict() for k,g in group}
pprint(d)

{'Apple': {'Green': {'4': [0.0, 10.0]},
           'Red': {'100': [0.0, 10.0], '101': [0.0, 10.0]},
           'Yellow': {'7': [0.0, 10.0]}},
 'Banana': {'Green': {'100': [0.0, 10.0]}, 'Red': {'100': [0.0, 10.0]}},
 'Orange': {'Green': {'4': [0.0, 10.0]}, 'Yellow': {'7': [0.0, 10.0]}}}