具有MultiIndex的DataFrame用于dict

Question

我有一个带有MultiIndex的数据帧.我想知道我是否以正确的方式创建了数据框(见下文).

             01.01  02.01  03.01  04.01
bar total1     40     52     18     11
    total2     36     85      5     92
baz total1     23     39     45     70
    total2     50     49     51     65
foo total1     23     97     17     97
    total2     64     56     94     45
qux total1     13     73     38      4
    total2     80      8     61     50

df.index.values 结果是:

array([('bar', 'total1'), ('bar', 'total2'), ('baz', 'total1'),
       ('baz', 'total2'), ('foo', 'total1'), ('foo', 'total2'),
       ('qux', 'total1'), ('qux', 'total2')], dtype=object)

df.index.get_level_values 结果是:

<bound method MultiIndex.get_level_values of MultiIndex(levels=[[u'bar', u'baz', u'foo', u'qux'], [u'total1', u'total2']],
           labels=[[0, 0, 1, 1, 2, 2, 3, 3], [0, 1, 0, 1, 0, 1, 0, 1]],names=[]

我最终希望将df转换为字典的字典,这样第一个字典键是['bar','baz','foo','qux']之一,值是日期,内部字典是'total1'和'totals2'作为键,值是df的整数.另一种解释是,例如,如果dict1是dict然后调用:

dict1['bar']

会导致输出:

{u'bar':{'01.01':{'total1':40,'total2':36},'02.01':{'total1':52,'total2':85},'03.01':{'total1':18,'total2':5},'04.01':{'total1':11,'total2':92} } }

为了达到这个目的,我需要改变的方式和内容是什么？这是一个索引问题吗？

Answer 1

将整个数据帧转换为字典尝试:

df.groupby(level=0).apply(lambda df: df.xs(df.name).to_dict()).to_dict()

{'bar': {'01.01': {'total1': 40, 'total2': 36},
  '02.01': {'total1': 52, 'total2': 85},
  '03.01': {'total1': 18, 'total2': 5},
  '04.01': {'total1': 11, 'total2': 92}},
 'baz': {'01.01': {'total1': 23, 'total2': 50},
  '02.01': {'total1': 39, 'total2': 49},
  '03.01': {'total1': 45, 'total2': 51},
  '04.01': {'total1': 70, 'total2': 65}},
 'foo': {'01.01': {'total1': 23, 'total2': 64},
  '02.01': {'total1': 97, 'total2': 56},
  '03.01': {'total1': 17, 'total2': 94},
  '04.01': {'total1': 97, 'total2': 45}},
 'qux': {'01.01': {'total1': 13, 'total2': 80},
  '02.01': {'total1': 73, 'total2': 8},
  '03.01': {'total1': 38, 'total2': 61},
  '04.01': {'total1': 4, 'total2': 50}}}

要转换一个特定列,请在将其转换为字典之前进行选择

df.groupby(level=0).apply(lambda df: df.xs(df.name)[colname].to_dict()).to_dict()