Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
146 views
in Technique[技术] by (71.8m points)

python - DataFrame with MultiIndex to dict

I have a dataframe with a MultiIndex. I am wondering whether I created the data frame in the correct manner (see below).

             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 results in:

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

df.index.get_level_values results in:

<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=[]

I am ultimately looking to transform the df into a dict of dictionaries such that the first dict key are one of ['bar','baz', 'foo','qux'] and values are the dates and the inner dictionary is made of 'total1' and 'totals2' as key and the values are the integers of the df. Alternative explanation, is for example if dict1 is the dict then calling:

dict1['bar']

would result in the output:

{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} } }

How and what would I need to alter in order to achieve this? Is this an indexing issue?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

For converting whole dataframe to dictionary Try:

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}}}

For converting one particular column, select before converting it to dictionary i.e

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

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...