如何从这个pandas多索引数据框中选择这种数据

时间:2018-04-19 10:51:29

标签: python pandas dataframe multi-index

我正在处理期货市场数据,这是一个多索引数据框的示例:

date_index = pd.date_range('2018-03-20', periods = 10)
contract = ['ZN1805', 'ZN1806', 'ZN1807']
price = ['open', 'close']
columns = pd.MultiIndex.from_product([contract, price], names=['contract', 'price'])
df1 = pd.DataFrame(data=np.random.randint(100, 150, (10, columns.shape[0])), index=date_index, columns=columns)

df2 = pd.DataFrame(columns=['contract', 'close'], index=df1.index)
# Set the data in contract column randomly here for illustration
df2.contract = np.random.choice(contract, 10)

以下是df1的样子,

df1
Out[357]: 
contract   ZN1805       ZN1806       ZN1807      
price        open close   open close   open close
2018-03-20    145   144    116   127    107   128
2018-03-21    116   143    114   103    114   148
2018-03-22    101   135    143   125    140   129
2018-03-23    106   139    100   127    116   100
2018-03-24    104   101    148   132    102   140
2018-03-25    125   141    106   136    128   134
2018-03-26    148   146    142   143    108   137
2018-03-27    110   123    128   128    124   127
2018-03-28    144   143    117   116    112   140
2018-03-29    143   114    115   105    124   118

df2将是:

df2
Out[364]: 
           contract close
2018-03-20   ZN1805   NaN
2018-03-21   ZN1807   NaN
2018-03-22   ZN1806   NaN
2018-03-23   ZN1807   NaN
2018-03-24   ZN1807   NaN
2018-03-25   ZN1806   NaN
2018-03-26   ZN1807   NaN
2018-03-27   ZN1806   NaN
2018-03-28   ZN1805   NaN
2018-03-29   ZN1807   NaN

我的问题是我如何诡异地'填写close df2的{​​{1}}列df1列,其date索引和contract值相同?

我试过了:

from pandas import IndexSlice as idx
df2['close'] = df1.loc[df2.index, idx[df2.contract.values.tolist(), 'close']]

但是我收到了一个错误:

UnsortedIndexError: 'MultiIndex Slicing requires the index to be fully lexsorted tuple len (2), lexsort depth (1)'

我知道我可以用迭代的方式来过滤每一行,但是有任何pythonic方法可以做到吗?

2 个答案:

答案 0 :(得分:2)

使用join创建的2列xs选择close级别,unstack重新塑造:

s = df1.xs('close', axis=1, level=1).unstack().rename('close')

df2 = (df2.drop('close', 1)
          .reset_index()
          .join(s, on=['contract', 'index'])
          .set_index('index')
          .rename_axis(None))

print (df2)
           contract  close
2018-03-20   ZN1805    124
2018-03-21   ZN1805    112
2018-03-22   ZN1807    118
2018-03-23   ZN1807    136
2018-03-24   ZN1805    103
2018-03-25   ZN1805    135
2018-03-26   ZN1805    138
2018-03-27   ZN1805    109
2018-03-28   ZN1805    129
2018-03-29   ZN1805    104

答案 1 :(得分:0)

@ jezrael的答案非常好,但如果对于那些不熟悉xs的人(像我一样),我会想出一个更复杂的方法来获得s

s= df1.loc[:, idx[:, 'close']]
s.columns = s.columns.droplevel(1)
s = s.unstack().rename('close')

当然,这个3行声明并不具有吸引力。 :d 然后我们可以用同样的方式得到df1:

df2 = (df2.drop('close', 1)
          .reset_index()
          .join(s, on=['contract', 'index'])
          .set_index('index')
          .rename_axis(None))