Pandas:创建一个以元组为键的字典

FaC*_*fee 2 python dictionary tuples pandas

鉴于这种DataFrame

import pandas as pd
first=[0,1,2,3,4]
second=[10.2,5.7,7.4,17.1,86.11]
third=['a','b','c','d','e']
fourth=['z','zz','zzz','zzzz','zzzzz']
df=pd.DataFrame({'first':first,'second':second,'third':third,'fourth':fourth})
df=df[['first','second','third','fourth']]

   first  second third fourth
0      0   10.20     a      z
1      1    5.70     b     zz
2      2    7.40     c    zzz
3      3   17.10     d   zzzz
4      4   86.11     e  zzzzz
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我可以创建一个字典,其中包含一系列列作为值,如下所示:

d = {df.loc[idx, 'first']: [df.loc[idx, 'second'], df.loc[idx, 'third']] for idx in range(df.shape[0])}
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但是我怎样才能创建一个字典,比如说,一个包含firstsecond作为键的元组呢?

结果将是:

In[1]:d
Out[1]: 
{(0,10.199999999999999): 'a',
 (1,5.7000000000000002): 'b',
 (2,7.4000000000000004): 'c',
 (3,17.100000000000001): 'd',
 (4,86.109999999999999): 'e'}
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PS:我怎样才能确保这pandas不会扰乱价值观?10.20 现在变成了 10.1999999999...

jez*_*ael 6

您需要创建然后MultiIndex调用:set_indexSeries.to_dict

a = df.set_index(['first','second']).third.to_dict()
print (a)
{(2, 7.4): 'c', (1, 5.7): 'b', (3, 17.1): 'd', (0, 10.2): 'a', (4, 86.11): 'e'}
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