【问题标题】:Pandas: create a dictionary with a tuple as keyPandas:创建一个以元组为键的字典
【发布时间】:2017-07-05 17:16:47
【问题描述】:

鉴于此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

我可以创建一个包含列列表作为值的字典,如下所示:

d = {df.loc[idx, 'first']: [df.loc[idx, 'second'], df.loc[idx, 'third']] for idx in range(df.shape[0])}

但是我怎样才能创建一个字典,比如一个包含 firstsecond 作为键的元组?

结果是:

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'}

PS:我如何确保pandas 不会混淆这些值? 10.20 现在变成了 10.1999999999...

【问题讨论】:

    标签: python pandas dictionary tuples


    【解决方案1】:

    您需要通过set_index 创建MultiIndex,然后调用Series.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'}
    

    【讨论】:

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