pandas :创建一个以元组为键的字典 [英] Pandas: create a dictionary with a tuple as key

问题描述

给出此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

我可以创建一个以列列表作为值的字典，像这样:

I can create a dictionary with a list of columns as values, like this:

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

但是如何创建一个包含以first和second作为键的元组的字典?

but how can I create a dictionary with, say, a tuple containing first and second as key?

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'}

PS:如何确定pandas不会弄乱值? 10.20现在变成了10.1999999999 ...

PS: and how could I make sure that pandas doesn't mess up with the values? 10.20 has now become 10.1999999999...

推荐答案

您需要通过MultiIndex .set_index.html"rel =" nofollow noreferrer> set_index ，然后调用

You need create MultiIndex by set_index and then call 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'}

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