Python many-to-one mapping (creating equivalence c

2019-02-06 14:15发布

问题:

I have a project of converting one database to another. One of the original database columns defines the row's category. This column should be mapped to a new category in the new database.

For example, let's assume the original categories are:parrot, spam, cheese_shop, Cleese, Gilliam, Palin

Now that's a little verbose for me, And I want to have these rows categorized as sketch, actor - That is, define all the sketches and all the actors as two equivalence classes.

>>> monty={'parrot':'sketch', 'spam':'sketch', 'cheese_shop':'sketch', 
'Cleese':'actor', 'Gilliam':'actor', 'Palin':'actor'}
>>> monty
{'Gilliam': 'actor', 'Cleese': 'actor', 'parrot': 'sketch', 'spam': 'sketch', 
'Palin': 'actor', 'cheese_shop': 'sketch'}

That's quite awkward- I would prefer having something like:

monty={ ('parrot','spam','cheese_shop'): 'sketch', 
        ('Cleese', 'Gilliam', 'Palin') : 'actors'}

But this, of course, sets the entire tuple as a key:

>>> monty['parrot']

Traceback (most recent call last):
  File "<pyshell#29>", line 1, in <module>
    monty['parrot']
KeyError: 'parrot'

Any ideas how to create an elegant many-to-one dictionary in Python?

Thanks,

Adam

回答1:

It seems to me that you have two concerns. First, how do you express your mapping originally, that is, how do you type the mapping into your new_mapping.py file. Second, how does the mapping work during the re-mapping process. There's no reason for these two representations to be the same.

Start with the mapping you like:

monty = { 
    ('parrot','spam','cheese_shop'): 'sketch', 
    ('Cleese', 'Gilliam', 'Palin') : 'actors',
}

then convert it into the mapping you need:

working_monty = {}
for k, v in monty.items():
    for key in k:
        working_monty[key] = v

producing:

{'Gilliam': 'actors', 'Cleese': 'actors', 'parrot': 'sketch', 'spam': 'sketch', 'Palin': 'actors', 'cheese_shop': 'sketch'}

then use working_monty to do the work.



回答2:

You could override dict's indexer, but perhaps the following simpler solution would be better:

>>> assoc_list = ( (('parrot','spam','cheese_shop'), 'sketch'), (('Cleese', 'Gilliam', 'Palin'), 'actors') )
>>> equiv_dict = dict()
>>> for keys, value in assoc_list:
    for key in keys:
        equiv_dict[key] = value


>>> equiv_dict['parrot']
'sketch'
>>> equiv_dict['spam']
'sketch'

(Perhaps the nested for loop can be compressed an impressive one-liner, but this works and is readable.)



回答3:

>>> monty={ ('parrot','spam','cheese_shop'): 'sketch', 
        ('Cleese', 'Gilliam', 'Palin') : 'actors'}

>>> item=lambda x:[z for y,z in monty.items() if x in y][0]
>>>
>>> item("parrot")
'sketch'
>>> item("Cleese")
'actors'

But let me tell you, It will be slow than normal one to one dictionary.



回答4:

If you want to have multiple keys pointing to the same value, i.e.

m_dictionary{('k1', 'k2', 'k3', 'k4'):1, ('k5', 'k6'):2} and access them as,

`print(m_dictionary['k1'])` ==> `1`.

Check this multi dictionary python module multi_key_dict. Install and Import it. https://pypi.python.org/pypi/multi_key_dict