invert dictionary with list &c

anton muhin antonmuhin.REMOVE.ME.FOR.REAL.MAIL at rambler.ru
Fri Nov 28 13:13:53 CET 2003


Des Small wrote:

> anton muhin <antonmuhin.REMOVE.ME.FOR.REAL.MAIL at rambler.ru> writes:
> 
> 
>>Des Small wrote:
>>
>>>Des Small <des.small at bristol.ac.uk> writes:
>>>
>>>
>>>>anton muhin <antonmuhin.REMOVE.ME.FOR.REAL.MAIL at rambler.ru> writes:
>>>
>>>[...]
>>
>>Or like this:
>>
>>def dict_update(iter, func, default, d):
>>     def process_element(d, e):
>>         d[e[0]] = func(d.get(e[0], default), *e[1:])
>>         return d
>>
>>     return reduce(process_element, iter, d)
>>
>>def count(l):
>>     return dict_update(l, lambda x: x + 1, 0, {})
> 
> 
> With these I get:
> 
> 
>>>>count(["yes", "yes", "no"])
> 
> 
> Traceback (most recent call last):
>   File "<stdin>", line 1, in ?
>   File "<stdin>", line 2, in count
>   File "<stdin>", line 5, in dict_update
>   File "<stdin>", line 3, in process_element
> TypeError: <lambda>() takes exactly 1 argument (3 given)
> 
> I tried to write something that would work on more generic arguments,
> but I couldn't do it without explicit type checking, so I gave up.
> 
> [...]
> 
> Des
> has now embalmed versions in his utils collection.
> 

My fault :( A small patch:

def count(l):
     return dict_update([(e,) for e in l], lambda x: x + 1, 0, {})

now

print count(['yes', 'yes', 'no'])

print count('aabbbc')

prints

{'yes': 2, 'no': 1}
{'a': 2, 'c': 1, 'b': 3}

regards,
anton.





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