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Re: Dictionaries with variable default.

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  Re: Dictionaries with variable default. Antoon Pardon <antoon.pardon@rece.vub.ac.be> - 2014-11-04 14:21 +0100
    Re: Dictionaries with variable default. Cousin Stanley <cousinstanley@gmail.com> - 2014-11-04 07:13 -0700

#80177 — Re: Dictionaries with variable default.

Op 03-11-14 om 12:09 schreef Chris Angelico:
> On Mon, Nov 3, 2014 at 10:04 PM, Antoon Pardon
> <antoon.pardon@rece.vub.ac.be> wrote:
>> Is it possible to have a default dictionary where the default is dependant
>> on the key?
>>
>> I was hoping something like this might work:
>>>>> m = defaultdict(lambda key: key+1)
>> But it obviously doesn't:
>>>>> m[3]
>> Traceback (most recent call last):
>>   File "<stdin>", line 1, in <module>
>> TypeError: <lambda>() takes exactly 1 argument (0 given)
> Yes, but not with the defaultdict. Instead, just implement __missing__:
>
> class KeyPlusOne(dict):
>     def __missing__(self, key): return key+1
>
> ChrisA

So How should I call this:

class ...dict(dict):
    def __init__(self, fun):
        self.fun = fun

    def __missing__(self, key):
       return self.fun(key)

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#80183

> So How should I call this:
> 
> class ...dict(dict):
>     def __init__(self, fun):
>         self.fun = fun
> 
> def __missing__(self, key):
>     return self.fun(key)

  I don't know how you should,
  but I tried the following 
  which seems to work .... 

class KeyPlusOne( dict ) :
    def __missing__( self , key ) : 
        return ( 2 * key ) + 1 

kp1 = KeyPlusOne()

d   = { }

for n in range( 11 ) :  
    d[ n ] = kp1[ n ]

print '\n   key :  value \n'

for key , value in d.iteritems() : 
    print '    %2d :  %2d' % ( key , value )


-- 
Stanley C. Kitching
Human Being
Phoenix, Arizona

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