CODEWITHSUNDEEP
python-2.7dictionarylambda
Abc
Abc

Reputation: 41

How to initialize key,value pair in dictionary with lambda expression?

I have a dictionary where key is a tuple (class,word). I want to initialize the value of dictionary for a particular class, irrespective of the word to a particular default value. Eg: dict = defaultdict(lambda : -1.0)

dict [('A', any word here)] = some constant value according to lambda expression

Here of the tuple ('A', any word here)- A is the class while "any word here" can be any random unknown word. This was my attempt, but am unable to figure out the correct way: dict = defaultdict(lambda tag, x: some_constant)).

How to I make x a placeholder that might come in its place?

Upvotes: 2

Views: 2397

Answers (1)

John1024
John1024

Reputation: 113924

As I understand it, you want a dictionary to have a default value that depends on part of the key. If so, I think this does what you want:

class MyDict(dict):
    def __missing__(self, key):
        class_, word = key
        self[key] = class_ # Replace this with what you want
        return self[key]

Now, MyDict behaves just as a normal dictionary except that missing values are filled in according to the custom function which you will have to adjust to your needs.

For example:

In [22]: d = MyDict()
In [23]: d[(int, 'delta')]
Out[23]: int    
In [24]: d[(str, 'gamma')]
Out[24]: str    
In [25]: d
Out[25]: {(str, 'gamma'): str, (int, 'delta'): int}

Be careful of one point: when you use the get attribute to a dictionary with a missing key, it will not call __missing__. Instead, it will return whatever you specify as the second argument to get and that defaults to None.

The problem with defaultdict

One might like defaultdict to behave as you want with: 

dict = defaultdict(lambda tag, x: some_constant))

However, this will not work because the defaultdict constructor does not accept arguments. With defaultdict, the default value must be the same regardless of key.

Upvotes: 3

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