Python dict with default value based on key

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I need a dictionary that is automatically filled with a default value for each accessed key that is missing. I've found defaultdict and some other ways to achieve this, but the problem in my case is that I want the default value for each key to be specific to the key itself.

For example, with defaultdict I can achieve something like this:

from collections import defaultdict
d = defaultdict(lambda: 5)
> d[1] = 3
> d[1]
> 3
> d[2]
> 5

But what if I need the default value for each accessed missing key to be for example key + 5? Something like:

from collections import defaultdict
d = defaultdict(lambda key: key + 5)  # <-- This does not work as defaultdict expects lambda function to be without any parameters
> d[1] = 3
> d[1]
> 3
> d[2]
> 7         <- Calculated from accessed key + 5 (2+5)
> d[5]
> 10        <- Calculated from accessed key + 5 (5+5)

Is there a clean, builtin way to achieve what I need? I know that I can create a subclass of dict and implement this specific functionality at __getitem__ level, but I want to avoid that if possible.

I couldn't find a solution in other answers, so sorry if it is still a duplicate.

2 Answers

I don't think there is a builtin way of doing this. However, instead of subclassing dict and change getitem, you can subclass defaultdict itself to tell __missing__() to call the default_factory with an argument (the key), rather than without args. It is still a custom class, not a builtin, but it should be quite efficient still.

from collections import defaultdict

class DefaultDict(defaultdict):
    def __missing__(self, key):
        return self.default_factory(key)

Then:

d = DefaultDict(lambda key: key + 5)

d[2]
# 7

You can also use a function like this

dic = {}
DEFAULT_VALUE = 5

def dict_get(item):
    try:
        return [dic[item]]
    except:
        dic[int(item)] = DEFAULT_VALUE + int(item)
        return DEFAULT_VALUE + int(item)

print(dict_get(10))
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