Yes, it is a hash mapping or hash table. You can read a description of python’s dict implementation, as written by Tim Peters, here. That’s why you can’t use something ‘not hashable’ as a dict key, like a list: >>> a = {} >>> b = [‘some’, ‘list’] >>> hash(b) Traceback (most recent call last): … Read more
TryGetValue will already assign the default value for the type to the dictionary, so you can just use: dictionary.TryGetValue(key, out value); and just ignore the return value. However, that really will just return default(TValue), not some custom default value (nor, more usefully, the result of executing a delegate). There’s nothing more powerful built into the … Read more
You could use get twice: example_dict.get(‘key1’, {}).get(‘key2’) This will return None if either key1 or key2 does not exist. Note that this could still raise an AttributeError if example_dict[‘key1’] exists but is not a dict (or a dict-like object with a get method). The try..except code you posted would raise a TypeError instead if example_dict[‘key1’] … Read more
You can write an object that behaves like a dict quite easily with ABCs (Abstract Base Classes) from the collections.abc module. It even tells you if you missed a method, so below is the minimal version that shuts the ABC up. from collections.abc import MutableMapping class TransformedDict(MutableMapping): “””A dictionary that applies an arbitrary key-altering function … Read more
Curly braces. Passing keyword arguments into dict(), though it works beautifully in a lot of scenarios, can only initialize a map if the keys are valid Python identifiers. This works: a = {‘import’: ‘trade’, 1: 7.8} a = dict({‘import’: ‘trade’, 1: 7.8}) This won’t work: a = dict(import=”trade”, 1=7.8) It will result in the following … Read more
In the particular case of a map the old options were only two: operator[] and insert (different flavors of insert). So I will start explaining those. The operator[] is a find-or-add operator. It will try to find an element with the given key inside the map, and if it exists it will return a reference … Read more
You can use a dict comprehension: {k: v for k, v in points.items() if v[0] < 5 and v[1] < 5} And in Python 2, starting from 2.7: {k: v for k, v in points.iteritems() if v[0] < 5 and v[1] < 5}
Dictionaries in Swift (and other languages) are not ordered. When you iterate through the dictionary, there’s no guarantee that the order will match the initialization order. In this example, Swift processes the “Square” key before the others. You can see this by adding a print statement to the loop. 25 is the 5th element of … Read more
dict.items() Does the trick.
Well, you can achieve that with Jackson, too. (and it seems to be more comfortable since you were considering using jackson). Use ObjectMapper‘s convertValue method: final ObjectMapper mapper = new ObjectMapper(); // jackson’s objectmapper final MyPojo pojo = mapper.convertValue(map, MyPojo.class); No need to convert into JSON string or something else; direct conversion does much faster.