Is it possible to initialize an empty OrderedDict in Python using a predefined sorting mechanism?

Question

All the examples that I could find ( in the documentation , etc.) define OrderedDicts, passing the data to the constructor. From the docs:

# regular unsorted dictionary d = {'banana': 3, 'apple':4, 'pear': 1, 'orange': 2} # dictionary sorted by key OrderedDict(sorted(d.items(), key=lambda t: t[0])) OrderedDict([('apple', 4), ('banana', 3), ('orange', 2), ('pear', 1)]) 

On the other hand, you can initialize an OrderedDict without providing any parameters to the constructors, which preserves the order in which key pairs and values ​​are added.

I am looking for some kind of construct that looks like the following, except without "d.items ()". In fact, I ask him to remember the mechanism without giving him an example that might seem crazy. Is my only option to “hack” this by providing the original “d” (below) with one element, or is there a better way?

 OrderedDict(sorted(d.items(), key=lambda t: t[0])) 

Thanks!

Answer 1

OrderedDict has only one sorting algorithm: insertion order. No matter what elements are added to the OrderedDict , this is the order of the OrderedDict .

If you want to use some other method, you can either write your dict subclass, or just a sort function that you can apply when order is really important.