Getting The Key Index In A Python Ordereddict?

I have a collections.OrderedDict with a list of key, value pairs. I would like to compute the index i such that the ith key matches a given value. For example: food = OrderedDict([

Solution 1:

Basically, no. OrderedDict gets its ability to look things up quickly by key name just by using a regular, unordered dict under the hood. The order information is stored separately in a doubly linked list. Because of this, there's no way to go directly from the key to its index. The order in an OrderedDict is mainly intended to be available for iteration; a key does not "know" its own order.

Solution 2:

As others have pointed out, an OrderedDict is just a dictionary that internally remembers what order entries were added to it. However, you can leverage its ability to look-up things quickly by storing the desired index along with the rest of the data for each entry. Here's what I mean:

from collections import OrderedDict

foods = [('beans', 33), ('rice', 44), ('pineapple', 55), ('chicken', 66)]
food = OrderedDict(((v[0], (v[1], i)) for i, v inenumerate(foods))) # saves iprint(food['rice'][1])  # --> 1print(food['chicken'][1])  # --> 3

Solution 3:

The OrderedDict is a subclass of dict which has the ability to traverse its keys in order (and reversed order) by maintaining a doubly linked list. So it does not know the index of a key. It can only traverse the linked list to find the items in O(n) time.

Perusing the source code may be the most satisfying way to confirm that the index is not maintained by OrderedDict. You'll see that no where is an index ever used or obtained.