Ranking values in a Dictionary (and taking care of ex-aequos correctly)

Question

I would like to rank the values in a dictionary.

For instance, I have this dictionary: {\"A\": 10, \"B: 3, \"C\": 8, \"D\": 3, \"E\": 2} The result should l

Answer 1

First sort the data in ascending order based on the number, like this

>>> data = {"A": 10, "B": 3, "C": 8, "D": 3, "E": 2}
>>> s_data = sorted(data.items(), key=lambda item: item[1])
>>> s_data
[('E', 2), ('D', 3), ('B', 3), ('C', 8), ('A', 10)]

Now, for every element processed,

• if it is not the same as the previous element then the rank should be incremented by the number of similar elements processed till now

• if it is the same, then simply count the current element as a similar element

To implement this, initialize few variables, like this

>>> rank, count, previous, result = 0, 0, None, {}

then keep checking if the current element is not equal to the previous element and if it is true, increment rank by the number of times the similar elements occured.

>>> for key, num in s_data:
...     count += 1
...     if num != previous:
...         rank += count
...         previous = num
...         count = 0
...     result[key] = rank

Now, result will have the result you wanted.

>>> result
{'D': 2, 'C': 4, 'E': 1, 'B': 2, 'A': 5}