Arrange elements with same count in alphabetical order

Question

Python Collection Counter.most_common(n) method returns the top n elements with their counts. However, if the counts for two elements is the same, how can I return the result sorted by alphabetical order?

For example: for a string like: BBBAAACCD, for the "2-most common" elements, I want the result to be for specified n = 2:

[('A', 3), ('B', 3), ('C', 2)]

and NOT:

[('B', 3), ('A', 3), ('C', 2)]

Notice that although A and B have the same frequency, A comes before B in the resultant list since it comes before B in alphabetical order.

[('A', 3), ('B', 3), ('C', 2)]

How can I achieve that?

Answer 1

Although this question is already a bit old i'd like to suggest a very simple solution to the problem which just involves sorting the input of Counter() before creating the Counter object itself. If you then call most_common(n) you will get the top n entries sorted in alphabetical order.

from collections import Counter

char_counter = Counter(sorted('ccccbbbbdaef'))
for char in char_counter.most_common(3):
  print(*char)

resulting in the output:

b 4
c 4
a 1

Answer 2

There are two issues here:

1. Include duplicates when considering top n most common values excluding duplicates.
2. For any duplicates, order alphabetically.

None of the solutions thus far address the first issue. You can use a heap queue with the itertools unique_everseen recipe (also available in 3rd party libraries such as toolz.unique) to calculate the nth largest count.

Then use sorted with a custom key.

from collections import Counter
from heapq import nlargest
from toolz import unique

x = 'BBBAAACCD'

c = Counter(x)
n = 2
nth_largest = nlargest(n, unique(c.values()))[-1]

def sort_key(x):
    return -x[1], x[0]

gen = ((k, v) for k, v in c.items() if v >= nth_largest)
res = sorted(gen, key=sort_key)

[('A', 3), ('B', 3), ('C', 2)]