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)]

Answer 3

I would first sort your output array in alphabetical order and than sort again by most occurrences which will keep the alphabetical order:

from collections import Counter
alphabetic_sorted = sorted(Counter('BBBAAACCD').most_common(), key=lambda tup: tup[0])
final_sorted = sorted(alphabetic_sorted, key=lambda tup: tup[1], reverse=True)
print(final_sorted[:3])

Output:

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

Answer 4

I would go for:

sorted(Counter('AAABBBCCD').most_common(), key=lambda t: (-t[1], t[0]))

This sorts count descending (as they are already, which should be more performant) and then sorts by name ascending in each equal count group

Answer 5

This is one of the problems I got in the interview exam and failed to do it. Came home slept for a while and solution came in my mind.

from collections import Counter


def bags(list):
    cnt = Counter(list)
    print(cnt)
    order = sorted(cnt.most_common(2), key=lambda i:( i[1],i[0]), reverse=True)
    print(order)
    return order[0][0]


print(bags(['a','b','c','a','b']))

Answer 6

from collections import Counter


s = 'qqweertyuiopasdfghjklzxcvbnm'

s_list = list(s)

elements = Counter(s_list).most_common()

print(elements)
alphabet_sort = sorted(elements, key=lambda x: x[0])
print(alphabet_sort)
num_sort = sorted(alphabet_sort, key=lambda x: x[1], reverse=True)
print(num_sort)

if you need to get slice:

print(num_sort[:3])

Answer 7

s = "BBBAAACCD"    
p = [(i,s.count(i)) for i in sorted(set(s))]

**If you are okay with not using the Counter.

Answer 8

from collections import Counter
print(sorted(Counter('AAABBBCCD').most_common(3)))

This question seems to be a duplicate How to sort Counter by value? - python