How to compare character frequency and return the most occurring character? - python-3.x

Question

I am trying to build a function which returns the most occurred character in a given string and it's working pretty nicely, but how do I return None if the characters have same frequency?
Like for input: 'abac'
Expected output is: 'a'
and for input: 'abab'
Expected output is: None
I have tried using a dictionary to store character frequency and then returning the element with largest value.
def most_occuring_char(str1):
count = {}
max = 0
c = ''
for char in str1:
if char in count.keys():
count[char]+=1
else:
count[char] = 1
for char in str1:
if max < count[char]:
max = count[char]
c = char
return c
I don't know how to check whether the count dictionary elements have same frequency.

Answer 1

You can do that counting with the dict using collections.Counter.
You basically only have to add a check to see if the maximum count is unique (if so, return the char with maximum number of occurrences) or not (if so, return None):
from collections import Counter
def most_occurring_char(string):
counter = Counter(string)
max_char_count = max(counter.values())
is_unique = len([char_count for char_count in counter.values() if char_count == max_char_count]) == 1
if is_unique:
char = [char for char, count in counter.items() if count == max_char_count][0]
return char
return None
# Tests
assert most_occurring_char('abac') == 'a'
assert most_occurring_char('abab') is None

Answer 2

Once you have a dictionary containing the counts of every character (after your first for loop), you can inspect this to determine whether certain counts are the same or not.
If you wish to return None only when all the character counts are the same, you could extract the values (i.e. the character counts) from your dictionary, sort them so they are in numerical order, and compare the first and last values. Since they are sorted, if the first and last values are the same, so are all the intervening values. This can be done using the following code:
count_values = sorted(count.values())
if count_values[0] == count_values[-1]: return None
If you wish to return None whenever there is no single most frequent character, you could instead compare the last value of the sorted list to the second last. If these are equal, there are two or more characters that occur most frequently. The code for this is very similar to the code above.
count_values = sorted(count.values())
if count_values[-1] == count_values[-2]: return None

Answer 3

Another possibility:
def most_occuring_char(s):
from collections import Counter
d = Counter(s)
k = sorted(d, key=lambda x:d[x], reverse=True)
if len(k) == 1: return k[0]
return None if len(k) == 0 or d[k[0]] == d[k[1]] else k[0]
#Test
print(most_occuring_char('abac')) #a
print(most_occuring_char('abab')) #None (same frequencies)
print(most_occuring_char('x')) #x
print(most_occuring_char('abcccba')) #c
print(most_occuring_char('')) #None (empty string)