Implement the function most_popular_character(my_string), which gets the string argument my_string and returns its most frequent letter. In case of a tie, break it by returning the letter of smaller ASCII value. Note that lowercase and uppercase letters are considered different (e.g., ‘A’ < ‘a’). You may assume my_string consists of English letters only, and is not empty.
Example 1: >>> most_popular_character("HelloWorld") >>> 'l'
Example 2: >>> most_popular_character("gggcccbb") >>> 'c'
Explanation: cee and gee appear three times each (and bee twice), but cee precedes gee lexicographically.
Hints (you may ignore these):
def most_popular_character(my_string):
char_count = {} # define dictionary
for c in my_string:
if c in char_count: #if c is in the dictionary:
char_count[c] = 1
else: # if c isn't in the dictionary - create it and put 1
char_count[c] = 1
sorted_chars = sorted(char_count) # sort the dictionary
char_count = char_count.keys() # place the dictionary in a list
max_per = 0
for i in range(len(sorted_chars) - 1):
if sorted_chars[i] >= sorted_chars[i+1]:
max_per = sorted_chars[i]
break
return max_per
my function returns 0 right now, and I think the problem is in the last for loop and if statement - but I can't figure out what the problem is..
If you have any suggestions on how to adjust the code it would be very appreciated!
Your dictionary didn't get off to a good start by you forgetting to add 1 to the character count, instead you are resetting to 1 each time.
Have a look here to get the gist of getting the maximum value from a dict: https://datagy.io/python-get-dictionary-key-with-max-value/
def most_popular_character(my_string):
# NOTE: you might want to convert the entire sting to upper or lower case, first, depending on the use
# e.g. my_string = my_string.lower()
char_count = {} # define dictionary
for c in my_string:
if c in char_count: #if c is in the dictionary:
char_count[c] += 1 # add 1 to it
else: # if c isn't in the dictionary - create it and put 1
char_count[c] = 1
# Never under estimate the power of print in debugging
print(char_count)
# max(char_count.values()) will give the highest value
# But there may be more than 1 item with the highest count, so get them all
max_keys = [key for key, value in char_count.items() if value == max(char_count.values())]
# Choose the lowest by sorting them and pick the first item
low_item = sorted(max_keys)[0]
return low_item, max(char_count.values())
print(most_popular_character("HelloWorld"))
print(most_popular_character("gggcccbb"))
print(most_popular_character("gggHHHAAAAaaaccccbb 12 3"))
Result:
{'H': 1, 'e': 1, 'l': 3, 'o': 2, 'W': 1, 'r': 1, 'd': 1}
('l', 3)
{'g': 3, 'c': 3, 'b': 2}
('c', 3)
{'g': 3, 'H': 3, 'A': 4, 'a': 3, 'c': 4, 'b': 2, ' ': 2, '1': 1, '2': 1, '3': 1}
('A', 4)
So: l and 3, c and 3, A and 4