how to find the most popular letter in a string that also has the lowest ascii value-CodePudding

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):

Build a dictionary mapping letters to their frequency;
Find the largest frequency;
Find the smallest letter having that frequency.

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!

CodePudding user response：

def most_popular_character(my_string):
   history_l = [l for l in my_string] #each letter in string
   char_dict = {} #creating dict
   for item in history_l: #for each letter in string
       char_dict[item] = history_l.count(item)

   return [max(char_dict.values()),min(char_dict.values())]

I didn't understand the last part of minimum frequency, so I make this function return a maximum frequency and a minimum frequency as a list!

CodePudding user response：

Use a Counter to count the characters, and use the max function to select the "biggest" character according to your two criteria.

>>> from collections import Counter
>>> def most_popular_character(my_string):
...     chars = Counter(my_string)
...     return max(chars, key=lambda c: (chars[c], -ord(c)))
...
>>> most_popular_character("HelloWorld")
'l'
>>> most_popular_character("gggcccbb")
'c'

Note that using max is more efficient than sorting the entire dictionary, because it only needs to iterate over the dictionary once and find the single largest item, as opposed to sorting every item relative to every other item.

CodePudding user response：

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