如何计算字符串的出现次数并只打印一次？

Question

I want my code to output each letter in the string only once and in alphabetical order, for example banana will output abn .

The catch is that I still need it to count the occurrences of each letter in the string, so the output should be as follows:

a occurs in the word banana a total of 3 times(s)
b occurs in the word banana a total of 1 time(s)
n occurs in the word banana a total of 2 time(s)
...

This is my code:

def letter_counter(string):
    stg = string.lower()
    stg = ''.join(sorted(stg))
    for i in stg:
        a = stg.count(i)
        print(f'the letter {i} appears in the word {string} {a} times')
            
letter_counter('banana')

And the current output is as follows:

the letter a appears in the word banana 3 times
the letter a appears in the word banana 3 times
the letter a appears in the word banana 3 times
the letter b appears in the word banana 1 times
the letter n appears in the word banana 2 times
the letter n appears in the word banana 2 times

Answer 1

You can use a Counter to easily count the letters for you:

from collections import Counter

def letter_counter(string):
    for letter, count in sorted(Counter(string.lower()).items()):
        print(f'the letter {letter} appears in the word {string} {count} times')

letter_counter("banana")

Gives:

the letter a appears in the word banana 3 times
the letter b appears in the word banana 1 times
the letter n appears in the word banana 2 times

Answer 2

For unique letters you can try to use set(). So something like for i in sorted(set(stg)):

Answer 3

The trick to remove duplicates is to make it a set :

def letter_counter(string):
    stg = string.lower()
    stg = ''.join(stg)
    for i in sorted(set(stg)):
        a = stg.count(i)
        print(f'the letter {i} appears in the word {string} {a} time{"" if a == 1 else "s"}')

letter_counter('banana')

prints out:

the letter a appears in the word banana 3 times
the letter b appears in the word banana 1 time
the letter n appears in the word banana 2 times

Note the move from sorted one line later. A set is unordered so the original sorted order is lost. Sorting it again, just before looping, sorts this out.