如何計算字符串的出現次數並只打印一次？

Question

我希望我的代碼只按字母順序輸出字符串中的每個字母一次，例如banana將輸出abn 。

問題是我仍然需要它來計算字符串中每個字母的出現次數，因此輸出應如下所示：

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)
...

這是我的代碼：

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')

當前輸出如下：

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

您可以使用Counter輕松地為您計算字母：

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")

給出：

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

對於獨特的字母，您可以嘗試使用 set()。 所以類似for i in sorted(set(stg)):

Answer 3

刪除重復項的技巧是使它成為一個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')

打印出來：

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

請注意從sorted后的一行移動。 set是無序的，因此原始排序順序丟失。 再次排序，就在循環之前，將其排序。