我希望我的代碼僅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，所以輸出應該如下：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')而當前的輸出如下：the letter a appears in the word banana 3 timesthe letter a appears in the word banana 3 timesthe letter a appears in the word banana 3 timesthe letter b appears in the word banana 1 timesthe letter n appears in the word banana 2 timesthe letter n appears in the word banana 2 times
2 回答
呼喚遠方
TA貢獻1856條經驗 獲得超11個贊
您可以使用 aCounter為您輕松計算字母：
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
夢里花落0921
TA貢獻1772條經驗 獲得超6個贊
刪除重復項的技巧是使其成為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
請注意稍后從sorted一行移動。Aset是無序的，因此原始排序順序丟失。在循環之前再次對其進行排序，將其排序。
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