如何计算字符串的出现次数并且只打印一次？

Question

我希望我的代码仅按字母顺序输出字符串中的每个字母一次，例如banana将输出abn.

问题是我仍然需要它来计算字符串中每个字母的出现次数，所以输出应该如下：

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)

...

这是我的代码：

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')

而当前的输出如下：

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 aCounter为您轻松计算字母：from collections import Counterdef 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")给出：the 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 times

Answer 2

删除重复项的技巧是使其成为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')打印出来：the letter a appears in the word banana 3 timesthe letter b appears in the word banana 1 timethe letter n appears in the word banana 2 times请注意稍后从sorted一行移动。Aset是无序的，因此原始排序顺序丢失。在循环之前再次对其进行排序，将其排序。