如何计算字符串的出现次数并只打印一次？

Question

我希望我的代码只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，因此输出应如下所示：

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)
...

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')

当前输出如下：

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Counter轻松地为您计算字母：

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")

给出：

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

对于独特的字母，您可以尝试使用 set()。 所以类似for i in sorted(set(stg)):

Answer 3

删除重复项的技巧是使它成为一个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 times
the letter b appears in the word banana 1 time
the letter n appears in the word banana 2 times

请注意从sorted后的一行移动。 set是无序的，因此原始排序顺序丢失。 再次排序，就在循环之前，将其排序。