找出句子或单词中重复次数最多的字符

对于一个简单的问题来说太复杂了，只需使用字典来计算出现次数：

sentence = input("enter a sentence or a word: ")
max = None
max_count = 0

letters = dict()

for letter in sentence:
    if not letter.isalpha():
        continue
    my_letter = letter.lower()
    count = 1
    if my_letter not in letters:
        letters[my_letter] = 1
        # count is already set to 1
    else:
        count = letters[my_letter] + 1
        letters[my_letter] = count
    if count > max_count:
        max_count = count
        max = my_letter

print(f'{max} is repeated the most at {max_count}')

使用集合。计数器来计算字母。使用列表推导式仅从小写句子中选择字母。请注意，按出现顺序仅返回 1 个最常出现的字符。因此，在下面的示例中，它打印但不打印。但是，如果您想要所有最常见的字母，那么访问这些元素是微不足道的。most_common(1)obCounter

from collections import Counter

sentence = input('Enter a sentence or a word: ')
letters = [c for c in sentence.lower() if c.isalpha()]
cnt = Counter(letters)
print(f"letters: occurrences: {cnt}")
print(f"most common: {cnt.most_common(1)}")

例：

输入：

Enter a sentence or a word: foo bbar.

输出：

letters: occurrences: Counter({'o': 2, 'b': 2, 'f': 1, 'a': 1, 'r': 1})
most common: [('o', 2)]

在考虑如何帮助修复当前代码时，我会考虑使用来帮助积累不同字母列表，并避免我们已经完成的工作。in

    if i in list_of_letters:
        continue
    else:
        list_of_letters.append(i)

这样一来，我们就可以在for循环的末尾删除一大块代码。

完整示例：

def remove(string):
    return "".join(string.split())

sentence = input('Enter a sentence or a word: ')
new = remove(sentence)
new = new.lower()

list_of_letters = []
most_letter = ""
most = 0

for i in new:
    ## ---------------
    ## if we already counted this letter we can skip doing so again
    ## ---------------
    if i in list_of_letters:
        continue
    ## ---------------

    ## ---------------
    ## ---------------
    ## If we get here, this is a new letter we have not seen before
    ## ---------------
    ## ---------------

    list_of_letters.append(i)

    ## ---------------
    ## Count how many times letter "i" appears in "new"
    ## ---------------
    rep = 0
    for n in new:
        if i == n:
            rep += 1
    ## ---------------

    ## ---------------
    ## If the prior count "rep" is more than are current max
    ## ---------------
    if rep > most:
        most_letter = i
        most = rep
    ## ---------------

    ## ---------------
    ## This count is less than some prior count
    ## we can ignore it
    ## ---------------
    #elif rep == most:
    #    for z in list_of_letters:
    #                if i == z:
    #                    break
    #                else:
    #                    list_of_letters.append(i)   
    ## ---------------

print(f"In the phrase '{sentence}'")
print(f"The most common letter was: '{most_letter}' found {most} times")
print(f"Distinct Letters: '{''.join(list_of_letters)}'")

如果你运行它并在出现提示时输入“hello world”，你将返回：

In the phrase 'hello world'
The most common letter was: 'l' found 3 times
Distinct Letters: 'helowrd'

替代方法：

您还可以查看许多简单的改进。例如，调用 to 将为您提供不同的字母，并计算该字母在 中出现的次数。list_of_letters = list(set(new))rep = new.count(i)inew

但是，要同时获得不同的字母列表及其计数，我们可以利用 .它将或多或少地为我们完成所有工作。collections.Counter

import collections

sentence = input('Enter a sentence or a word: ')

## -----------
## make someone else do the work for us
## -----------
counted_letters = collections.Counter(sentence.lower().replace(" ", ""))
## -----------

list_of_letters = list(counted_letters.keys())
most_letter, most = counted_letters.most_common(1)[0]

print(f"In the phrase '{sentence}'")
print(f"The most common letter was: '{most_letter}' found {most} times")
print(f"Distinct Letters: '{''.join(list_of_letters)}'")