Question

这是对我的编码。我必须写出一个方案,让用户能够展示和展示在扼杀中最经常出现的特性。我不能利用最高或计票职能。除了这一例子外,我完全失去了我的书。

我试图这样做:

# lst = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]


def main():


    count = 0

    my_sentence = input("Enter a sentence: ")

    for ch in my_sentence:
        if ch ==  T  or ch ==  t :
            count += 1
    

    print( The letter T appears , count,  times. )


main()

找到具体信函是可行的,但对于我来说,我甚至不能开始理解,在任何判决中,我会找到最常用的文字。我可能会 st。我在十小时后试图解决这一问题,我可以再想一想。

Answer 1

利用字典计算和比较每封信的频率:

def most_frequent_character(sentence):
    character_frequency_dictionary = {} # make new dictionary to keep track of a character s frequency

    for character in sentence:
        # if we have not encountered this character before, set the count of the character to 1
        if character not in character_frequency_dictionary:
            character_frequency_dictionary[character] = 1

        # else, if we have see then character, increment the count of that character
        else:
            character_frequency_dictionary[character] += 1

    # now we have a dict of chars and their frequency, so we need to now see which is the most frequent

    # default values
    max_freq = 0
    most_freq_char =   

    # compare each characters frequency to find the most frequent
    for character in character_frequency_dictionary:
        frequency = character_frequency_dictionary[character]

        if frequency > max_freq:
            max_freq = frequency
            most_freq_char = character

    return most_freq_char


print(most_frequent_character( abcdeaa )) # a
print(most_frequent_character( AAaaddd )) # d (not checking case)
print(most_frequent_character( abcffff )) # f
print(most_frequent_character( iiiiiii )) # i
print(most_frequent_character( abcdefh )) # a (first found)

如有必要,如果你需要这种对案件不敏感的话,将体力转换为下级(使用方位)。

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