字符串中最常见的字符 [英] Most common character in a string

问题描述

编写一个函数，该函数将由字母
字符组成的字符串作为输入参数，并返回最常见的字符。
忽略空格，即不要将任何空格都视为字符。
请注意，此处的大小写无关紧要，即小写
字符等于大写字符。如果
之间出现平局，则某些字符返回计数最高的最后一个字符

Write a function that takes a string consisting of alphabetic characters as input argument and returns the most common character. Ignore white spaces i.e. Do not count any white space as a character. Note that capitalization does not matter here i.e. that a lower case character is equal to a upper case character. In case of a tie between certain characters return the last character that has the most count

更新的代码

def most_common_character (input_str):
    input_str = input_str.lower()
    new_string = "".join(input_str.split())
    print(new_string)
    length = len(new_string)
    print(length)
    count = 1
    j = 0
    higher_count = 0
    return_character = ""
    for i in range(0, len(new_string)):
        character = new_string[i]
        while (length - 1):
            if (character == new_string[j + 1]):
                count += 1
            j += 1
            length -= 1    
            if (higher_count < count):
                higher_count = count
    return (character)     

#Main Program
input_str = input("Enter a string: ")
result = most_common_character(input_str)
print(result)

上面是我的代码。我收到字符串索引超出范围的错误，我不明白为什么。代码也只检查第一个字符的出现，我对如何继续到下一个字符并采用最大计数感到困惑。

The above is my code. I am getting an error of string index out of bound which I can't understand why. Also the code only checks the occurrence of first character I am confused about how to proceed to the next character and take the maximum count?

运行代码时出现的错误：

> Your answer is NOT CORRECT Your code was tested with different inputs.
> For example when your function is called as shown below:
> 
> most_common_character ('The cosmos is infinite')
> 
> ############# Your function returns ############# e The returned variable type is: type 'str'
> 
> ######### Correct return value should be ######## i The returned variable type is: type 'str'
> 
> ####### Output of student print statements ###### thecosmosisinfinite 19

推荐答案

实际上您的代码几乎是正确的。您需要将 count ， j ， length 移到其中您的 for i在范围（0，len（new_string））中，因为您需要在每次迭代中重新开始，并且如果计数大于 higher_count ，则需要将该 character 保存为 return_character 并返回它而不是 character ，它总是字符串的最后一个字符，因为 character = new_string [i] 。

Actually your code is almost correct. You need to move count, j, length inside of your for i in range(0, len(new_string)) because you need to start over on each iteration and also if count is greater than higher_count, you need to save that charater as return_character and return it instead of character which is always last char of your string because of character = new_string[i].

我不明白您为什么使用 j + 1 和而length-1 。纠正它们之后，它现在也涵盖了平局情况。

I don't see why have you used j+1 and while length-1. After correcting them, it now covers tie situations aswell.

def most_common_character (input_str):
    input_str = input_str.lower()
    new_string = "".join(input_str.split())
    higher_count = 0
    return_character = ""

    for i in range(0, len(new_string)):
        count = 0
        length = len(new_string)
        j = 0
        character = new_string[i]
        while length > 0:
            if (character == new_string[j]):
                count += 1
            j += 1
            length -= 1    
            if (higher_count <= count):
                higher_count = count
                return_character = character
    return (return_character) 

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