计算一个字符的频率，直到引入一个新字符 - python答案

【问题标题】：Counting the frequency of a character until a new character is introduced - python计算一个字符的频率，直到引入一个新字符 - python
【发布时间】：2021-01-15 17:24:33
【问题描述】：

我正在编写一个程序来根据用户输入计算字母的频率，直到“！”被介绍了。以下是我的程序：

list1=[] 

character = "" 
while character != '!' :
      character = input()
      list1.append(character)

result=[]
for alphabet in ['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'] :
     if list1.count(alphabet) > 0:
            result.append(alphabet)
            result.append(list1.count(alphabet))

 print(result)

但是，显然，在输入新字符之前，我应该计算频率。例如，如果输入是 (aabc)，那么我的程序应该计算两个 'a'，然后转到 'b'。

在继续使用新字母表之前，我是否可以修改循环以计算频率？

【问题讨论】：

您需要的搜索词组是“run-length encoding”。

标签： python list while-loop word-frequency

【解决方案1】：

d = {}
while True:
    temp = input().strip()
    if temp == '!':
        break
    if temp.islower() is False:
        print("Invalid character")
        continue
    if temp not in d:
        d[temp] = 1
    else:
        d[temp] += 1

我用过字典

d 将包含执行后的结果。

【讨论】：

【解决方案2】：


    import string
    
    alpha = string.printable[10:36]
    # [a-z]
    
    
    # return a dictionary consists of keys of 26 alphabets and values of frequency
    def detect(string):
        di = {}
        for k in alpha:
            di[k] = string.count(k)
        return di
    
    
    text = ""
    
    print('Please input the text, end with a exclamation mark !')
    while text.count('!') == 0:
        text += input()
    
    
    # get rid of any after exclamation mark
    main_text = text.split('!')[0]
    print('result is: ' + str(detect(main_text)))

如果您想包含大写字母：


    import string
    
    alpha = string.printable[10:62]
    # [a-Z]
    
    
    # return a dictionary consists of keys of 26 alphabets and values of frequency
    def detect(string):
        di = {}
        for k in alpha:
            di[k] = string.count(k)
        return di
    
    
    text = ""
    
    print('Please input the text, end with a exclamation mark !')
    while text.count('!') == 0:
        text += input()
    
    
    # get rid of any after exclamation mark
    main_text = text.split('!')[0]
    print('result is: ' + str(detect(main_text)))

如果你不想和string模块打交道，只要是识别utf-8字符就统计每个输入的频率：


    def detect(string):
        di = {}
        for k in string:
            di[k] = string.count(k)
        return di
    
    
    text = ""
    
    print('Please input the text, end with a exclamation mark !')
    while text.count('!') == 0:
        text += input()
    
    
    # get rid of any after exclamation mark
    main_text = text.split('!')[0]
    print('result is: ' + str(detect(main_text)))

最后，如果你只想计算低 26 个字母：


    import string
    
    alpha = string.printable[10:36]
    # [a-z]
    
    def detect(info):
        di = {}
        for k in info:
            if k in alpha:
                di[k] = info.count(k)
        return di
    
    
    text = ""
    
    print('Please input the text, end with a exclamation mark !')
    while text.count('!') == 0:
        text += input()
    
    
    # get rid of any after exclamation mark
    main_text = text.split('!')[0]
    print('result is: ' + str(detect(main_text)))

【讨论】：