计算字符的频率直到引入新的字符-python

如何解决计算字符的频率直到引入新的字符-python

我正在编写一个程序来根据用户输入来计算字母的频率，直到“！”为止。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，我应该已经计算过频率，直到在输入中插入新字符为止。 例如，如果输入为（aabc），则我的程序应将两个“ a”相加，然后移至“ b”。

无论如何，在继续使用新字母之前，我是否需要修改循环以计数频率？

解决方法

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。

,

    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)))

如果要包含大写字母：

    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)))

如果您不想处理string模块，并且只要能识别utf-8字符，就计算每个输入的频率：

    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，如果您只想计算低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)))