Counting the frequency of a character until a new character is introduced - python

Question

I was writing a program to count the frequency of alphabets based on the user input until "!" was introduced. The following is my program:

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)

However, apparently, I should have counted the frequency until a new character has been plugged in the input. For instance, if the input is (aabc), then my program should count the two 'a's then move on to 'b'.

Is there anyway for me to modify my loop to count the frequency before moving on to a new alphabet?

Answer 1

    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)))

if you want to include capitalised alphabets:

    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)))

if you don't want to deal with string module, and count the frequency of every input as long as it is recognised utf-8 character:

    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)))

finally, if you only want to count lower 26 alphabet:

    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)))

Answer 2

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

I've used a dictionary

d will contain the result after execution.