[SOLVED] Letter Counter Intermediate Python

see this code, I am checking that If keys or letters are avalible then just update the value by 1 , else make a key of name that letter and make it's values 1 https://code.sololearn.com/c4B7Zb7voT80/?ref=app

Here is what i did ,it works(it would help) text = input() dict = {} #your code goes here for i in text: if i not in dict: dict[i]=1 else: dict[i]+=1 print(dict)

Bianca S You could also use dictionary comprehension here: a = input() print({x:a.count(x) for x in set(a)}) # Hope this helps

Hasnae Bouhmady your solution is the best !

""" you could also use the built-in Counter class from 'collections' module, wich return a dict-like, wich can be easily converted to a real dict: """ from collections import Counter text = input() print(dict(Counter(text))) # https://pymotw.com/2/collections/counter.html

Try this code I tried it using dictionary comprehension text = input() dic = {I: text.count(i) for i in text } print(dic)

What do dict={} do?

@saad Khan dict={} is an empty dictionary ,you need to declare it first and then you can fill in it whith keys and values

Calvin Thomas technically your code is doing what is wanted but they want in order according to order of input letters so that's why it doesn't really works.

# These three solutions for help: #solving number 1: text = input() #your code goes here from collections import Counter c = Counter(text) print(dict(c)) #solving number 2: word = input('Enter word: ') d = dict() for i in word: d[i] = word.count(i) print(d) #solving number 3: word = input('Enter word: ') print({letter: word.count(letter) for letter in word})

text = input() dict = {} for x in text: dict[x]=text.count(x) print(dict) #your code goes here

for letters in text: if letters in dict.keys(): dict[letters] += 1 else: dict[letters] = 1 print(dict)