Counting how many times each letter appears in the string

Question

I'm trying to count how many times each letter appears in the string and then display it in a list eg. The word "hello" would be: {h:1, e:1, l:2, o:2} This is my code.

    text = input()
    dict = {}
    
    for k in range(len(text)):
      if text[k] in dict:
         count +=1
         dict.update({text[k]:count})
      else:
         count=1
         dict.update({text[k]:count})        


    print(dict)

The code works on smaller words but not for words that are >10 Can someone please point out what is wrong or what I am missing.

You're resetting the count each time, but the count depends on the current value inside your dict[text[k]] - it does not depend on the variable you keep using for counting. You can instead do dict[text[k]] += 1 in your upper check, and dict[text[k]] = 1 in your lower else. However, you can do this by doing from collections import Counter and cnt = Counter(text); no need to go through the text step by step. Another thing - do not name your dict dict - dict is an internally defined function that creates a dictionary. — MatsLindh
– MatsLindh, Commented Jun 27, 2022 at 19:50
Examples that don't work are words like incomprehensibilities, Tergiversation — Marcus
– Marcus, Commented Jun 27, 2022 at 19:56
It's not about the length, it's about letter order. Try ababba, it's fun! — Nikolaj Š.
– Nikolaj Š., Commented Jun 27, 2022 at 19:57
@MarkRansom, there it is, I can distinctly see count = 1 =) — Nikolaj Š.
– Nikolaj Š., Commented Jun 27, 2022 at 20:01

Tomer Ariel · Accepted Answer · 2022-06-27 19:51:12Z

2

You can use the following code to count letter frequencies:

d = {}
for letter in text:
    d[letter] = d.get(letter, 0) + 1
print(d)

answered Jun 27, 2022 at 19:51

Tomer Ariel

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3 Comments

Nikolaj Š. Over a year ago

OP is definitely learning to code, so it's more useful to point out their mistake, instead of using magic class to do everything for them =)

Marcus Over a year ago

Can you please explain what the code does new to python.

Nikolaj Š. Over a year ago

@DanielHao: "When learning a new language…" — I'm pretty sure he learns his first language.

devdude19289 · Accepted Answer · 2022-06-27 20:05:51Z

0

I've changed your code around a bit that seems to work:

trans = {}
def letterCounter(num):
    for ch in num.lower():
        if ch in trans.keys():
            trans[ch] += 1;
        else:
            trans[ch] = 1;
            
    print('Counts letters shown in the input:')
    for x in sorted(trans):
        print(f"Letter \"{x}\": {trans[x]}")

try:
    usr = input('Please insert your input & press ENTER to count its letters: ')
    letterCounter(usr)
except ValueError:
    print('Input must contain a word or sentence')

I know this is a little bit different than what you asked for, but this seems to work very well, and I used www.cologic.dev to do it. It uses code completion to help you learn and code more efficiently.

answered Jun 27, 2022 at 20:05

devdude19289

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1 Comment

devdude19289 Over a year ago

@Marcus np. use LAUNCH50 when signing up, and let me know if you need more help with code

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