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For example, I am trying to count the amount of times a letter appears in my input. Also, I'm trying to use the tools I've learned so far. I haven't learned the .count tools in the course I'm taking so far.

1st I run this code:

any_word = input()
char_amount = {}

for i in any_word:
    char_amount[i] = len(i)

print(char_amount)

My input is: hello

The result is this: I get every time is below and my if statements fail to update the key for 'l' to 2.

{'h': 1, 'e': 1, 'l': 1, 'o': 1}

My if statements fail to update the key for 'l' to 2. I cant figure out the logic to an if statement to add 1 to the 'l' key because the duplicate is recognized in every key and the result is below. I assume i need another variable but i can't think up the logic. Thanks for any help:

{'h': 2, 'e': 2, 'l': 2, 'o': 2}
Chadeau
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    Does this answer your question? [how to count the frequency of letters in text excluding whitespace and numbers?](https://stackoverflow.com/questions/60941943/how-to-count-the-frequency-of-letters-in-text-excluding-whitespace-and-numbers) – Daniel Hao Sep 04 '22 at 16:41
  • Or maybe this - https://stackoverflow.com/questions/40985203/ It only take <2 min. to search... ;-) – Daniel Hao Sep 04 '22 at 16:49

4 Answers4

2

You should really be using the Counter class from the collections module:

from collections import Counter

value = input('Enter some text: ')

counter = Counter(value)

print(counter)

Thus, for an input of 'hello' you'd get:

Counter({'l': 2, 'h': 1, 'e': 1, 'o': 1})

Counter subclasses dict so you can access it just like any other dictionary

DarkKnight
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1

Use char_amount[i] = char_amount.get(i, 0) + 1 - this will try to get key i from dict char_amount, if not found returns 0. Then increases the value of this key by 1

any_word = input()
char_amount = {}

for i in any_word:
    char_amount[i] = char_amount.get(i, 0) + 1

print(char_amount)

Prints:

hello
{'h': 1, 'e': 1, 'l': 2, 'o': 1}
Andrej Kesely
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    OP is clearly a beginner but for the sake of completeness it's worth mentioning that the Counter class from the *collections* module is the better option – DarkKnight Sep 04 '22 at 16:15
  • @Vlad Can you post the answer with `collections.Counter`? It's always good to see alternatives - I'll upvote it :) – Andrej Kesely Sep 04 '22 at 16:17
  • Why is the Counter class from the collections module is the better option? – Chadeau Sep 05 '22 at 00:00
1

Without collections Module:

word = input()
char_amount = {}

for char in word:
    if char not in char_amount:
        char_amount[char] = 1
    else:
        char_amount[char] += 1

print(char_amount)

Using collections module:

from collections import Counter

print(Counter(input()))
George Muscat
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  • is the collections counter faster than doing this with if else statements? – Chadeau Sep 04 '22 at 17:54
  • @Chadeau I haven't tested each method. However, they should take similar time, with collections.Counter being much clearer with many other useful builtin methods. – George Muscat Sep 05 '22 at 16:28
1

use built in counter

from collections import Counter

my_counter = Counter("hello")
print(my_counter)

Counter({'l': 2, 'h': 1, 'e': 1, 'o': 1})
Lucas M. Uriarte
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