Some letters are not getting counted in the below code in Python-CodePudding

I am trying to write a code to count letters in a string and store them as key value pair in a separate dictionary with key as the letter and value as the total count of that respective letter.

def count_letters(text):
  result = {}
  counter = 0
  # Go through each letter in the text
  for letter in text:
    if letter.islower() or letter.isupper():
      # Check if the letter needs to be counted or not
      if letter not in result:
        counter = 0
      # Add or increment the value in the dictionary
      counter  = 1
      result[letter.lower()] = counter
  return result


print(count_letters("This is a sentence."))
# Should be {'t': 2, 'h': 1, 'i': 2, 's': 3, 'a': 1, 'e': 3, 'n': 2, 'c': 1}

Could anyone please help me out what is going wrong in the code as I am not able to debug it.

I am getting Output as below:

{'t': 2, 'h': 1, 'i': 2, 's': 2, 'a': 1, 'e': 2, 'n': 4, 'c': 1}

CodePudding user response：

I just add second condition to your code and it should work now.

def count_letters(text):
  result = {}
  counter = 0
  # Go through each letter in the text
  for letter in text:
    if letter.islower() or letter.isupper():
      # Check if the letter needs to be counted or not
      if letter not in result:
        counter = 0
      else:
        counter=result[letter.lower()]
      # Add or increment the value in the dictionary
      counter  = 1
      result[letter.lower()] = counter
  return result


print(count_letters("This is a sentence."))
# Should be {'t': 2, 'h': 1, 'i': 2, 's': 3, 'a': 1, 'e': 3, 'n': 2, 'c': 1}

output now:

{'t': 2, 'h': 1, 'i': 2, 's': 3, 'a': 1, 'e': 3, 'n': 2, 'c': 1}

Condition added :

else:
        counter=result[letter.lower()]

EDIT: Because you are doing dictionary just with lower letters you maybe want to count upper letters as lower in sentence. If you want You need to edit the code to

if letter.lower() not in result:

Otherwise the UPPER letters in for example middle of sentence will NOT be counted.

CodePudding user response：

def count_letters(text):
  result = {}
  counter = 0
  # Go through each letter in the text
  for letter in text:
    if letter.islower() or letter.isupper():
      # Check if the letter needs to be counted or not
      if letter not in result:
        counter = 0
      # Add or increment the value in the dictionary
      counter  = 1
      result[letter.lower()] = counter
  return result

print(count_letters("This is a sentence."))

In your code the problem is with counter variable. As it resets its value if new letter come in result dict and does not store the count for previous letters.

Counter in your code is working like this:

loop letter counter
1 T 1
2 h 1
3 i 1
4 s 1
5 i 2
6 s 2
7 a 1
8 s 2 - remain same here as 6 line
9 e 1
10 n 1
11 t 2
12 e 3
13 n 4
14 c 1
15 e 2

Above problem could be solved by directly making changes in the dictionary rather then using any another variable

def count_letters(text):
  result = {}
  # Go through each letter in the text
  for letter in text:
    if letter.islower() or letter.isupper():
      # Check if the letter needs to be counted or not
      if letter not in result:
        result[letter.lower()] = 0
      # Add or increment the value in the dictionary
      result[letter.lower()]  = 1
  return result


print(count_letters("This is a sentence."))

Here the output: {'t': 2, 'h': 1, 'i': 2, 's': 3, 'a': 1, 'e': 3, 'n': 2, 'c': 1}

I hope it clear your doubt

CodePudding user response：

Might suggest using Counter from collections:

import string
from collections import Counter
custom_string = "This is a sentence."
storage = Counter(custom_string.lower().translate(str.maketrans('', '', string.punctuation)))

where .translate(str.maketrans('', '', string.punctuation)) strips punctuation marks. Output of storage variable:

Counter({'s': 3, ' ': 3, 'e': 3, 't': 2, 'i': 2, 'n': 2, 'h': 1, 'a': 1, 'c': 1})