how to count the frequency of letters in text excluding whitespace and numbers?

Question

Use a dictionary to count the frequency of letters in the input string. Only letters should be counted, not blank spaces, numbers, or punctuation. Upper case should be considered the same as lower case. For example, count_letters("This is a sentence.") should return {'t': 2, 'h': 1, 'i': 2, 's': 3, 'a': 1, 'e': 3, 'n': 2, 'c': 1}

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

    print(count_letters("AaBbCc"))
    # Should be {'a': 2, 'b': 2, 'c': 2}

    print(count_letters("Math is fun! 2+2=4"))
    # Should be {'m': 1, 'a': 1, 't': 1, 'h': 1, 'i': 1, 's': 1, 'f': 1, 'u': 1, 'n': 1}

    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}

Okay, so what part of the problem are you having difficulty with? — Karl Knechtel, Mar 31 '20 at 01:15

score 8 · Answer 1 · answered Jul 22 '20 at 03:33

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

print(count_letters("AaBbCc"))
# Should be {'a': 2, 'b': 2, 'c': 2}

print(count_letters("Math is fun! 2+2=4"))
# Should be {'m': 1, 'a': 1, 't': 1, 'h': 1, 'i': 1, 's': 1, 'f': 1, 'u': 1, 'n': 1}

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}

score 7 · Accepted Answer · answered Mar 31 '20 at 01:20

7

This should work:

>>> from collections import Counter
>>> from string import ascii_letters
>>> def count_letters(s) :
...     filtered = [c for c in s.lower() if c in ascii_letters]
...     return Counter(filtered)
... 
>>> count_letters('Math is fun! 2+2=4')
Counter({'a': 1, 'f': 1, 'i': 1, 'h': 1, 'm': 1, 'n': 1, 's': 1, 'u': 1, 't': 1})
>>>

answered Mar 31 '20 at 01:20

lenik

23,228
4
34
43

Thanks for reminding about `ascii_letters`. I was actually looking for something like this – Alex.Kh Mar 31 '20 at 01:28
1

@YahiaNaeem which part do you have a problem with ? – lenik Mar 31 '20 at 01:35
I didn't study the OOP yet this question assume that I don't know this classes – Yahia Naeem Mar 31 '20 at 01:45
@YahiaNaeem where did you see any classes in my answer ?? – lenik Mar 31 '20 at 01:47
No,sorry I meant modules . – Yahia Naeem Mar 31 '20 at 02:11

Shobhit Umrao · Answer 3 · 2020-05-12T19:41:46.510

So I got your question,

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

  print(count_letters("AaBbCc"))

  # Should be {'a': 2, 'b': 2, 'c': 2}

  print(count_letters("Math is fun! 2+2=4"))
  # Should be {'m': 1, 'a': 1, 't': 1, 'h': 1, 'i': 1, 's': 1, 'f': 1, 'u': 
  1, 'n': 1}

  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}

score 1 · Answer 4 · edited Oct 09 '21 at 11:05

def count_letters(text):
  result = {}
  for letter in text:
    #check if it alphabet or something else
    # Check if the letter needs to be counted or not
    if letter.isalpha():
      result[letter.lower()]=result.get(letter.lower(),0)+1
    # Add or increment the value in the dictionary
  return result

print(count_letters("AaBbCc"))
# Should be {'a': 2, 'b': 2, 'c': 2}

print(count_letters("Math is fun! 2+2=4"))
# Should be {'m': 1, 'a': 1, 't': 1, 'h': 1, 'i': 1, 's': 1, 'f': 1, 'u': 1, 'n': 1}

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}

PAGADALA VAMSI Krishna · Answer 5 · 2020-06-29T05:26:41.587

def count_letters(text):
  result = {}
  # Go through each letter in the text
  text=text.lower()
  for letter in text:
    if letter in 'abcdefghijklmnopqrtsuvwxyz':

      if letter in result:
        result[letter]+=1
      
      else:
        result[letter]=1
  return result

print(count_letters("AaBbCc"))
# Should be {'a': 2, 'b': 2, 'c': 2}

print(count_letters("Math is fun! 2+2=4"))
# Should be {'m': 1, 'a': 1, 't': 1, 'h': 1, 'i': 1, 's': 1, 'f': 1, 'u': 1, 'n': 1}

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}

score 1 · Answer 6 · edited Jul 06 '20 at 02:04

The ".isalpha()" method comes in very handy. See how I solved it:

def count_letters(text):
  result = {}
  # Go through each letter in the text
  for letter in text:
    # Check if the letter needs to be counted or not
    
    if letter.isalpha() and letter not in result:
      letter=letter.lower()
      result[letter]=1
    elif letter.isalpha()==False:
      pass
    else:
      result[letter]+=1

    ___
    # Add or increment the value in the dictionary
    ___
  return result

print(count_letters("AaBbCc"))
# Should be {'a': 2, 'b': 2, 'c': 2}

print(count_letters("Math is fun! 2+2=4"))
# Should be {'m': 1, 'a': 1, 't': 1, 'h': 1, 'i': 1, 's': 1, 'f': 1, 'u': 1, 'n': 1}

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}

score 1 · Answer 7 · answered Feb 27 '21 at 14:25

you can use the .isalpha() method to check if the character is an alphabet letter.

Then use the .get method to return the value from a specific key.

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

score 1 · Answer 8 · answered Jan 23 '22 at 22:39

if letter.isalpha() and letter != " ": checks if the letter is only a letter, not a number, a punctuation or a blank space.

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

Nikhil Tiwari · Answer 9 · 2020-05-10T13:29:44.053

0

def count_letters(text):
 r = {}
  # Go through each letter in the text
 for letter in text:
    # Check if the letter needs to be counted or not
     for letter in text:
# Check if the letter needs to be counted or not
 for alp in letter:
  # To check the letters are alphabets or not
  if alp in ascii_letters:
   #Converting into lowercase as python is case sensitive
    if alp.lower() in r and alp!=' ':
      r[alp.lower()]+=1
    elif alp.lower() not in r and alp!=' ':
      r[alp.lower()]=1
  # Add or increment the value in the dictionary
  return r

print(count_letters("AaBbCc"))
print(count_letters("Math is fun! 2+2=4"))
print(count_letters("This is a sentence."))

edited May 10 '20 at 13:29

answered May 07 '20 at 17:02

Nikhil Tiwari

1
3

Code dumps without any explanation are rarely helpful. Stack Overflow is about learning, not providing snippets to blindly copy and paste. Please [edit] your question and explain how it works better than what the OP provided. – ChrisGPT was on strike May 07 '20 at 19:29
Also, please _never_ use single-space indentation. It is very hard to read, and in a language like Python where whitespace is significant it's _insane_. – ChrisGPT was on strike May 07 '20 at 19:30

score 0 · Answer 10 · answered May 17 '20 at 11:10

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

print(count_letters("AaBbCc"))
# Should be {'a': 2, 'b': 2, 'c': 2}

print(count_letters("Math is fun! 2+2=4"))
# Should be {'m': 1, 'a': 1, 't': 1, 'h': 1, 'i': 1, 's': 1, 'f': 1, 'u': 1, 'n': 1}

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}

score 0 · Answer 11 · answered May 17 '20 at 11:16

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

print(count_letters("AaBbCc"))
# Should be {'a': 2, 'b': 2, 'c': 2}

print(count_letters("Math is fun! 2+2=4"))
# Should be {'m': 1, 'a': 1, 't': 1, 'h': 1, 'i': 1, 's': 1, 'f': 1, 'u': 1, 'n': 1}

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}

score 0 · Answer 12 · edited Dec 05 '21 at 21:16

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

print(count_letters("AaBbCc"))
# Should be {'a': 2, 'b': 2, 'c': 2}

print(count_letters("Math is fun! 2+2=4"))
# Should be {'m': 1, 'a': 1, 't': 1, 'h': 1, 'i': 1, 's': 1, 'f': 1, 'u': 1, 'n': 1}

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}

get function would seem to be something new to beginers but you can go on and research on it — Rushabh, Jun 02 '20 at 20:55

score 0 · Answer 13 · edited Mar 01 '21 at 03:14

0

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

print(count_letters("AaBbCc"))
# Should be {'a': 2, 'b': 2, 'c': 2}

print(count_letters("Math is fun! 2+2=4"))
# Should be {'m': 1, 'a': 1, 't': 1, 'h': 1, 'i': 1, 's': 1, 'f': 1, 'u': 1, 'n': 1}

edited Mar 01 '21 at 03:14

SuperStormer

4,997
5
25
35

answered Jun 02 '20 at 22:40

abdala

1

1

Please take a look at the following post which explains how to properly format code blocks https://meta.stackoverflow.com/questions/251361/how-do-i-format-my-code-blocks – NotAName Jun 02 '20 at 23:54
Can you include inline code span so the code is readable & verifiable. – Usama Abdulrehman Jun 03 '20 at 01:16

sen-sensei · Answer 14 · 2020-07-14T17:03:57.370

Well, in the second comment it says that you should : # Check if the letter needs to be counted or not

You did not check this. You checked only if the letter is in your dictionary named 'result', which is needed to add characters that don't exist already in the new dictionary. In this case, your program takes a blank space as a "letter". If we were to change the 'letter' variable as 'element', things would be more clear.

if letter.isalpha() == True:

You should use the above line to check only for the letters. For each element of the input text, this block would be skipped when the a non-letter character is met(blank space, number of special character).

All in all, your script should look like this:

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

print(count_letters("AaBbCc"))
# Should be {'a': 2, 'b': 2, 'c': 2}

print(count_letters("Math is fun! 2+2=4"))
# Should be {'m': 1, 'a': 1, 't': 1, 'h': 1, 'i': 1, 's': 1, 'f': 1, 'u': 1, 'n': 1}

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}

score 0 · Answer 15 · answered Aug 04 '20 at 10:40

This is what i did

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

print(count_letters("AaBbCc"))
# Should be {'a': 2, 'b': 2, 'c': 2}
print(count_letters("Math is fun! 2+2=4"))
# Should be {'m': 1, 'a': 1, 't': 1, 'h': 1, 'i': 1, 's': 1, 'f': 1, 'u': 1, 'n': 1}
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}

score 0 · Answer 16 · edited Nov 09 '20 at 12:38

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


    
  return result

print(count_letters("AaBbCc"))
# Should be {'a': 2, 'b': 2, 'c': 2}

print(count_letters("Math is fun! 2+2=4"))
# Should be {'m': 1, 'a': 1, 't': 1, 'h': 1, 'i': 1, 's': 1, 'f': 1, 'u': 1, 'n': 1}

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}

score 0 · Answer 17 · answered Feb 13 '21 at 01:15

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

print(count_letters("AaBbCc"))
# Should be {'a': 2, 'b': 2, 'c': 2}

print(count_letters("Math is fun! 2+2=4"))
# Should be {'m': 1, 'a': 1, 't': 1, 'h': 1, 'i': 1, 's': 1, 'f': 1, 'u': 1, 'n': 1}

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}

score 0 · Answer 18 · answered Mar 21 '21 at 22:12

    def count_letters(text):
  result = {}
  # Go through each letter in the text
  text=text.casefold()
  for letter in text:
    # Check if the letter needs to be counted or not
    if (letter.isalpha()): 

    # Add or increment the value in the dictionary
      result.update({letter:text.count(letter)})
  return result

print(count_letters("AaBbCc"))
# Should be {'a': 2, 'b': 2, 'c': 2}

print(count_letters("Math is fun! 2+2=4"))
# Should be {'m': 1, 'a': 1, 't': 1, 'h': 1, 'i': 1, 's': 1, 'f': 1, 'u': 1, 'n': 1}

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}

score 0 · Answer 19 · answered Jul 04 '21 at 17:32

0

def count_letters(text):
  result = {}
  for letter in text.lower():
    if letter.isalpha():
      lettercount = text.lower().count(letter)
      result[letter] = lettercount
  return result

answered Jul 04 '21 at 17:32

Mayank Raj Vaid

1
1

score 0 · Answer 20 · answered Sep 12 '21 at 17:28

def count_letters(text):
    result = {}
    text = text.lower().replace(" ","")
    text = text.replace("", " ").split()
    dummy = []
    dummy1 = ""

    for x in text: # remove non-letter
        if x.isalpha():
            dummy.append(x)
            dummy1 = "".join(dummy)

    for letter in dummy1:
        if letter not in result:
            result[letter] = 0
        result[letter] += 1
    return result

print(count_letters("AaBbCc"))
print(count_letters("Math is fun! 2+2=4"))
print(count_letters("This is a sentence."))

score 0 · Answer 21 · answered Oct 09 '21 at 09:38

0

def  count_letters(text):
    result = {}
    for letter in text.lower():
        if letter == " " or letter.isalpha() == False:
            continue:
        if letter not in result:
            result[letter] = 0
        result[letter] += 1
    return result

answered Oct 09 '21 at 09:38

olawale oladeji

1
1

3

As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Community Oct 09 '21 at 09:40
`letter.isalpha()` already handles the whitespace check (`" ".isalpha()` is `False`) – Gino Mempin Oct 09 '21 at 11:07

score 0 · Answer 22 · answered Dec 05 '21 at 03:16

def count_letters(text):
  result = {}
  # Go through each letter in the text
  for letter in text.lower():
    # Check if the letter needs to be counted or not
     if letter.isalpha() and letter not in result:
       result[letter] = text.lower().count(letter)
  return result

print(count_letters("AaBbCc"))
# Should be {'a': 2, 'b': 2, 'c': 2}

print(count_letters("Math is fun! 2+2=4"))
# Should be {'m': 1, 'a': 1, 't': 1, 'h': 1, 'i': 1, 's': 1, 'f': 1, 'u': 1, 'n': 1}

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}

score 0 · Answer 23 · answered Feb 22 '22 at 08:45

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

print(count_letters("AaBbCc"))
# Should be {'a': 2, 'b': 2, 'c': 2}

print(count_letters("Math is fun! 2+2=4"))
# Should be {'m': 1, 'a': 1, 't': 1, 'h': 1, 'i': 1, 's': 1, 'f': 1, 'u': 1, 'n': 1}

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}

I ran your code and getting this output: ```>>> print(count_letters("AaBbCc")) {'a': 0, 'b': 0, 'c': 0} >>> print(count_letters("Math is fun! 2+2=4")) {'m': 0, 'a': 1, 't': 1, 'h': 1, 'i': 1, 's': 1, 'f': 1, 'u': 1, 'n': 1} >>> print(count_letters("This is a sentence.")) {'t': 0, 'h': 1, 'i': 2, 's': 3, 'a': 1, 'e': 3, 'n': 2, 'c': 1}``` — user11717481, Feb 25 '22 at 23:42

how to count the frequency of letters in text excluding whitespace and numbers?

23 Answers23

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