[Solved] Counting Letter Frequency in a String (Python)

I am trying to count the occurrences of each letter of a word

word = input("Enter a word")

Alphabet=['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']

for i in range(0,26):
    print(word.count(Alphabet[i]))

This currently outputs the number of times each letter occurs including the ones that don’t.

How do I list the letters vertically with the frequency alongside it, e.g., like the following?

word=”Hello”

H 1

E 1

L 2

O 1

Solution #1:

from collections import Counter
counts=Counter(word) # Counter({'l': 2, 'H': 1, 'e': 1, 'o': 1})
for i in word:
    print(i,counts[i])

Try using Counter, which will create a dictionary that contains the frequencies of all items in a collection.

Otherwise, you could do a condition on your current code to print only if word.count(Alphabet[i]) is greater than 0, though that would be slower.

Respondent: LMc

Solution #2:

def char_frequency(str1):
    dict = {}
    for n in str1:
        keys = dict.keys()
        if n in keys:
            dict[n] += 1
        else:
            dict[n] = 1
    return dict
print(char_frequency('google.com'))
Respondent: Uzam Hashmi

Solution #3:

As Pythonista said, this is a job for collections.Counter:

from collections import Counter
print(Counter('cats on wheels'))

This prints:

{'s': 2, ' ': 2, 'e': 2, 't': 1, 'n': 1, 'l': 1, 'a': 1, 'c': 1, 'w': 1, 'h': 1, 'o': 1}
Respondent: duhaime

Solution #4:

s = input()
t = s.lower()

for i in range(len(s)):
    b = t.count(t[i])
    print("{} -- {}".format(s[i], b))
Respondent: Swetank

Solution #5:

Following up what LMc said, your code was already pretty close to functional. You just needed to post-process the result set to remove ‘uninteresting’ output. Here’s one way to make your code work:

#!/usr/bin/env python
word = raw_input("Enter a word: ")

Alphabet = [
    'a','b','c','d','e','f','g','h','i','j','k','l','m',
    'n','o','p','q','r','s','t','u','v','w','x','y','z'
]

hits = [
    (Alphabet[i], word.count(Alphabet[i]))
    for i in range(len(Alphabet))
    if word.count(Alphabet[i])
]

for letter, frequency in hits:
    print letter.upper(), frequency

But the solution using collections.Counter is much more elegant/Pythonic.

Respondent: evadeflow

Solution #6:

An easy and simple solution without a library:

string = input()
f = {}
for i in string:
  f[i] = f.get(i,0) + 1
print(f)

Here is the link for get(): https://docs.quantifiedcode.com/python-anti-patterns/correctness/not_using_get_to_return_a_default_value_from_a_dictionary.html

Respondent: soheshdoshi

Solution #7:

If using libraries or built-in functions is to be avoided then the following code may help:

s = "aaabbc"  # Sample string
dict_counter = {}  # Empty dict for holding characters
                   # as keys and count as values
for char in s:  # Traversing the whole string
                # character by character
    if not dict_counter or char not in dict_counter.keys(): # Checking whether the dict is
                                                            # empty or contains the character
        dict_counter.update({char: 1}) # If not then adding the
                                       # character to dict with count = 1
    elif char in dict_counter.keys(): # If the character is already
                                      # in the dict then update count
        dict_counter[char] += 1
for key, val in dict_counter.items(): # Looping over each key and
                                      # value pair for printing
    print(key, val)

Output:

a 3
b 2
c 1

Solution #8:

For future references: When you have a list with all the words you want, lets say wordlistit’s pretty simple

for numbers in range(len(wordlist)):
    if wordlist[numbers][0] == 'a':
        print(wordlist[numbers])
Respondent: Stanley

Solution #9:

Another way could be to remove repeated characters and iterate only on the unique characters (by using set()) and then counting the occurrence of each unique character (by using str.count())

def char_count(string):
    freq = {}
    for char in set(string):
        freq[char] = string.count(char)
    return freq


if __name__ == "__main__":
    s = "HelloWorldHello"
    print(char_count(s))
    # Output: {'e': 2, 'o': 3, 'W': 1, 'r': 1, 'd': 1, 'l': 5, 'H': 2}
Respondent: Syed Rafay

Solution #10:

It might make sense to include all letters of the alphabet. For example, if you’re interested in calculating the cosine difference between word distributions you typically require all letters.

You can use this method:

from collections import Counter 

def character_distribution_of_string(pass_string):
  letters = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
  chars_in_string = Counter(pass_string)
  res = {}
  for letter in letters:
    if(letter in chars_in_string):
      res[letter] = chars_in_string[letter]
    else: 
      res[letter] = 0 
  return(res)

Usage:

character_distribution_of_string("This is a string that I want to know about")

Full Character Distribution

{'a': 4,
 'b': 1,
 'c': 0,
 'd': 0,
 'e': 0,
 'f': 0,
 'g': 1,
 'h': 2,
 'i': 3,
 'j': 0,
 'k': 1,
 'l': 0,
 'm': 0,
 'n': 3,
 'o': 3,
 'p': 0,
 'q': 0,
 'r': 1,
 's': 3,
 't': 6,
 'u': 1,
 'v': 0,
 'w': 2,
 'x': 0,
 'y': 0,
 'z': 0}

You can extract the character vector easily:

list(character_distribution_of_string("This is a string that I want to know about").values())

giving…

[4, 1, 0, 0, 0, 0, 1, 2, 3, 0, 1, 0, 0, 3, 3, 0, 0, 1, 3, 6, 1, 0, 2, 0, 0, 0]
Respondent: Cybernetic

Solution #11:

Initialize an empty dictionary and iterate over every character of the word. If the current character in present in the dictionary, increment its value by 1, and if not, set its value to 1.

word="Hello"
characters={}
for character in word:
    if character in characters:
        characters[character] += 1
    else:
        characters[character] =  1
print(characters)
Respondent: Saurabh

Solution #12:

import string
word = input("Enter a word:  ")
word = word.lower()

Alphabet=list(string.ascii_lowercase)
res = []

for i in range(0,26): 
    res.append(word.count(Alphabet[i]))

for i in range (0,26):
    if res[i] != 0:
        print(str(Alphabet[i].upper()) + " " + str(res[i]))
Respondent: Anaam

Solution #13:

def string(n):
    a=list()
    n=n.replace(" ","")
    for i in  (n):
        c=n.count(i)
        a.append(i)
        a.append(c)
        y=dict(zip(*[iter(a)]*2))
    print(y)

string(“Lets hope for better life”)
#Output:{‘L’: 1, ‘e’: 5, ‘t’: 3, ‘s’: 1, ‘h’: 1, ‘o’: 2, ‘p’: 1, ‘f’: 2, ‘r’: 2, ‘b’: 1, ‘l’: 1, ‘i’: 1}
(if u notice in output 2 L-letter one uppercase and other lowercase..if u want them together look for the code below)

In the output it remove repeated characters ,Drop empty spaces and iterate only on the unique characters.
IF you want to count of both uppercase and lowercase together the:

def string(n):
    n=n.lower() #either use (n.uperr()) 
    a=list()
    n=n.replace(" ","")
    for i in  (n):
        c=n.count(i)
        a.append(i)
        a.append(c)
        y=dict(zip(*[iter(a)]*2))
    print(y)

string(“Lets hope for better life”)
#output:{‘l’: 2, ‘e’: 5, ‘t’: 3, ‘s’: 1, ‘h’: 1, ‘o’: 2, ‘p’: 1, ‘f’: 2, ‘r’: 2, ‘b’: 1, ‘i’: 1}

Respondent: Pratiksha Bhokate

The answers/resolutions are collected from stackoverflow, are licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0 .

Leave a Reply

Your email address will not be published.