# [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.

## 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
``````

## 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}
``````

## Solution #4:

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

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

## 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.

## 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)
``````

## 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 `wordlist`it’s pretty simple

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

## 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}
``````

## 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]
``````

## 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)
``````

## 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]))
``````

## 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}

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