How to create txt frequency counter with all letters (a-z) in python 3

Question

I have a text file named textf that looks something like the following:

rxgmgcwbd c qcyurr bkxgmq, lwrg grru rrwxtam rwgzwt am quyam cv avrrgdwkxgcr.iwxbdamcz xdalguj qarc ram av vcmfwgmgum. yw'g

I want to do a frequency count for each letter in the text file but I want it with the condition that if a letter does not appear in the text, it should have a key:value pair with value 0. For example if z was not in the text it should look something like 'z': 0 and so on for all letters (a to z). I did the following code:

import string  
from collections import Counter 
with open("textf.txt") as tf: 
    letter = tf.read()
letter_count = Counter(letter.translate(str.maketrans('','',string.punctuation)))
print("Frequency count of letter:","\n",letter_count)

But the output looks something like this:

Counter({' ': 110, 'r': 12, 'c': 88, 'a': 55, 'g': 57, 'w': 76, 'm': 76, 'x': 72, 'u': 70, 'q': 41, 'y': 40, 'j': 36, 'l': 32, 'b': 18, 'd': 28, 'v': 27, 'k': 22, 't': 19, 'f': 18, 'z': 16, 'i': 7})

I am trying to make it so that the space count ' ': 110 is not shown and that I have all the letters(a-z) and when the letter does not appear in the text that my result prints something like 'n': 0 and so on. Any ideas or suggestions of how I could make this possible?

Answer 1

One way to do this is to make a normal dict from your Counter, using the lowercase letters as the keys of the new dict. We use the dict.get method to supply a default value of zero for missing letters.

import string  
from collections import Counter 

letter = "rxgmgcwbd c qcyurr bkxgmq, lwrg grru rrwxtam rwgzwt am quyam cv avrrgdwkxgcr.iwxbdamcz xdalguj qarc ram av vcmfwgmgum. yw'g"

letter_count = Counter(letter.translate(str.maketrans('','',string.punctuation)))
letter_count = {k: letter_count.get(k, 0) for k in string.ascii_lowercase}
print("Frequency count of letter:\n", letter_count)

output

Frequency count of letter:
 {'a': 9, 'b': 3, 'c': 8, 'd': 4, 'e': 0, 'f': 1, 'g': 12, 'h': 0, 'i': 1, 'j': 1, 'k': 2, 'l': 2, 'm': 10, 'n': 0, 'o': 0, 'p': 0, 'q': 4, 'r': 14, 's': 0, 't': 2, 'u': 5, 'v': 4, 'w': 9, 'x': 6, 'y': 3, 'z': 2}

If you do this in Python 3.6+ you get the side-benefit that the new dict is alphabetically sorted (although that behaviour is currently just an implementation detail that should not be relied upon).

---

As user2357112 mentions in the comments, we don't need to use letter_count.get(k, 0), since a Counter automatically returns zero if we try to read the value of a non-existent key. So that dict comprehension can be changed to

letter_count = {k: letter_count[k] for k in string.ascii_lowercase}

Answer 2

You can do this like so:

x = "rxgmgcwbd c qcyurr bkxgmq, lwrg grru rrwxtam rwgzwt am quyam cv avrrgdwkxgcr.iwxbdamcz xdalguj qarc ram av vcmfwgmgum. yw'g"

import string

freq = {i:0 for i in string.ascii_lowercase}
for i in x:
    if i in freq:
        freq[i] += 1

You can also replace the for-loop with a dictionary-comprehension (though it's inefficient for what we are trying to do since it uses count - but added as a way just for reference):

freq = {i:x.count(i) for i in freq}

This will give as a result:

{'a': 9, 'c': 8, 'b': 3, 'e': 0, 'd': 4, 'g': 12, 'f': 1, 'i': 1, 'h': 0, 'k': 2, 'j': 1, 'm': 10, 'l': 2, 'o': 0, 'n': 0, 'q': 4, 'p': 0, 's': 0, 'r': 14, 'u': 5, 't': 2, 'w': 9, 'v': 4, 'y': 3, 'x': 6, 'z': 2}