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(az) 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

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}

Answer 2

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 3

You can initialise your Counter() with a dictionary. In this case a dictionary comprehension is used to initialize all the lowercase letters to zero.

Using update() with the letter will then add to these existing values:

from collections import Counter 

letter = "hello world "
letter_counts = Counter({l:0 for l in string.ascii_lowercase})
letter_counts.update(letter.translate(str.maketrans('','',string.punctuation + ' ')))

print(letter_counts)

Giving you:

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

To get rid of the space, add it to the punctuation string.

Answer 4

How about

import string
from collections import defaultdict

row="rxgmgcwbd c qcyurr bkxgmq, lwrg grru rrwxtam rwgzwt am quyam cv avrrgdwkxgcr.iwxbdamcz xdalguj qarc ram av vcmfwgmgum. yw'g"
letters = string.ascii_lowercase
stats = defaultdict(list)
for l in letters:
    stats[l]=0
for l in row:
    if l.isalpha():
        stats[l]+=1

Answer 5

You can use dict.fromkeys to init dictionary with default 0 value for missing letters. And then update this dictionary:

import string

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

letter_count = dict.fromkeys(string.ascii_lowercase, 0)
for c in x:
    if c in string.ascii_lowercase:
        letter_count[c] += 1
print letter_count

Output:

{'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}

Answer 6

... that my result prints something like ...

The other answers focus on choosing a different data structure, but to me it sounds like you already chose the right data structure, Counter , and just want to display the result nicely. So something like this would do:

display_str = "{" + ", ".join("'{}': {}".format(x, letter_count[x]) for x in string.ascii_lowercase) + "}"
print("Frequency count of letter:", display_str, sep="\n")