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