Using regular expression, list all the letters that follows a vowel according to their occurrence fr-CodePudding

How can I find consonants letters that came after the vowels in words of string and count the frequency

str = 'car regular double bad '
result19 = re.findall(r'\b\w*[aeiou][^ aeiou]\w*\b' , str)
print(result19) #doesn't work

Expected output

letter r count  = 2
letter b count = 1
letter d count = 1

CodePudding user response：

I am not sure whether this is what you want or not, but it might help as an answer and not a comment.

I think you are on the right track, but you need a few modifications and other lines to achieve the excepted:

import re
myStr = 'car regular double bad '
result19 = re.findall(r'[aeiou][^aeiou\s] ' , myStr)
myDict = {}
for value in result19:
  if not value[1] in myDict:
    myDict[value[1]] = 0
  myDict[value[1]]  = 1
myDict

This will result in a dictionary containing the values and the number the have appeared:

{'b': 1, 'd': 1, 'g': 1, 'l': 1, 'r': 2}

For having a better output you can use a for loop to print each key and its value:

for chr, value in myDict.items():
  print(chr, "->", value)

Output

r -> 2
g -> 1
l -> 1
b -> 1
d -> 1

CodePudding user response：

Your pattern \b\w*[aeiou][^ aeiou]\w*\b matches zero or more repetitions of a word character using \w* and only matches a single occurrence of [aeiou][^ aeiou] in the "word"

If you want to match all consonant letters based on the alphabet a-z after a vowel, you can match a single occurrence of [aeiou] and use a capture group matching a single consonant.

Then make use of re.findall to return a list of the group values.

import re

txt = 'car regular double bad '
lst = re.findall(r'[aeiou]([b-df-hj-np-tv-z])', txt)
dct = {c: lst.count(c) for c in lst}

print(dct)

Output

{'r': 2, 'g': 1, 'l': 1, 'b': 1, 'd': 1}

If you want to match a non whitespace char other than a vowel after matching a vowel, you can use this pattern [aeiou]([^\saeiou])

Note that the l is also in the output as it comes after the u in ul