如何在 python 中编写正则表达式以查找所有两个辅音相邻的单词

Question

I tried this code but does not work

string2 = "eat black  eat eaten  name 1234 stop  double " //sample of string
result62 = re.findall(r'\b\w+[^ 0-9][^ aeiouAEIOU]{2}\b', string2)
print(result62)

Answer 1

I would use the following regex:

\b[a-z]*[b-df-hj-np-tv-z]{2}[a-z]*\b

This looks for a

• word break \b
• some number of letters [az]*
• 2 consonants [b-df-hj-np-tv-z]{2}
• some number of letters [az]*
• a word break

I would look for two consonants specifically to avoid having to worry about matching (for example) fuel.

In python:

string2 = "eat black  eat eaten  name 1234 stop  double fuel."
result62 = re.findall(r'\b[a-z]*[b-df-hj-np-tv-z]{2}[a-z]*\b', string2, re.I)

Use the re.I flag to avoid having to specify both upper and lower case letters in the character classes.

Result (for your data):

['black', 'stop', 'double']

Answer 2

\b\w*(?:(?![aeiou0-9])\w){2}\w*\b

You can try this.See demo.

https://regex101.com/r/36VzAk/1

Answer 3

This should work:

string2 = "eat black  eat eaten  name 1234 stop  double "
result62 = re.findall(r'\b(?=[a-z]*(?:(?![aeiou])[a-z]){2})[a-z]*\b', string2)
print(result62)

prints:

['black', 'stop', 'double']

Answer 4

Assuming you want to retrieve the whole words having two consonants next to each other, and ignoring digits, you could use this regex (\w*[^aeiou\d\s]{2,}\w*) .

• \w* will look for any word character, zero or more times
• [^aeiou\d\s]{2,} will look for at least two consecutive consonants (any non-digit, non-whitespace, non-vowels characters)
• \w* will again look for any word character, zero or more times

import re

my_string = "eat black eat eaten  name 1234 stop double"
my_re = re.compile(r"(\w*[^aeiou\d\s]{2,}\w*)", re.I)

matches = my_re.findall(my_string)

for match in matches:
    print(match)

Outputs

black
stop
double