grep consonants in regular expression

Question

I want to use grep to find words that have a specific number of consonants. But using the following command will give me consecutive consonants, How do i modify the command so that non-consecutive consonants are also OK?

grep -iE '[^aeiouy]{3}' filename

Answer 1

Something like this should work

grep -iE '\b[bcdfghjklmnpqrstvwxz][aeiouy]*?[bcdfghjklmnpqrstvwxz][aeiouy]*?[aeiouy][bcdfghjklmnpqrstvwxz]\b'

That will match [a consonant][any number of vowels][consonant][vowel(s)][consonant]

Answer 2

grep -iP '^(?=(.*[^aeiouy]){3}).*$' filename

您可以尝试一下。这将选择在字符串中任何位置至少包含3辅音的单词。使用P或perl模式。