正则表达式从字符串中删除eed

Question

I am trying to replace 'eed' and 'eedly' with 'ee' from words where there is a vowel before either term ('eed' or 'eedly') appears.

So for example, the word indeed would become indee because there is a vowel ('i') that happens before the 'eed'. On the other hand the word 'feed' would not change because there is no vowel before the suffix 'eed' .

I have this regex: (?i)([aeiou]([aeiou])*[e{2}][d]|[dly]\\\\b) You can see what is happening with this here .

As you can see, this is correctly identifying words that end with 'eed' , but it is not correctly identifying 'eedly' .

Also, when it does the replace, it is replacing all words that end with 'eed' , even words like feed which it should not remove the eed

What should I be considering here in order to make it correctly identify the words based on the rules I specified?

Answer 1

You can use:

str = str.replaceAll("(?i)\\b(\\w*?[aeiou]\\w*)eed(?:ly)?", "$1ee");

Updated RegEx Demo

\\\\b(\\\\w*?[aeiou]\\\\w*) before eed or eedly makes sure there is at least one vowel in the same word before this.

To expedite this regex you can use negated expression regex:

\\b([^\\Waeiou]*[aeiou]\\w*)eed(?:ly)?

RegEx Breakup:

\\b                 # word boundary
(                   # start captured group #`
   [^\\Waeiou]*     # match 0 or more of non-vowel and non-word characters
   [aeiou]          # match one vowel
   \\w*             # followed by 0 or more word characters
)                   # end captured group #`
eed                 # followed by literal "eed"
(?:                 # start non-capturing group
   ly               # match literal "ly"
)?                  # end non-capturing group, ? makes it optional

Replacement is:

"$1ee" which means back reference to captured group #1 followed by "ee"

Answer 2

find dly before finding d. otherwise your regex evaluation stops after finding eed.

(?i)([aeiou]([aeiou])*[e{2}](dly|d))