replaceAll（）删除所有带有这些异常的标点符号

Question

I am trying to get a String[] using a .txt file, and I need to remove all punctuation with some exceptions. Here is my code:

replaceAll("[^a-zA-Z ]", "");

exceptions: 1.hyphen(s) that are inside a word. 2.Get rid of the words that contain digits 3.Get rid of the words contains two punctuation at the end and the beginning

Answer 1

[^a-zA-Z ] is a character class. This means that it will match only one character and in this case will match anything that is not az, AZ or a whitespace.

If you want to match words, you need to use character classes with quantifiers for example +. If you want to match different patterns you need to apply the or logical operator | .

Knowing this, you may now match words that ends with one or more number or that has a number in the middle [^a-zA-Z ][0-9]+|[^a-zA-Z ]+[0-9] . I'll leave it to you as an exercise to apply it for your three cased since this sounds like a school assignment.

Answer 2

I have very complicated regex but it works.

\S*\d+\S*|\p{Punct}{2,}\S*|\S*\p{Punct}{2,}|[\p{Punct}&&[^-]]+|(?<![a-z])\-(?![a-z])

Explanation:

Match this alternative «\S*\d+\S*»
   Match a single character that is NOT a “whitespace character” «\S*»
      Between zero and unlimited times, as many times as possible, giving back as needed (greedy) «*»
   Match a single character that is a “digit” «\d+»
      Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
   Match a single character that is NOT a “whitespace character” «\S*»
      Between zero and unlimited times, as many times as possible, giving back as needed (greedy) «*»
Or match this alternative «\p{Punct}{2,}\S*»
   Match a character from the POSIX character class “punct” «\p{Punct}{2,}»
      Between 2 and unlimited times, as many times as possible, giving back as needed (greedy) «{2,}»
   Match a single character that is NOT a “whitespace character” «\S*»
      Between zero and unlimited times, as many times as possible, giving back as needed (greedy) «*»
Or match this alternative «\S*\p{Punct}{2,}»
   Match a single character that is NOT a “whitespace character” «\S*»
      Between zero and unlimited times, as many times as possible, giving back as needed (greedy) «*»
   Match a character from the POSIX character class “punct” «\p{Punct}{2,}»
      Between 2 and unlimited times, as many times as possible, giving back as needed (greedy) «{2,}»
Or match this alternative «[\p{Punct}&&[^-]]+»
   Match a single character present in the list below «[\p{Punct}&&[^-]]+»
      Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
      A character from the POSIX character class “punct” «\p{Punct}»
      Except the literal character “-” «&&[^-]»
Or match this alternative «(?<![a-z])\-(?![a-z])»
   Assert that it is impossible to match the regex below with the match ending at this position (negative lookbehind) «(?<![a-z])»
      Match a single character in the range between “a” and “z” «[a-z]»
   Match the character “-” literally «\-»
   Assert that it is impossible to match the regex below starting at this position (negative lookahead) «(?![a-z])»
      Match a single character in the range between “a” and “z” «[a-z]»

Example:

String text ="a-b ab--- - ---a --- , ++++ ?%# $22 43 4zzv";

String rx = "(?i)\\S*\\d+\\S*|\\p{Punct}{2,}\\S*|\\S*\\p{Punct}{2,}|[\\p{Punct}&&[^-]]+|(?<![a-z])\\-(?![a-z])";

String result = text.replaceAll(rx, " ").trim();

System.out.println(result);

Code above will print:

a-b