Regex in Java - parsing a string array

Question

I have a string array like this:

    String tweetString = ExudeData.getInstance().filterStoppingsKeepDuplicates(tweets.text);
    // get array of words and split
    String[] wordArray = tweetString.split(" ");

After I split the array, I print the following:

System.out.println(Arrays.toString(wordArray));

And the output I get is:

[new, single, fallin, dropping, days, artwork, hueshq, production, iseedaviddrums, amp, bigearl7, mix, reallygoldsmith, https, , , t, co, dk5xl4cicm, https, , , t, co, rvqkum0dk7]

What I want is to remove all the instances of commas, https, and single letters like 't' (after using split method above). So I want to end up with this:

[new, single, fallin, dropping, days, artwork, hueshq, production, iseedaviddrums, amp, bigearl7, mix, reallygoldsmith, co, dk5xl4cicm, https, co, rvqkum0dk7]

I've tried doing replaceAll like this:

String sanitizedString = wordArray.replaceAll("\\s+", " ").replaceAll(",+", ",");

But that just gave me the same initial output with no changes. Any ideas?

Easier to answer if you provide the original input. Note that `split` itself takes a regex. You might want to start with that. — Mena, Mar 12 '19 at 14:31
@Mena - yes I want to parse out the strings I mentioned above after I used the split method if you read my actual question — Hana, Mar 12 '19 at 14:35
It doesn't seem like you have actual commas in your output array, only empty strings. — RealSkeptic, Mar 12 '19 at 14:36
Also, `wordArray` is an array, so you can't possibly have used `replaceAll` on it - it wouldn't have compiled. — RealSkeptic, Mar 12 '19 at 14:38
Use this to first replace/remove all multiple matching words with "" https://stackoverflow.com/questions/1326682/java-replacing-multiple-different-substring-in-a-string-at-once-or-in-the-most and then split — Optional, Mar 12 '19 at 14:42

Youcef LAIDANI · Accepted Answer · 2019-03-12T14:51:35.973

2

If you are using Java 8

String[] result = Arrays.stream(tweetString.split("\\s+"))
            .filter(s -> !s.isEmpty())
            .toArray(String[]::new);

What I want is to remove all the instances of commas, https, and single letters like 't'

In this case you can make multiple filters like @Andronicus do or with matches and some regex like so :

String[] result = Arrays.stream(tweetString.split("\\s+"))
            .filter(s -> !s.matches("https|.|\\s+"))
            .toArray(String[]::new);

edited Mar 12 '19 at 14:51

answered Mar 12 '19 at 14:37

Youcef LAIDANI

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I'm using Java 11 – Hana Mar 12 '19 at 14:39
@Hana what work in Java 8 work in Java 11 except depricated, so my code work in both of them – Youcef LAIDANI Mar 12 '19 at 14:41
1

THANK YOU! This one worked for me :) the second result you posted was what I was looking for – Hana Mar 12 '19 at 14:59

score 1 · Answer 2 · answered Mar 12 '19 at 14:40

1

You can do something like this:

String[] filtered = Arrays
    .stream(tweetString.split("[ ,]"))
    .filter(str -> str.length() > 1)
    .filter(str -> !str.equals("http"))

answered Mar 12 '19 at 14:40

Andronicus

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score 1 · Answer 3 · answered Mar 12 '19 at 14:46

Based on my comment here is quick solution. (Enhance the regex with all your keywords)

 private static void replaceFromRegex(final String text ) {
    String result = text.replaceAll("https($|\\s)| (?<!\\S)[^ ](?!\\S)","");
      System.out.println(result);
  }

and then test

  public static void main(String []args) throws Exception{
      replaceFromRegex("new single fallin dropping, , https");
     }

Note: This is just sample and you will have to enhance regex to consider starting word (e.g string starting with https and then space, etc)

Regex in Java - parsing a string array

3 Answers3