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
– Mena, Commented Mar 12, 2019 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
– Hana, Commented Mar 12, 2019 at 14:35
It doesn't seem like you have actual commas in your output array, only empty strings. — RealSkeptic
– RealSkeptic, Commented Mar 12, 2019 at 14:36
Also, wordArray is an array, so you can't possibly have used replaceAll on it - it wouldn't have compiled. — RealSkeptic
– RealSkeptic, Commented Mar 12, 2019 at 14:38

Youcef LAIDANI · Accepted Answer · 2019-03-12 14:51:35Z

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, 2019 at 14:51

answered Mar 12, 2019 at 14:37

Youcef LAIDANI

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3 Comments

Hana Over a year ago

I'm using Java 11

Youcef LAIDANI Over a year ago

@Hana what work in Java 8 work in Java 11 except depricated, so my code work in both of them

Hana Over a year ago

THANK YOU! This one worked for me :) the second result you posted was what I was looking for

Andronicus · Accepted Answer · 2019-03-12 14:40:06Z

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, 2019 at 14:40

Andronicus

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Comments

Optional · Accepted Answer · 2019-03-12 14:46:15Z

1

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)

answered Mar 12, 2019 at 14:46

Optional

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