Removing numbers and characters from a string using pattern

Question

So I have a list of words, like 50,000 of them, and I want to remove certain numbers and letters from them. Specifically, I want to remove anything that has a number from 0-99 followed by either an E or Z, so for example: 4E, 11Z, 11E, 20Z, etc

The words that I want to remove them from look like this:-

6S,9,12S-trimethyl-2E,4E,8E,10E-tetradecatetraenoic acid
7Z,14Z-eicosadienoic acid
13,17,21,25-tetramethyl-5Z-hexacosenoic acid
CDP-DG(18:1(11Z)/22:6(4Z,7Z,10Z,13Z,16Z,19Z))
PC(20:4(5Z,8Z,11Z,14Z)/17:2(9Z,12Z))

As you can see the thing I want to remove appears in different ways in the words (as in within a bracket or after a hyphen etc). So far, I've done:

public class EZConfig {

    public static void main(String[] args) throws IOException{

     BufferedReader br = new BufferedReader(new FileReader("C:/Users/colles-a-l-kxc127/Dropbox/PhD/Java/MetabolitesCompiled/src/commonNames"));

        try {

            StringBuilder sb = new StringBuilder();
            String line = br.readLine();

            while (line != null) {

                if(line.contains("[0-99][E|Z]")){

                    System.out.println(line + " TRUE");
                }
                else{
                    System.out.println(line);
                }

                line = br.readLine();
            }

        } finally {
            br.close();
        }
    }
}

Just to see if I can pick up the number/E or Z annotations but I can't seem. I need to basically script something that will remove all those annotations from my list of words. Anyone know what I can do in order to achieve this?

As a side note -- [0-99] doesn't match any number between 0 and 99. It matches any digit, then 9, if I recall correctly, but the syntax you're looking for is [0-9]+, which will match one or more digits in a row. — anon
– anon, Commented Dec 18, 2014 at 13:32

Community · Accepted Answer · 2017-05-23 12:28:27Z

3

You cannot pass a regular expression to String.contains - or rather, it will be treated as literal.

I would use this draft solution:

// declare as constant somewhere
static final Pattern MY_PATTERN = Pattern.compile("\\d+[EZ]");

Then, instead of your if(line.contains("[0-99][E|Z]")){ statement, you can use:

if (MY_PATTERN.matcher(line).find()) {

On the long run, if you're removing that from your words, you probably want to use:

line = line.replaceAll("\\d+[EZ]", "");

Edit

As newbiedoodle mentions (hadn't noticed), the character class [0-99] will not match a range between 0 and 99.

If you need to limit your digits to < 100, you can use \\d{1,2} instead of the more generic \\d+.

Notes

To remove [optional] parenthesis surrounding the pattern, an optional hyphen starting it and an optional comma ending it as well, you can use the following idiom: "-?\\(?\\d+[EZ]\\)?,?".

Note that parenthesis need to be double escaped in this context.

edited May 23, 2017 at 12:28

CommunityBot

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answered Dec 18, 2014 at 13:27

Mena

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1 Comment

user2062207 Over a year ago

Ah that's it! Can't accept your answers until two minutes for some reason but yeah that's perfect. Just out of curiosity, say if I also wanted to remove any brackets, hyphens or commas associated with it, so for example (11E) or -11Z,9E- etc, how could I incorporate this into the pattern or the .replaceAll method?

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