Java Search an arraylist of strings for words that are in another arraylist

Question

As the title says, I have an arraylist of over 10000 words and I wish to use another arraylist of words where I chose the words to search for. Cant explain it more simple than that but there's the code you may catch on to what I am trying to achieve. Basically, one arraylist of a lot of words then another arraylist of 5 or so words that will check if those words appear in the long arraylist of words.

//TO DO: Profile the search method

    try {
        Scanner input = new Scanner(new File("textFile.txt"));
        int reps = 100;


        List<String> list = new ArrayList();

         List<String> searchValues = new ArrayList();
         searchValues.add("You");
         searchValues.add("and");
         searchValues.add("So");
         searchValues.add("we");
         searchValues.add("important");

        while (input.hasNext()) {
            list.add(input.next());
        }
        input.close();

        System.out.println("Amount of words in a .txt file: " + list.size());


        //Start to time the method
        long start = System.currentTimeMillis();
        for (int i = 0; i < reps; i++) {

            for (int j = 0; j < list.size(); j++) {
                //List value = index.search(list.get(j));
                List value = index.search(list.get(j));


            }
        }
        long end = System.currentTimeMillis();
        System.out.println("Time Taken: " + (end - start) + "ms");


    } catch (IOException exc) {
        System.out.println("File does not exist");
        exc.printStackTrace();
        System.exit(1);
    }

Will take a look now. Thank you.

Aleksandar Zoric
– Aleksandar Zoric

2015-11-13 14:00:59 +00:00
Commented Nov 13, 2015 at 14:00 — Aleksandar Zoric
– Aleksandar Zoric, Commented Nov 13, 2015 at 14:00

Andy Turner · Accepted Answer · 2015-11-13 14:00:55Z

3

list.retainAll(searchValues)

will remove all of the values from list which are not in searchValues. (Javadoc)

answered Nov 13, 2015 at 14:00

Andy Turner

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Comments

pandaadb · Accepted Answer · 2015-11-13 14:12:11Z

1

retainAll internally calls contains() which will iterate over the second list, so you get O(n x m) time complexity searching.

You could search for words by using a hashmap, inserting all entries and finding the ones which occured more than once. This should give you 2 x O(n) + O(m).

However, that would likely not matter if your second list only has 5 objects.

answered Nov 13, 2015 at 14:12

pandaadb

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Mohammed Aouf Zouag · Accepted Answer · 2015-11-13 14:10:34Z

0

You can also use :

list.stream().filter(s -> ! searchValues.contains(s));

This will return a stream consisting of the elements that don't exist in the searchValues list (considering that you are working with Java 8).

edited Nov 13, 2015 at 14:10

answered Nov 13, 2015 at 14:05

Mohammed Aouf Zouag

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