Generating random words in Java?

Question

I wrote up a program that can sort words and determine any anagrams. I want to generate an array of random strings so that I can test my method's runtime.

public static String[] generateRandomWords(int numberOfWords){
String[] randomStrings = new String[numberOfWords];
Random random = Random();
    return null;
}

(method stub)

I just want lowercase words of length 1-10. I read something about generating random numbers, then casting to char or something, but I didn't totally understand. If someone can show me how to generate random words, then I should easily be able to just use a for loop to insert the words into the array. Thanks!

Brimful

random words based on what source? (e.g. random numbers + ???? = random words) – Pungy

Answer 1

Do you need actual English words, or just random strings that only contain letters a-z?

If you need actual English words, the only way to do it is to use a dictionary, and select words from it at random.

If you don't need English words, then something like this will do:

public static String[] generateRandomWords(int numberOfWords)
{
    String[] randomStrings = new String[numberOfWords];
    Random random = new Random();
    for(int i = 0; i < numberOfWords; i++)
    {
        char[] word = new char[random.nextInt(8)+3]; // words of length 3 through 10. (1 and 2 letter words are boring.)
        for(int j = 0; j < word.length; j++)
        {
            word[j] = (char)('a' + random.nextInt(26));
        }
        randomStrings[i] = new String(word);
    }
    return randomStrings;
}

Gawen

Thanks! One thing you need to to is cast 'a' to char or else the compiler freaks out at you. Thanks everyone! – Brimful

No, you don't need to cast 'a' to char, but the result of the addition (since this is automatically int). So it should be in fact word[j] = (char)('a' + random.nextInt(26)); – Dermatoglyphics

Thanks for the correction, it's been a while since I worked with raw characters in Java. – Gawen

Haha, no problem...it took like 10secs to fix. – Brimful

Random letters aren't words – Prolate

Answer 2

RandomStringUtils from commons-lang

Selway

New link up, no longer dead. – Perforated

Answer 3

Why generating random words? When you can use some dictionaries.

Chamade

I am trying to get an average runtime, so randomly generated words should give me the best average. Also, this is for a school assignment, so I must conform to what my teacher wants... – Brimful

you should have specified that, and tagged your question as "homework" – Chamade

Oh, sorry I didn't even know that was a category, but I suppose that would make perfect sense. I will be sure to do that next time. – Brimful

Answer 4

If you want to generate random words of a given length, you'll either need an algorithm to determine if a given string is a word (hard), or access to a word list of all the words in a given language (easy). If it helps, here's a list of every word in the Scrabble dictionary.

Once you have a list of all words in a language, you can load those words into an ArrayList or other linear structure. You can then generate a random index into that list to get the random word.

Argentiferous

Answer 5

You can call this method for each word you want to generate. Note that the probability of generating anagrams should be relatively low though.

String generateRandomWord(int wordLength) {
    Random r = new Random(); // Intialize a Random Number Generator with SysTime as the seed
    StringBuilder sb = new StringBuilder(wordLength);
    for(int i = 0; i < wordLength; i++) { // For each letter in the word
        char tmp = 'a' + r.nextInt('z' - 'a'); // Generate a letter between a and z
        sb.append(tmp); // Add it to the String
    }
    return sb.toString();
}

Genesis

+1 but I would suggest using StringBuilder instead of doing out += ... – Brotherson

@Brotherson Wouldn't the compiler optimize his algorithm to use StringBuilder? – Rascality

@Rascality I'm pretty sure it wouldn't get optimized. – Mandamandaean

@Willi You are correct. Since the String concat is in a loop, the compiler will not optimize the loop operations. The compiler will try to optimize trivial String concats. Source #1532961 – Rascality

@Rascality U., @Willi: The compiler uses the StringBuilder, but he creates for each += a new one. This line looks after compiling like this: out = new StringBuilder().append(out).append(r.nextInt('z'-'a')+'a').toString(); – By the way, you should here convert to char again, since int + char = int, and thus your loop will append decimal numbers to the StringBuilder. – Dermatoglyphics

Answer 6

If you want random words without using a dictionary...

1. Make a list of all the letters you want possible in your words
2. Generate a random index to pick out a letter from the list
3. Repeat until you have your desired word length

Repeat these steps for the number of words you want to generate.

Rascality

Answer 7

 public static void main(final String[] args) throws IOException {
        final var dict   = getDictionary();
        final var random = ThreadLocalRandom.current();
        final var word   = dict.get(random.nextInt(0, dict.size()));
        System.out.println(word);
    }

    private static List<String> getDictionary() throws IOException {
        Path filePath = Paths.get("/usr/share/dict/words");
        return Files.readAllLines(filePath);
    }

Lanthanide

or /usr/dict/words, see en.wikipedia.org/wiki/Words_(Unix) – Lanthanide

As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – Absinthe