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!

random words based on what source? (e.g. random numbers + ???? = random words) — Jason S, Feb 10 '11 at 00:09

David Yaw · Accepted Answer · 2011-02-10T01:01:39.363

25

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;
}

edited Feb 10 '11 at 01:01

answered Feb 10 '11 at 00:17

David Yaw

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Thanks! One thing you need to to is cast 'a' to char or else the compiler freaks out at you. Thanks everyone! – Mr_CryptoPrime Feb 10 '11 at 00:27
1

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));` – Paŭlo Ebermann Feb 10 '11 at 00:36
Thanks for the correction, it's been a while since I worked with raw characters in Java. – David Yaw Feb 10 '11 at 01:02
Haha, no problem...it took like 10secs to fix. – Mr_CryptoPrime Feb 10 '11 at 06:47
Random letters aren't words – Rafael Pizao Mar 06 '22 at 03:34

score 12 · Answer 2 · edited Aug 29 '15 at 18:31

12

RandomStringUtils from commons-lang

edited Aug 29 '15 at 18:31

Damian

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answered Feb 10 '11 at 00:13

Aravind Yarram

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New link up, no longer dead. – Chris - Jr Jun 16 '16 at 17:02

Yanick Rochon · Answer 3 · 2018-07-20T01:12:10.470

5

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

edited Jul 20 '18 at 01:12

answered Feb 10 '11 at 00:11

Yanick Rochon

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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... – Mr_CryptoPrime Feb 10 '11 at 00:14
3

you should have specified that, and tagged your question as "homework" – Yanick Rochon Feb 10 '11 at 00:19
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. – Mr_CryptoPrime Feb 10 '11 at 06:46

templatetypedef · Answer 4 · 2020-06-20T23:08:45.580

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.

Argote · Answer 5 · 2011-02-10T00:56:06.323

2

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();
}

edited Feb 10 '11 at 00:56

answered Feb 10 '11 at 00:13

Argote

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+1 but I would suggest using `StringBuilder` instead of doing `out += ...` – casablanca Feb 10 '11 at 00:15
@casablanca Wouldn't the compiler optimize his algorithm to use StringBuilder? – Kin Feb 10 '11 at 00:18
@Kin I'm pretty sure it wouldn't get optimized. – whiskeysierra Feb 10 '11 at 00:21
@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 http://stackoverflow.com/questions/1532461/stringbuilder-vs-string-concatenation-in-tostring-in-java – Kin Feb 10 '11 at 00:29
1

@Kin 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. – Paŭlo Ebermann Feb 10 '11 at 00:42

score 1 · Answer 6 · answered Feb 10 '11 at 00:13

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

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

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

Generating random words in Java?

6 Answers6

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