创建仅包含某些字符的String []

Question

I am trying to create a String[] which contains only words that comprise of certain characters. For example I have a dictionary containing a number of words like so:

arm army art as at attack attempt attention attraction authority automatic awake baby back bad bag balance

I want to narrow the list down so that it only contains words with the characters a , b and g . Therefore the list should only contain the word 'bag' in this example. Currently I am trying to do this using regexes but having never used them before I can't seem to get it to work. Here is my code:

public class LetterJugglingMain {
public static void main(String[] args) {
    String dictFile = "/Users/simonrhillary/Desktop/Dictionary(3).txt";
    fileReader fr = new fileReader();
    fr.openFile(dictFile);
    String[] dictionary = fr.fileToArray();
    String regx = "able";
    String[] newDict = createListOfValidWords(dictionary, regx);
    printArray(newDict);
}

public static String[] createListOfValidWords(String[] d, String regex){
    List<String> narrowed = new ArrayList<String>();
    for(int i = 0; i<d.length; i++){
        if(d[i].matches(regex)){
            narrowed.add(d[i]);
            System.out.println("added " + d[i]);
        }
    }
    String[] narrowArray = narrowed.toArray(new String[0]);
    return narrowArray;
}

however the array returned is always empty unless the String regex is the exact word! Any ideas? I can post more code if needed...I think I must be trying to initialise the regex wrong. The narrowed down list must contain ONLY the characters from the regex.

Answer 1

The regex able will match only the string "able" . However, if you want a regular expression to match either character of a , b , l or e , the regex you're looking for is [able] (in brackets). If you want words containing several such characters, add a + for repeating the pattern: [able]+ .

Answer 2

Frankly, I'm not an expert in regexes, but I don't think it's the best tool to do what you want. I would use a method like the following:

public boolean containsAll(String s, Set<Character> chars) {
    Set<Character> copy = new HashSet<Character>();
    for (int i = 0; i < s.length() && copy.size() < chars.size(); i++) {
        char c = s.charAt(i);
        if (chars.contains(c)) {
            copy.add(c);
        }
    }
    return copy.size() == chars.size();
}

Answer 3

The OP wants words that contain every character. Not just one of them. And other characters are not a problem.

If this is the case, I think the simiplest way would be to loop through the entire string, character by character, and check to see if it contains all of the characters you want. Keep flags to check and see if every character has been found.

If this isn't the case.... :

Try using the regex:

^[able]+$

Here's what it does:

^ matches the beginning of the string and $ matches the end of the string. This makes sure that you're not getting a partial match.

[able] matches the characters you want the string to consist of, in this case a , b , l , and e . + Makes sure that there are 1 or more of these characters in the string.

Note: This regex will match a string that contains these 4 letters. For example, it will match:

able, albe, aeble, aaaabbblllleeee

and will not match

qable, treatable, and abled.

Answer 4

A sample regex that filters out words that contains at least one occurrence of all characters in a set. This will match any English word (case-insensitive) that contains at least one occurrence of all the characters a, b, g:

(?i)(?=.*a)(?=.*b)(?=.*g)[a-z]+

Example of strings that match would be bag , baggy , grab .

Example of strings that don't match would be big , argument , nothing .

The (?i) means turns on case-insensitive flag.

You need to append as many (?=.*<character>) as the number of characters in the set, for each of the characters.

I assume a word only contains English alphabet, so I specify [az] . Specify more if you need space, hyphen, etc.

I assume matches(String regex) method in String class, so I omitted the ^ and $ .

The performance may be bad , since in the worst case (the characters are found at the end of the words), I think that the regex engine may go through the string for around n times where n is the number of characters in the set. It may not be an actual concern at all, since the words are very short, but if it turns out that this is a bottleneck, you may consider doing simple looping.