What is an efficient way to replace many characters in a string?

Question

String handling in Java is something I\'m trying to learn to do well. Currently I want to take in a string and replace any characters I find.

Here is my current inef

Answer 1

What i see being inefficient is that you are gonna check again characters that have already been replaced, which is useless.

I would get the charArray of the String instance, iterate over it, and for each character spam a series of if-else like this:

char[] array = word.toCharArray();
for(int i=0; i<array.length; ++i){
    char currentChar = array[i];
    if(currentChar.equals('é'))
        array[i] = 'e';
    else if(currentChar.equals('ö'))
        array[i] = 'o';
    else if(//...
}

Answer 2

I doubt, that you can speed up the 'character replacement' at all really. As for the case of regular expression replacement, you may compile the regexs beforehand

Answer 3

I just implemented this utility class that replaces a char or a group of chars of a String. It is equivalent to bash tr and perl tr///, aka, transliterate. I hope it helps someone!

package your.package.name;

/**
 * Utility class that replaces chars of a String, aka, transliterate.
 * 
 * It's equivalent to bash 'tr' and perl 'tr///'.
 *
 */
public class ReplaceChars {

    public static String replace(String string, String from, String to) {
        return new String(replace(string.toCharArray(), from.toCharArray(), to.toCharArray()));
    }

    public static char[] replace(char[] chars, char[] from, char[] to) {

        char[] output = chars.clone();
        for (int i = 0; i < output.length; i++) {
            for (int j = 0; j < from.length; j++) {
                if (output[i] == from[j]) {
                    output[i] = to[j];
                    break;
                }
            }
        }
        return output;
    }

    /**
     * For tests!
     */
    public static void main(String[] args) {

        // Example from: https://en.wikipedia.org/wiki/Caesar_cipher
        String string = "THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG";
        String from = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
        String to = "XYZABCDEFGHIJKLMNOPQRSTUVW";

        System.out.println();
        System.out.println("Cesar cypher: " + string);
        System.out.println("Result:       " + ReplaceChars.replace(string, from, to));
    }
}

This is the output:

Cesar cypher: THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG
Result:       QEB NRFZH YOLTK CLU GRJMP LSBO QEB IXWV ALD

Answer 4

You could create a table of String[] which is Character.MAX_VALUE in length. (Including the mapping to lower case)

As the replacements got more complex, the time to perform them would remain the same.

private static final String[] REPLACEMENT = new String[Character.MAX_VALUE+1];
static {
    for(int i=Character.MIN_VALUE;i<=Character.MAX_VALUE;i++)
        REPLACEMENT[i] = Character.toString(Character.toLowerCase((char) i));
    // substitute
    REPLACEMENT['á'] =  "a";
    // remove
    REPLACEMENT['-'] =  "";
    // expand
    REPLACEMENT['æ'] = "ae";
}

public String convertWord(String word) {
    StringBuilder sb = new StringBuilder(word.length());
    for(int i=0;i<word.length();i++)
        sb.append(REPLACEMENT[word.charAt(i)]);
    return sb.toString();
} 

Answer 5

My suggestion would be:

• Convert the String to a char[] array
• Run through the array, testing each character one by one (e.g. with a switch statement) and replacing it if needed
• Convert the char[] array back to a String

I think this is probably the fastest performance you will get in pure Java.

EDIT: I notice you are doing some changes that change the length of the string. In this case, the same principle applies, however you need to keep two arrays and increment both a source index and a destination index separately. You might also need to resize the destination array if you run out of target space (i.e. reallocate a larger array and arraycopy the existing destination array into it)

Answer 6

My implementation is based on look up table.

public static String convertWord(String str) {
    char[] words = str.toCharArray();
    char[] find = {'á','é','ú','ý','ð','ó','ö','æ','þ','-','.',
            '/'};
    String[] replace = {"a","e","u","y","d","o","o","ae","th"};
    StringBuilder out = new StringBuilder(str.length());
    for (int i = 0; i < words.length; i++) {
        boolean matchFailed = true;
        for(int w = 0; w < find.length; w++) {
            if(words[i] == find[w]) {
                if(w < replace.length) {
                    out.append(replace[w]);
                }
                matchFailed = false;
                break;
            }
        }
        if(matchFailed) out.append(words[i]);
    }
    return out.toString();
}