Is there a way to get rid of accents and convert a whole string to regular letters?

Question

Is there a better way for getting rid of accents and making those letters regular apart from using String.replaceAll() method and replacing letters one by one?

Answer 1

Depending on the language, those might not be considered accents (which change the sound of the letter), but diacritical marks

https://en.wikipedia.org/wiki/Diacritic#Languages_with_letters_containing_diacritics

"Bosnian and Croatian have the symbols č, ć, đ, š and ž, which are considered separate letters and are listed as such in dictionaries and other contexts in which words are listed according to alphabetical order."

Removing them might be inherently changing the meaning of the word, or changing the letters into completely different ones.

Answer 2

I suggest Junidecode . It will handle not only 'Ł' and 'Ø', but it also works well for transcribing from other alphabets, such as Chinese, into Latin alphabet.

Answer 3

EDIT: If you're not stuck with Java <6 and speed is not critical and/or translation table is too limiting, use answer by David. The point is to use Normalizer (introduced in Java 6) instead of translation table inside the loop.

While this is not "perfect" solution, it works well when you know the range (in our case Latin1,2), worked before Java 6 (not a real issue though) and is much faster than the most suggested version (may or may not be an issue):

    /**
 * Mirror of the unicode table from 00c0 to 017f without diacritics.
 */
private static final String tab00c0 = "AAAAAAACEEEEIIII" +
    "DNOOOOO\u00d7\u00d8UUUUYI\u00df" +
    "aaaaaaaceeeeiiii" +
    "\u00f0nooooo\u00f7\u00f8uuuuy\u00fey" +
    "AaAaAaCcCcCcCcDd" +
    "DdEeEeEeEeEeGgGg" +
    "GgGgHhHhIiIiIiIi" +
    "IiJjJjKkkLlLlLlL" +
    "lLlNnNnNnnNnOoOo" +
    "OoOoRrRrRrSsSsSs" +
    "SsTtTtTtUuUuUuUu" +
    "UuUuWwYyYZzZzZzF";

/**
 * Returns string without diacritics - 7 bit approximation.
 *
 * @param source string to convert
 * @return corresponding string without diacritics
 */
public static String removeDiacritic(String source) {
    char[] vysl = new char[source.length()];
    char one;
    for (int i = 0; i < source.length(); i++) {
        one = source.charAt(i);
        if (one >= '\u00c0' && one <= '\u017f') {
            one = tab00c0.charAt((int) one - '\u00c0');
        }
        vysl[i] = one;
    }
    return new String(vysl);
}

Tests on my HW with 32bit JDK show that this performs conversion from àèéľšťč89FDČ to aeelstc89FDC 1 million times in ~100ms while Normalizer way makes it in 3.7s (37x slower). In case your needs are around performance and you know the input range, this may be for you.

Enjoy :-)

Answer 4

I think the best solution is converting each char to HEX and replace it with another HEX. It's because there are 2 Unicode typing:

Composite Unicode
Precomposed Unicode

For example "Ồ" written by Composite Unicode is different from "Ồ" written by Precomposed Unicode. You can copy my sample chars and convert them to see the difference.

In Composite Unicode, "Ồ" is combined from 2 char: Ô (U+00d4) and ̀ (U+0300)
In Precomposed Unicode, "Ồ" is single char (U+1ED2)

I have developed this feature for some banks to convert the info before sending it to core-bank (usually don't support Unicode) and faced this issue when the end-users use multiple Unicode typing to input the data. So I think, converting to HEX and replace it is the most reliable way.

Answer 5

As of 2011 you can use Apache Commons StringUtils.stripAccents(input) (since 3.0):

    String input = StringUtils.stripAccents("Tĥïŝ ĩš â fůňķŷ Šťŕĭńġ");
    System.out.println(input);
    // Prints "This is a funky String"

Note:

The accepted answer (Erick Robertson's) doesn't work for Ø or Ł. Apache Commons 3.5 doesn't work for Ø either, but it does work for Ł. After reading the Wikipedia article for Ø, I'm not sure it should be replaced with "O": it's a separate letter in Norwegian and Danish, alphabetized after "z". It's a good example of the limitations of the "strip accents" approach.

Answer 6

I have faced the same issue related to Strings equality check, One of the comparing string has ASCII character code 128-255.

i.e., Non-breaking space - [Hex - A0] Space [Hex - 20]. To show Non-breaking space over HTML. I have used the following spacing entities. Their character and its bytes are like &emsp is very wide space[ ]{-30, -128, -125}, &ensp is somewhat wide space[ ]{-30, -128, -126}, &thinsp is narrow space[ ]{32} , Non HTML Space {}

String s1 = "My Sample Space Data", s2 = "My Sample Space Data";
System.out.format("S1: %s\n", java.util.Arrays.toString(s1.getBytes()));
System.out.format("S2: %s\n", java.util.Arrays.toString(s2.getBytes()));

Output in Bytes:

S1: [77, 121, 32, 83, 97, 109, 112, 108, 101, 32, 83, 112, 97, 99, 101, 32, 68, 97, 116, 97] S2: [77, 121, -30, -128, -125, 83, 97, 109, 112, 108, 101, -30, -128, -125, 83, 112, 97, 99, 101, -30, -128, -125, 68, 97, 116, 97]

Use below code for Different Spaces and their Byte-Codes: wiki for List_of_Unicode_characters

String spacing_entities = "very wide space,narrow space,regular space,invisible separator";
System.out.println("Space String :"+ spacing_entities);
byte[] byteArray = 
    // spacing_entities.getBytes( Charset.forName("UTF-8") );
    // Charset.forName("UTF-8").encode( s2 ).array();
    {-30, -128, -125, 44, -30, -128, -126, 44, 32, 44, -62, -96};
System.out.println("Bytes:"+ Arrays.toString( byteArray ) );
try {
    System.out.format("Bytes to String[%S] \n ", new String(byteArray, "UTF-8"));
} catch (UnsupportedEncodingException e) {
    e.printStackTrace();
}

• ➩ ASCII transliterations of Unicode string for Java. unidecode

  String initials = Unidecode.decode( s2 );
  

• ➩ using Guava: Google Core Libraries for Java.

  String replaceFrom = CharMatcher.WHITESPACE.replaceFrom( s2, " " );
  

  For URL encode for the space use Guava laibrary.

  String encodedString = UrlEscapers.urlFragmentEscaper().escape(inputString);
  

• ➩ To overcome this problem used String.replaceAll() with some RegularExpression.

  // \p{Z} or \p{Separator}: any kind of whitespace or invisible separator.
  s2 = s2.replaceAll("\\p{Zs}", " ");
  
  
  s2 = s2.replaceAll("[^\\p{ASCII}]", " ");
  s2 = s2.replaceAll(" ", " ");
  

• ➩ Using java.text.Normalizer.Form. This enum provides constants of the four Unicode normalization forms that are described in Unicode Standard Annex #15 — Unicode Normalization Forms and two methods to access them.

  s2 = Normalizer.normalize(s2, Normalizer.Form.NFKC);
  

---

Testing String and outputs on different approaches like ➩ Unidecode, Normalizer, StringUtils.

String strUni = "Tĥïŝ ĩš â fůňķŷ Šťŕĭńġ Æ,Ø,Ð,ß";

// This is a funky String AE,O,D,ss
String initials = Unidecode.decode( strUni );

// Following Produce this o/p: Tĥïŝ ĩš â fůňķŷ Šťŕĭńġ Æ,Ø,Ð,ß
String temp = Normalizer.normalize(strUni, Normalizer.Form.NFD);
Pattern pattern = Pattern.compile("\\p{InCombiningDiacriticalMarks}+");
temp = pattern.matcher(temp).replaceAll("");

String input = org.apache.commons.lang3.StringUtils.stripAccents( strUni );

---

Using Unidecode is the best choice, My final Code shown below.

public static void main(String[] args) {
    String s1 = "My Sample Space Data", s2 = "My Sample Space Data";
    String initials = Unidecode.decode( s2 );
    if( s1.equals(s2)) { //[ , ] %A0 - %2C - %20 « http://www.ascii-code.com/
        System.out.println("Equal Unicode Strings");
    } else if( s1.equals( initials ) ) {
        System.out.println("Equal Non Unicode Strings");
    } else {
        System.out.println("Not Equal");
    }

}