Get a Character Values from a Map and add all to a String

Question

I was wondering how i can get these values and put it together in a new String

class Rna {


    static Map<Character, Character> map = (Map<Character, Character>) Map.of(
            'G', 'C',
            'C', 'G',
            'T', 'A',
            'A', 'U'
    );

    String transcribe(String dnaStrand) {
        return dnaStrand.chars()
                .map(c -> map.get(c))
                .collect(StringBuilder::new, StringBuilder::appendCodePoint ,StringBuilder::append)
                .toString();
    }
}

transcribe should return an String like "AUGC". Now i`m getting this exception.

Answer 1

I think the problem is that you use the String.chars() method which returns a stream of ints but you cannot expect autoboxing of an int to a Character.

The map doesn't contain Integer keys, so always will return null for the get.

As a solution you could try to cast c to a char or even create a Character instance manually:

.map(c -> map.get((char)c)) or .map(c -> map.get(Character.valueof((char)c))

Answer 2

char obsolete

The char primitive type, and its wrapper class Character, have been legacy since Java 2. As a 16-bit value, these types are physically incapable of representing most characters.

Avoid char/Character types.

Code points

Instead, use Unicode code point integer numbers to work with individual characters.

Your code is mixing char values with code point code. Do not mix the types. Avoid char, use only code points.

Unfortunately, Java lacks a type for code points. So we use the int/Integer with several methods awkwardly spread out across various classes including String, StringBuilder, and Character.

Map< Integer , Integer > transMap = 
    Map.of(
        "G".codePointAt(0) , "C".codePointAt(0) ,
        "C".codePointAt(0) , "G".codePointAt(0) ,
        "T".codePointAt(0) , "A".codePointAt(0) ,
        "A".codePointAt(0) , "U".codePointAt(0) 
    )
;

The transcription method swaps integers. Then collect these integers using StringBuilder#appendCodePoints.

By the way, your transcribe method could be made more general by taking an parameter of type CharSequence rather than String. CharSequence is an interface implemented by String. The only method we need to call on the argument is codePoints. That method is required by CharSequence.

String transcribe( final CharSequence dnaStrand ) {
    return 
        dnaStrand
        .codePoints()
        .map( codePoint -> transMap.get( codePoint ) )
        .collect(
            StringBuilder :: new, 
            StringBuilder :: appendCodePoint, 
            StringBuilder :: append 
        )
        .toString() ;
}

Answer 3

Here is one way to do it. The method is called with a string of keys from your base/pair map. Any characters in that string map to null are filtered out. This presumes a map instance visible to the method. Otherwise it will need to be provided as an argument.

• chars() provides a stream of characters
• get the mapping for current character and convert to String
• filter out any null values.
• join String characters into long strand and return.

static String transcribe(String dnaStrand) {
    return dnaStrand.chars()
            .mapToObj(c -> String.valueOf(map.get((char)c)))
            .filter(str->!str.equals("null"))
            .collect(Collectors.joining());
}

System.out.println(transcribe("GCTA"));

will return and print

CGAU

Answer 4

Method get() of the Map interface accepts an argument of Object type as a parameter. Hence you can pass to it an instance of pretty much any type. Not necessarily of the type that matches the type the key. In such a case map will return null.

And that's basically what you are doing. Method chars() of the String class returns IntStream a stream primitive int values (note there is NO such thing as CharStream in java). Attempt to get a value from a Map<Character, Character> result in null because the map will look for the object of type Integer, and obviously, there will be no match.

Instead, you have to cast int to char explicitly and it'll be autoboxed into Character wrapper type.

Also, note that for any given string that contains any other character apart from uppercase 'G', 'C', 'T', 'A' your code will fail with NullPointerException.

If a fail-fast implementation wasn't your intent as a workaround you can use map.getOrDefault() instead of map.get().

Method getOrDefault() returns a default value instead of null for any key that isn't represented in the map.

For instance, using getOrDefault() you can preserve only characters that are contained in the map and eliminate all others by providing a default value that will be removed from the final string (like that getOrDefault((char) c, '!')), and then apply replace("!", "") to the result).

Another possible option that can be achieved with getOrDefault()is to change only those characters in the string that are contained in the map whilst others will remain intact as it's shown in the example below.

    public String transcribe(String dnaStrand) {
        return dnaStrand.chars()
                .map(c -> map.getOrDefault((char) c, (char) c))
                .collect(StringBuilder::new, 
                        StringBuilder::appendCodePoint, 
                        StringBuilder::append)
                .toString();
    }

Answer 5

If you want to obtain map value, you can not have a String like "AUGC", may be "UAGC"? And you can read more about Collection, Map and LinkedHashMap. If you will try this code, all will be work.

public class Main {
    public static void main(String[] args) {


        Map<Character, Character> map = new LinkedHashMap();
            map.put('G', 'C');
        map.put( 'C', 'G');
                map.put( 'T', 'A');
                        map.put( 'A', 'U');


Test test = new Test();
        System.out.println(test.transcribe(map));
    }
}
public class Test {

    String transcribe(Map<Character, Character> map) {
        StringBuilder stringBuilder = new StringBuilder();
        for (Character ch : map.values()) {
            stringBuilder.append(ch);
        }
        return String.valueOf(stringBuilder.reverse());
    }
}