Java equivalent of Python dictionary

Question

Is there a cleaner way for doing this ?

public void change(char x)
{
    if(x == 'a')
    {
       return 'b';
    }
    else if(x == 'b')
    {
       return 'a';
    }
    else if(x == 't')
    {
       return 'r';
    }
    else if(x == 'r')
    {
       return 't';
    }
    return 'z';
}

I was wondering if there was anything built into Java so that I can swap pairs of characters like a dictionary in Python or is this the best way to do this?

That looks quite simple and readable to me. Shorter is not always better. — JB Nizet, Feb 21 '14 at 22:31
@Rp-, using a Map for this translation would be overkill. Using a two dimensional array might make the code shorter, though. — Timo Geusch, Feb 21 '14 at 22:32
since your case is only about 4 chars, then what you have is the best it can get, just use `switch` instead of`if-else-if` style as mentioned in the answers. — Moha the almighty camel, Feb 21 '14 at 22:54
very very easy to read :-) return c=='a'?'b':c=='b'?'a':c=='t'?'r':c=='r'?'t':'z'; — Leo, Feb 21 '14 at 23:01
Possible duplicate of http://stackoverflow.com/questions/1540673/java-equivalent-to-python-dictionaries — Raedwald, Feb 24 '14 at 08:36

Johannes H. · Answer 1 · 2014-02-21T22:37:06.923

12

Well, it's not making it much shorter, but at least nicer: you could use the switch statement instead of your if-elseif-elseif-... construct.

public char change(char x)
{
    switch (x) {
        case 'a':
            return 'b';
        case 'b':
            return 'a';
        case 't':
            return 'r';
        case 'r':
            return 't';
        default:
            return 'z';
    }
}

edited Feb 21 '14 at 22:37

answered Feb 21 '14 at 22:32

Johannes H.

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what if you had so many pairs ? what if there were new pairs dynamically created at run time ? – Moha the almighty camel Feb 21 '14 at 22:34
1

"like dictionary in python" How is this like a Python Dictionary? – jeremyjjbrown Feb 21 '14 at 22:35
2

you need a `char` return type there :) – Reimeus Feb 21 '14 at 22:35
@Mhd.Tahawi No such requirement is specified in the quesiton. If those are true, one should consider jeremyjjbrown's solution. – Johannes H. Feb 21 '14 at 22:35
I cut "like dictionary in python" out of the question. – jeremyjjbrown Feb 21 '14 at 22:36
@jeremyjjbrown It's not linke one, I agree. But for such simple task, I just don't consider using a Map a solution I'd recommend. – Johannes H. Feb 21 '14 at 22:36
chains of if / else if tend to obscure what is being done and require longer time for the reader to comprehend what is being done ... and result in sets if jumps that often tend to be longer. – ErstwhileIII Feb 21 '14 at 23:01

score 12 · Accepted Answer · edited Feb 21 '14 at 23:26

12

One might try the following:

public char change(char x) {
    final String dictionary = "abtr";
    final String transform = "bart";

    int loc= dictionary.indexOf(x);
    if (loc < 0) {
        return 'z';
    } else {
        return transform.charAt(loc);
    }
}

edited Feb 21 '14 at 23:26

Mike-O

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answered Feb 21 '14 at 22:55

ErstwhileIII

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The "default" value is `"z"` for the author, not `x` itself. I have edited your answer. – Maxime Lorant Feb 21 '14 at 22:57
Nice one! I really like that one, it's short, readable and has little overhead. The only drawback is, that Java has to loop over both strings, which might be a performance issue for large replacement maps. – Johannes H. Feb 21 '14 at 22:57
Well a loop over dictionary, but calculated position for transform. – ErstwhileIII Feb 21 '14 at 22:59
Erm... right. So loop only over one string, agreed ;) – Johannes H. Feb 21 '14 at 23:04

jeremyjjbrown · Answer 3 · 2014-02-21T22:38:46.670

7

"so that i can swap pairs of characters like dictionary in python"

Just use a Map. It's just like a Python Dictionary.

Map<Character, Character> likeADict = new HashMap<Character, Character>();
likeADict.put('a','b');
likeADict.get('a');

You can add and remove values from it at runtime.

edited Feb 21 '14 at 22:38

answered Feb 21 '14 at 22:32

jeremyjjbrown

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I consider this an overkill. A `HashMap` has quite some overhead, it's not a particulary small structure... for something that simple, I wouldn't wast the memory and lookup time. BUt, agreed: It's the most flexible solution one can do. – Johannes H. Feb 21 '14 at 22:34
1

He still needs to check for `null` as return value to return the default (in his case `z`) – Lord M-Cube Feb 21 '14 at 22:35
@Mhd.Tahawi No, once is enough. Although the put is idempotent so he could put each pair x times. – jeremyjjbrown Feb 21 '14 at 22:41
@jeremyjjbrown, the pairs he has in his question are symmetric. if `a` swaps with `b` then `b` should swap with `a`. so I guess he need to insert the pair and its inverse to work , right ? in order for this to function as a two way map. or no ? – Moha the almighty camel Feb 21 '14 at 22:50
@Mhd.Tahawi Sure that's true for any of these solutions. – jeremyjjbrown Feb 21 '14 at 22:51

knightsb · Answer 4 · 2014-02-22T01:41:10.630

2

another cool way is using enum, that way adding new swap value is easy, just add a mapping into the enum

public enum swap{
    a('b'),
    b('a'),
    t('r'),     
    r('t'),
    z('z');

    private swap(char c){
        this.value=c;
    }
    public final char value;
}

and the use is:

public char change(char x)
{
   swap swapObject = swap.valueOf(Character.toString(x));
   return swapObject.value;
}

edited Feb 22 '14 at 01:41

answered Feb 21 '14 at 22:59

knightsb

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This has numerous errors. (1) Does not compile. `swap.valueOf(x)` should be `swap.valueOf(Character.toString(x))`. (2) `change` needs to return something. (3) The OP's code changed _all_ characters other than a, b, t, r, to z, and this code doesn't do that. In fact, it doesn't work at all unless `x` is one of those five. (4) This approach only works when all the characters to be swapped are letters. – ajb Feb 21 '14 at 23:46
i have changed the code, thanks.i do agree that in case we don't know all the cases we deal with or if we want a default value this approach is not good enough but for this scenario it seems sufficient. – knightsb Feb 22 '14 at 01:46

score 0 · Answer 5 · answered Feb 21 '14 at 22:42

You could use a Map, E.G:

public char change(char x) {
    Map <Character, Character> dictionary = new HashMap<Character, Character>() {{
        put('a', 'b');
        put('b', 'a');
        put('t', 'r');
        put('r', 't');
    }};

    if (dictionary.containsKey(x)) {
        return dictionary.get(x);
    } else {
        return 'z';
    }
}

However I wouldn't bother if its just a few characters. And I would probably make the dictionary a class field.

score -1 · Answer 6 · answered Feb 21 '14 at 22:43

-1

The dictionary technic is more flexible but not shorter (like every thing in java). Here is a complete example.

package example;

import java.util.TreeMap;

public class Dictionary {

    static final TreeMap<Character,Character> myDict = new TreeMap();

    static Character choose(TreeMap<Character,Character> dict, 
                            Character x, Character def) {
        Character result = dict.get(x);
        if (result==null) result = def;
        return result;
    }

    public static void main(String[] args) {
        myDict.put('a','b');
        myDict.put('b','a');
        myDict.put('r','t');
        System.out.println(choose(myDict,'a','z'));
        System.out.println(choose(myDict,'b','z'));
        System.out.println(choose(myDict,'r','z'));
        System.out.println(choose(myDict,'X','z'));
    }
}

answered Feb 21 '14 at 22:43

huckfinn

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Why use a TreeMap, which is O(log n), instead of a HashMap, which is O(1), when you don't care about the order of the keys? Also, it should be `new TreeMap<>()`. – JB Nizet Feb 21 '14 at 22:45
I like TreeMap .. yes HashMap is also OK.. and new TreeMap<> is possible but not nessesary OR? – huckfinn Feb 21 '14 at 22:50
4

@huckfinn "I like" is not a particular good reason for a bad design choice... You do realize that different `Map` (as well as different `List`) implementations have different characteristics for a reason? – Johannes H. Feb 21 '14 at 22:52
You "like" it? Any good reason? The diamond avoids using raw types and getting unchecked warnings from the compiler. – JB Nizet Feb 21 '14 at 22:52
Please stop pitching pennies, dimonds and map types.. Is the solution OK? – huckfinn Feb 21 '14 at 23:00

Java equivalent of Python dictionary

6 Answers6