I have a String[] with values like so:
public static final String[] VALUES = new String[] {"AB","BC","CD","AE"};
Given String s, is there a good way of testing whether VALUES contains s?
Arrays.asList(yourArray).contains(yourValue)
Warning: this doesn't work for arrays of primitives (see the comments).
Since java-8 you can now use Streams.
String[] values = {"AB","BC","CD","AE"};
boolean contains = Arrays.stream(values).anyMatch("s"::equals);
To check whether an array of int, double or long contains a value use IntStream, DoubleStream or LongStream respectively.
Example
int[] a = {1,2,3,4};
boolean contains = IntStream.of(a).anyMatch(x -> x == 4);
Concise update for Java SE 9
Reference arrays are bad. For this case we are after a set. Since Java SE 9 we have Set.of.
private static final Set<String> VALUES = Set.of(
"AB","BC","CD","AE"
);
"Given String s, is there a good way of testing whether VALUES contains s?"
VALUES.contains(s)
O(1).
The right type, immutable, O(1) and concise. Beautiful.*
Original answer details
Just to clear the code up to start with. We have (corrected):
public static final String[] VALUES = new String[] {"AB","BC","CD","AE"};
This is a mutable static which FindBugs will tell you is very naughty. Do not modify statics and do not allow other code to do so also. At an absolute minimum, the field should be private:
private static final String[] VALUES = new String[] {"AB","BC","CD","AE"};
(Note, you can actually drop the new String[]; bit.)
Reference arrays are still bad and we want a set:
private static final Set<String> VALUES = new HashSet<String>(Arrays.asList(
new String[] {"AB","BC","CD","AE"}
));
(Paranoid people, such as myself, may feel more at ease if this was wrapped in Collections.unmodifiableSet - it could then even be made public.)
(*To be a little more on brand, the collections API is predictably still missing immutable collection types and the syntax is still far too verbose, for my tastes.)
You can use ArrayUtils.contains from Apache Commons Lang
public static boolean contains(Object[] array, Object objectToFind)
Note that this method returns false if the passed array is null.
There are also methods available for primitive arrays of all kinds.
Example:
String[] fieldsToInclude = { "id", "name", "location" };
if ( ArrayUtils.contains( fieldsToInclude, "id" ) ) {
// Do some stuff.
}
Just simply implement it by hand:
public static <T> boolean contains(final T[] array, final T v) {
for (final T e : array)
if (e == v || v != null && v.equals(e))
return true;
return false;
}
Improvement:
The v != null condition is constant inside the method. It always evaluates to the same Boolean value during the method call. So if the input array is big, it is more efficient to evaluate this condition only once, and we can use a simplified/faster condition inside the for loop based on the result. The improved contains() method:
public static <T> boolean contains2(final T[] array, final T v) {
if (v == null) {
for (final T e : array)
if (e == null)
return true;
}
else {
for (final T e : array)
if (e == v || v.equals(e))
return true;
}
return false;
}
Four Different Ways to Check If an Array Contains a Value
Using List:
public static boolean useList(String[] arr, String targetValue) {
return Arrays.asList(arr).contains(targetValue);
}
Using Set:
public static boolean useSet(String[] arr, String targetValue) {
Set<String> set = new HashSet<String>(Arrays.asList(arr));
return set.contains(targetValue);
}
Using a simple loop:
public static boolean useLoop(String[] arr, String targetValue) {
for (String s: arr) {
if (s.equals(targetValue))
return true;
}
return false;
}
Using Arrays.binarySearch():
The code below is wrong, it is listed here for completeness. binarySearch() can ONLY be used on sorted arrays. You will find the result is weird below. This is the best option when array is sorted.
public static boolean binarySearch(String[] arr, String targetValue) {
return Arrays.binarySearch(arr, targetValue) >= 0;
}
Quick Example:
String testValue="test";
String newValueNotInList="newValue";
String[] valueArray = { "this", "is", "java" , "test" };
Arrays.asList(valueArray).contains(testValue); // returns true
Arrays.asList(valueArray).contains(newValueNotInList); // returns false
If the array is not sorted, you will have to iterate over everything and make a call to equals on each.
If the array is sorted, you can do a binary search, there's one in the Arrays class.
Generally speaking, if you are going to do a lot of membership checks, you may want to store everything in a Set, not in an array.
For what it's worth I ran a test comparing the 3 suggestions for speed. I generated random integers, converted them to a String and added them to an array. I then searched for the highest possible number/string, which would be a worst case scenario for the asList().contains().
When using a 10K array size the results were:
Sort & Search : 15
Binary Search : 0
asList.contains : 0
When using a 100K array the results were:
Sort & Search : 156
Binary Search : 0
asList.contains : 32
So if the array is created in sorted order the binary search is the fastest, otherwise the asList().contains would be the way to go. If you have many searches, then it may be worthwhile to sort the array so you can use the binary search. It all depends on your application.
I would think those are the results most people would expect. Here is the test code:
import java.util.*;
public class Test {
public static void main(String args[]) {
long start = 0;
int size = 100000;
String[] strings = new String[size];
Random random = new Random();
for (int i = 0; i < size; i++)
strings[i] = "" + random.nextInt(size);
start = System.currentTimeMillis();
Arrays.sort(strings);
System.out.println(Arrays.binarySearch(strings, "" + (size - 1)));
System.out.println("Sort & Search : "
+ (System.currentTimeMillis() - start));
start = System.currentTimeMillis();
System.out.println(Arrays.binarySearch(strings, "" + (size - 1)));
System.out.println("Search : "
+ (System.currentTimeMillis() - start));
start = System.currentTimeMillis();
System.out.println(Arrays.asList(strings).contains("" + (size - 1)));
System.out.println("Contains : "
+ (System.currentTimeMillis() - start));
}
}
Instead of using the quick array initialisation syntax too, you could just initialise it as a List straight away in a similar manner using the Arrays.asList method, e.g.:
public static final List<String> STRINGS = Arrays.asList("firstString", "secondString" ...., "lastString");
Then you can do (like above):
STRINGS.contains("the string you want to find");
With Java 8 you can create a stream and check if any entries in the stream matches "s":
String[] values = {"AB","BC","CD","AE"};
boolean sInArray = Arrays.stream(values).anyMatch("s"::equals);
Or as a generic method:
public static <T> boolean arrayContains(T[] array, T value) {
return Arrays.stream(array).anyMatch(value::equals);
}
You can use the Arrays class to perform a binary search for the value. If your array is not sorted, you will have to use the sort functions in the same class to sort the array, then search through it.
ObStupidAnswer (but I think there's a lesson in here somewhere):
enum Values {
AB, BC, CD, AE
}
try {
Values.valueOf(s);
return true;
} catch (IllegalArgumentException exc) {
return false;
}
Actually, if you use HashSet<String> as Tom Hawtin proposed you don't need to worry about sorting, and your speed is the same as with binary search on a presorted array, probably even faster.
It all depends on how your code is set up, obviously, but from where I stand, the order would be:
On an unsorted array:
HashSet
asList
sort & binary
On a sorted array:
HashSet
Binary
asList
So either way, HashSet for the win.
Developers often do:
Set<String> set = new HashSet<String>(Arrays.asList(arr));
return set.contains(targetValue);
The above code works, but there is no need to convert a list to set first. Converting a list to a set requires extra time. It can as simple as:
Arrays.asList(arr).contains(targetValue);
or
for (String s : arr) {
if (s.equals(targetValue))
return true;
}
return false;
The first one is more readable than the second one.
If you have the google collections library, Tom's answer can be simplified a lot by using ImmutableSet (http://google-collections.googlecode.com/svn/trunk/javadoc/com/google/common/collect/ImmutableSet.html)
This really removes a lot of clutter from the initialization proposed
private static final Set<String> VALUES = ImmutableSet.of("AB","BC","CD","AE");
In Java 8 use Streams.
List<String> myList =
Arrays.asList("a1", "a2", "b1", "c2", "c1");
myList.stream()
.filter(s -> s.startsWith("c"))
.map(String::toUpperCase)
.sorted()
.forEach(System.out::println);
One possible solution:
import java.util.Arrays;
import java.util.List;
public class ArrayContainsElement {
public static final List<String> VALUES = Arrays.asList("AB", "BC", "CD", "AE");
public static void main(String args[]) {
if (VALUES.contains("AB")) {
System.out.println("Contains");
} else {
System.out.println("Not contains");
}
}
}
Using a simple loop is the most efficient way of doing this.
boolean useLoop(String[] arr, String targetValue) {
for(String s: arr){
if(s.equals(targetValue))
return true;
}
return false;
}
Courtesy to Programcreek
the shortest solution
the array VALUES may contain duplicates
since Java 9
List.of(VALUES).contains(s);
Use the following (the contains() method is ArrayUtils.in() in this code):
ObjectUtils.java
public class ObjectUtils {
/**
* A null safe method to detect if two objects are equal.
* #param object1
* #param object2
* #return true if either both objects are null, or equal, else returns false.
*/
public static boolean equals(Object object1, Object object2) {
return object1 == null ? object2 == null : object1.equals(object2);
}
}
ArrayUtils.java
public class ArrayUtils {
/**
* Find the index of of an object is in given array,
* starting from given inclusive index.
* #param ts Array to be searched in.
* #param t Object to be searched.
* #param start The index from where the search must start.
* #return Index of the given object in the array if it is there, else -1.
*/
public static <T> int indexOf(final T[] ts, final T t, int start) {
for (int i = start; i < ts.length; ++i)
if (ObjectUtils.equals(ts[i], t))
return i;
return -1;
}
/**
* Find the index of of an object is in given array, starting from 0;
* #param ts Array to be searched in.
* #param t Object to be searched.
* #return indexOf(ts, t, 0)
*/
public static <T> int indexOf(final T[] ts, final T t) {
return indexOf(ts, t, 0);
}
/**
* Detect if the given object is in the given array.
* #param ts Array to be searched in.
* #param t Object to be searched.
* #return If indexOf(ts, t) is greater than -1.
*/
public static <T> boolean in(final T[] ts, final T t) {
return indexOf(ts, t) > -1;
}
}
As you can see in the code above, that there are other utility methods ObjectUtils.equals() and ArrayUtils.indexOf(), that were used at other places as well.
For arrays of limited length use the following (as given by camickr). This is slow for repeated checks, especially for longer arrays (linear search).
Arrays.asList(...).contains(...)
For fast performance if you repeatedly check against a larger set of elements
An array is the wrong structure. Use a TreeSet and add each element to it. It sorts elements and has a fast exist() method (binary search).
If the elements implement Comparable & you want the TreeSet sorted accordingly:
ElementClass.compareTo() method must be compatable with ElementClass.equals(): see Triads not showing up to fight? (Java Set missing an item)
TreeSet myElements = new TreeSet();
// Do this for each element (implementing *Comparable*)
myElements.add(nextElement);
// *Alternatively*, if an array is forceably provided from other code:
myElements.addAll(Arrays.asList(myArray));
Otherwise, use your own Comparator:
class MyComparator implements Comparator<ElementClass> {
int compareTo(ElementClass element1; ElementClass element2) {
// Your comparison of elements
// Should be consistent with object equality
}
boolean equals(Object otherComparator) {
// Your equality of comparators
}
}
// construct TreeSet with the comparator
TreeSet myElements = new TreeSet(new MyComparator());
// Do this for each element (implementing *Comparable*)
myElements.add(nextElement);
The payoff: check existence of some element:
// Fast binary search through sorted elements (performance ~ log(size)):
boolean containsElement = myElements.exists(someElement);
If you don't want it to be case sensitive
Arrays.stream(VALUES).anyMatch(s::equalsIgnoreCase);
Try this:
ArrayList<Integer> arrlist = new ArrayList<Integer>(8);
// use add() method to add elements in the list
arrlist.add(20);
arrlist.add(25);
arrlist.add(10);
arrlist.add(15);
boolean retval = arrlist.contains(10);
if (retval == true) {
System.out.println("10 is contained in the list");
}
else {
System.out.println("10 is not contained in the list");
}
Check this
String[] VALUES = new String[]{"AB", "BC", "CD", "AE"};
String s;
for (int i = 0; i < VALUES.length; i++) {
if (VALUES[i].equals(s)) {
// do your stuff
} else {
//do your stuff
}
}
Arrays.asList() -> then calling the contains() method will always work, but a search algorithm is much better since you don't need to create a lightweight list wrapper around the array, which is what Arrays.asList() does.
public boolean findString(String[] strings, String desired){
for (String str : strings){
if (desired.equals(str)) {
return true;
}
}
return false; //if we get here… there is no desired String, return false.
}
Use below -
String[] values = {"AB","BC","CD","AE"};
String s = "A";
boolean contains = Arrays.stream(values).anyMatch(v -> v.contains(s));
Use Array.BinarySearch(array,obj) for finding the given object in array or not.
Example:
if (Array.BinarySearch(str, i) > -1)` → true --exists
false --not exists
Try using Java 8 predicate test method
Here is a full example of it.
import java.util.Arrays;
import java.util.List;
import java.util.function.Predicate;
public class Test {
public static final List<String> VALUES =
Arrays.asList("AA", "AB", "BC", "CD", "AE");
public static void main(String args[]) {
Predicate<String> containsLetterA = VALUES -> VALUES.contains("AB");
for (String i : VALUES) {
System.out.println(containsLetterA.test(i));
}
}
}
http://mytechnologythought.blogspot.com/2019/10/java-8-predicate-test-method-example.html
https://github.com/VipulGulhane1/java8/blob/master/Test.java
Create a boolean initially set to false. Run a loop to check every value in the array and compare to the value you are checking against. If you ever get a match, set boolean to true and stop the looping. Then assert that the boolean is true.
As I'm dealing with low level Java using primitive types byte and byte[], the best so far I got is from bytes-java https://github.com/patrickfav/bytes-java seems a fine piece of work
You can check it by two methods
A) By converting the array into string and then check the required string by .contains method
String a = Arrays.toString(VALUES);
System.out.println(a.contains("AB"));
System.out.println(a.contains("BC"));
System.out.println(a.contains("CD"));
System.out.println(a.contains("AE"));
B) This is a more efficent method
Scanner s = new Scanner(System.in);
String u = s.next();
boolean d = true;
for (int i = 0; i < VAL.length; i++) {
if (VAL[i].equals(u) == d)
System.out.println(VAL[i] + " " + u + VAL[i].equals(u));
}
I am trying to make kind of highscores in Java.
Basically I want a hashmap to hold the double value (so index starts from the highest double, so it's easier for me to sort highscores) and then the second value will be the client object, like this:
private HashMap<Double, TempClient> players = new HashMap<Double, TempClient>();
And to insert a new value:
TempClient client = new TempClient(kills, rank, deaths, name);
this.players.put(client.getKdr(), client);
Now, of course I can't iterate through the hashmap because it gets the list item by key, not index.
How can I iterate through a hashmap? or any good ideas for my case?
I tried it in a Foo class:
Output:
0.5
0.6
0.9
0.1
2.5
Code:
public class Foo {
public static void main(String[] args) {
HashMap<Double, String> map = new LinkedHashMap<Double, String>();
map.put(0.5, "hey");
map.put(0.6, "hey1");
map.put(0.9, "hey2");
map.put(0.1, "hey425");
map.put(2.5, "hey36");
for (Double lol : map.keySet()) {
System.out.println(lol);
}
}
}
You can iterate like this.
for (Double k : players.keySet())
{
TempClient p = players.get(k);
// do work with k and p
}
If you want to keep keys sorted, use e.g. a TreeMap.
If you want to keep the keys in the order you inserted
them in there, use e.g. a LinkedHashMap.
The best way is to iterate through hashmap is using EntrySet.
for (Map.Entry<Double, TempClient> entry : map.entrySet()) {
Double key= entry.getKey();
TempClient value= entry.getValue();
// ...
}
You'd be better off making your TempClient objects implement Comparable, adding them to a list, and then just using Collections.sort().
Since you can't sort items in a HashMap, nor you can sort them by value in a TreeMap you could use a TreeSet with a custom class:
class Score implements Comparable<Score>
{
final Player player;
final int score;
Score(Player player, int score) {
this.player = player;
this.score = score;
}
public int compareTo(Score other) {
return Integer.compare(this.score, other.score);
}
public int hashCode() { return player.hashCode(); }
public boolean equals(Object o) { return this.player.equals(...); }
}
TreeSet<Score> scores = new TreeSet<Score>();
score.add(new Score(player, 500));
for (Score s : scores) {
..
}
This will have both the advantages:
it will be iterable
it will keep scores automatically sorted
It should work easily with consistente between equals, hashCode and compareTo but maybe you should tweak something (since it's untested code).
I have an arraylist defined whose elements are say, [man, animal, bird, reptile]. The elements in the arraylist are non-mandatory. The list can even be empty.
I always need to give the output as [animal,man,reptile,bird]. Means, the order of the elements are to be maintained. Is there any way of doing in arraylist?
I thought I can do like
for (String listElement: customList) { //custom list variable holds all elements
if (listElement.equalsIgnoreCase("animal"){
newList.add(0, listElement);
} else if("man") {
newlist.add(1, listElement);
}
But I would want to know the best practice of doing. Can someone please help me on this?
You can define a custom sorting and use it to order your array (save the comparator somewhere, so you don't have to instantiate it many times):
List<String> definedOrder = // define your custom order
Arrays.asList("animal", "man", "reptile", "bird");
Comparator<String> comparator = new Comparator<String>(){
#Override
public int compare(final String o1, final String o2){
// let your comparator look up your car's color in the custom order
return Integer.valueOf(definedOrder.indexOf(o1))
.compareTo(Integer.valueOf(definedOrder.indexOf(o2)));
}
};
Collections.sort(myList, comparator);
Use a custom comparator:
Collections.sort(customList, comparator);
int i = 0;
for (String temp : customList) {
System.out.println("customList " + ++i + " : " + temp);
}
Custom comparator below:
public static Comparator<String> comparator = new Comparator<String>() {
public int compare(String str1, String str2) {
return orderOf(str1) - orderOf(str2);
}
private int orderOf(String name) {
return ((List)Arrays.asList("animal", "man", "reptile", "bird")).indexOf(name);
}
};
You can use Collections.sort(yourArrayList, new CustomComparator());
You need to create your own comparator for this, though, but it is easy.
public class CustomComparator implements Comparator<YourType>{
#Override
public int compare(YoyrType o1, YoyrType o2) {
// logic for ordering the list
}
}
arraylist is ordered.
maybe you want to insert element into the list.
could you create a new list every time when you have to insert and copy each of them?
I am told to have a list of words sorted by length, and those that have the same length are sorted by alphabetical order. This is what I have for the method that does that so far.
public static void doIt(BufferedReader r, PrintWriter w) throws IOException {
TreeMap<String, Integer> s = new TreeMap<String, Integer>();
ArrayList<Integer> count = new ArrayList<Integer>();
String line;
int length;
while ((line = r.readLine()) != null) {
length = line.length();
s.put(line, length);
if (!count.contains(length)){
count.add(length);
}
}
Collections.sort(count);
System.out.println(count);
}
My mindset was to use a TreeMap to keep the String, and the length of the word as the key. I also have an ArrayList that keeps track of all the word's lengths without any duplicates, it's then sorted.
I was hoping to somehow call on the TreeMap for the key value of 5, which would list all the words with 5 letters in it.
I was wondering if I'm on the right track? I've been playing around for over an hour and can't seem to figure out what I should do after this. Am I approaching this from the right angle?
you want to use a string comparator that compares by length 1st. like so:
public class LengthFirstComparator implements Comparator<String> {
#Override
public int compare(String o1, String o2) {
if (o1.length()!=o2.length()) {
return o1.length()-o2.length(); //overflow impossible since lengths are non-negative
}
return o1.compareTo(o2);
}
}
then you could simply sort your Strings by calling Collections.sort(yourStringList, new LengthFirstComparator());
The easiest way would be to write a Comparator<String>. The Comparator<String> would receive two words, and compare them. If the first was shorter than the second, it should return -1. If the second was shorter than the first, it would return 1. If they are the same length, it should call the default String compareTo method. You can then simply sort your list using this custom Comparator.
You can do that using simple List. Try following code.
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;
/**
*
* #author Masudul Haque
*/
public class LengthSort {
public static void main(String[] args) {
List<String> list=new ArrayList<>();
list.add("cowa");
list.add("cow");
list.add("aow");
Collections.sort(list, new Comparator<String>() {
#Override
public int compare(String o1, String o2) {
if(o1.length()>o2.length()){
return 1;
}else{
return o1.compareTo(o2);
}
}
});
System.out.println(list);
}
}
By far the easiest and best way is to write a custom comparator as other answers say.
But to do it a similar way you were attempting would be to make the length the key and rather then having a single string as the value have a list of all the words of that length. So a map of the form
Map<Integer,List<String>>
You could then call the key of any length and return a sorted list of words like this
Collections.sort(yourMap.get(theLength))
BUT far more complicated then just using a comparator
You can use Java 8's lamba utilities to make concise functions that prevent the clutter of using comparator classes, like so:
Collections.sort(words, (string1, string2) -> Integer.compare(string1.length(), string2.length());
-Example taken from Effective Java by Joshua Bloch
If you have a sentence like - Apple and grape are not vegetables and told to sort based on the length and if two or more words are equal then if it has to be sorted alphabetically then the code is as below:
public class ExampleDemo {
public static void main(String[] args) {
String s = "Apple and grape are not vegetables";
ExampleDemo e = new ExampleDemo();
e.display(s);
}
public void display(String str) {
String[] st = str.split(" ");
List<String> list = new ArrayList<>();
for(String word: st){
list.add(word);
}
System.out.println("Before sorting: " + list);
Comparator<String> comparator = (s1,s2) -> Integer.compare(s1.length(), s2.length());
Collections.sort(list, comparator);
System.out.println("After sorting: " + list);
}
}
Output:
Before sorting: [Apple, and, grape, are, not, vegetables]
After sorting: [and, are, not, Apple, grape, vegetables]
I have a list of Strings. I want to evaluate each string based on a function that returns a double. Then I want the first 5 strings, based on their calculated values. If there are fewer than 5, I want all of them (in order). Let's say the strings are chemical compounds and the function computes the mass. The function is computationally expensive; I need to evaluate it once per string. (I'm just making up data here, though.)
H2O => 18.5
C12H11O22 => 109.1
HeNe => 32.0
H2SO4 => 54.37
HCl => 19.11
4FeO3 => 82.39
Xe6 => 281.9
The program should return the first five strings arranged in order by their respective values. For this sample data: H20, HCl, HeNe, H2SO4, 4FeO3. Actually, I don't really care about the order; I just need the five lowest in any order.
I thought about how I'd do this in Perl. It's just a few lines:
foreach $s (#str) {
$strmap{$s} = f($s);
}
#sorted = sort { $strmap{$a} <=> $strmap{$b} } keys %strmap;
return #sorted[0, 4]
But I need to do it in Java. And it's driving me crazy.
First I tried populating a HashMap<String, Double>, then using Collections.sort with a custom comparator, just like the Perl version. But scoping on the Comparator prevented it from referring to the HashMap to look up the values.
Then I tried a TreeMap<String, Double>, but it only sorts by key and no amount of coercing could get it to order the entries by value.
So I tried a TreeMap<Double, String>. It will discard entries with the same Double. However, the likelihood of having Strings that map to the same Double is low, so I pressed forward. Adding the entries to the TreeMap is no problem, but I ran into issues trying to extract the values from it.
TreeMap supplies a method called subMap, but its parameters are the keys that delimit the subset. I don't know what they are; I just want the first five of them. So I tried using the values method to get all the values out of the TreeMap, hoping they'd be in order. Then I can just get the first ten.
ArrayList<String> strs = (ArrayList<String>)(treemap.values());
return new ArrayList<String>(strs.subList(0, 5));
Nope. Runtime error: cannot cast TreeMap$Values to ArrayList.
List<String> strs = (List<String>)(treemap.values());
return new ArrayList<String>(strs.subList(0, 5));
Same. Runtime error trying to do the cast. OK, let's just assign to a Collection...
Collection<String> strs = treemap.values();
return new ArrayList<String>(strs.subList(0, 5));
Sorry, subList isn't a method of Collection.
Collection<String> strs = treemap.values();
ArrayList<String> a = new ArrayList<String>(strs);
return new ArrayList<String>(a.subList(0, 5));
Finally, something that works! But two extra data structures just to get the first five elements? And I'm not too wild about using Double as the key for TreeMap.
Is there a better solution?
I don't think you'll get more compact than the three lines above, not in Java.
Apart from that, I have the impression that a Map as a data structure is the wrong choice in the first place, since you do not seem to need by-string lookups (UNLESS you want in some way deal with multiple occurences of strings, but you didn't say so). An alternative approach would be to declare your own comparable data record class:
private static class Record implements Comparable<Record> {
// public final fields ok for this small example
public final String string;
public final double value;
public Record(String string, double value) {
this.string = string;
this.value = value;
}
#Override
public int compareTo(Record other) {
// define sorting according to double fields
return Double.compare(value, other.value);
}
}
// provide size to avoid reallocations
List<Record> records = new ArrayList<Record>(stringList.size());
for(String s : stringList)
records.add(new Record(s, calculateFitness(s));
Collections.sort(records); // sort according to compareTo method
int max = Math.min(10, records.size()); // maximum index
List<String> result = new ArrayList<String>(max);
for(int i = 0; i < max; i++)
result.add(records.get(i).string);
return result;
This is now much more verbose than the three lines above (this is Java, after all), but also includes the code that would be required to insert the key/value pairs into the map.
Would something like the following work for you?
Note that I've assumed you don't require the double value other than to sort the data.
public static void main(String[] args) throws Exception {
List<String> data = new ArrayList<>(Arrays.asList("t", "h", "i", "s", "i", "s", "t", "e", "s", "t", "d", "a", "t", "a"));
Collections.sort(data, new Comparator<String>() {
#Override
public int compare(String o1, String o2) {
double o1Value = evaluate(o1);
double o2Value = evaluate(o2);
return Double.compare(o1Value, o2Value);
}
});
List<String> result = data.subList(0, 10); // Note the end point is exclusive
for (String s : result) {
System.out.println(s);
}
}
private static double evaluate(String s) {
return s.codePointAt(0); // Nonsense, I know
}
This example prints:
a
a
d
e
h
i
i
s
s
s
Why don't you just create a class to combine the String, Double and function that does the calculation - something like:
public Thing implements Comparable<Thing>
{
private String s;
private Double d;
public Thing(String s)
{
this.s = s;
this.d = calculateDouble(s);
}
public String getString()
{
return this.s;
}
public Double getDouble()
{
return this.d;
}
public int compareTo(Thing other)
{
return getDouble().compareTo(other.getDouble());
}
public Double calculateDouble(String s)
{
...
}
}
Then all you need is a List<Thing>, Collections.sort and List.subList.