We Keep Coding

How can I sort a list based on another list values in Java [duplicate]

I've seen several other questions similiar to this one but I haven't really been able to find anything that resolves my problem.
My use case is this: user has a list of items initially (listA). They reorder the items and want to persist that order (listB), however, due to restrictions I'm unable persist the order on the backend so I have to sort listA after I retrieve it.
So basically, I have 2 ArrayLists (listA and listB). One with the specific order the lists should be in (listB) and the other has the list of items (listA). I want to sort listA based on listB.

Using Java 8:
Collections.sort(listToSort,
Comparator.comparing(item -> listWithOrder.indexOf(item)));
or better:
listToSort.sort(Comparator.comparingInt(listWithOrder::indexOf));

Collections.sort(listB, new Comparator<Item>() {
public int compare(Item left, Item right) {
return Integer.compare(listA.indexOf(left), listA.indexOf(right));
}
});
This is quite inefficient, though, and you should probably create a Map<Item, Integer> from listA to lookup the positions of the items faster.
Guava has a ready-to-use comparator for doing that: Ordering.explicit()

Let's say you have a listB list that defines the order in which you want to sort listA. This is just an example, but it demonstrates an order that is defined by a list, and not the natural order of the datatype:
List<String> listB = Arrays.asList("Sunday", "Monday", "Tuesday", "Wednesday",
"Thursday", "Friday", "Saturday");
Now, let's say that listA needs to be sorted according to this ordering. It's a List<Item>, and Item has a public String getWeekday() method.
Create a Map<String, Integer> that maps the values of everything in listB to something that can be sorted easily, such as the index, i.e. "Sunday" => 0, ..., "Saturday" => 6. This will provide a quick and easy lookup.
Map<String, Integer> weekdayOrder = new HashMap<String, Integer>();
for (int i = 0; i < listB.size(); i++)
{
String weekday = listB.get(i);
weekdayOrder.put(weekday, i);
}
Then you can create your custom Comparator<Item> that uses the Map to create an order:
public class ItemWeekdayComparator implements Comparator<Item>
{
private Map<String, Integer> sortOrder;
public ItemWeekdayComparator(Map<String, Integer> sortOrder)
{
this.sortOrder = sortOrder;
}
#Override
public int compare(Item i1, Item i2)
{
Integer weekdayPos1 = sortOrder.get(i1.getWeekday());
if (weekdayPos1 == null)
{
throw new IllegalArgumentException("Bad weekday encountered: " +
i1.getWeekday());
}
Integer weekdayPos2 = sortOrder.get(i2.getWeekday());
if (weekdayPos2 == null)
{
throw new IllegalArgumentException("Bad weekday encountered: " +
i2.getWeekday());
}
return weekdayPos1.compareTo(weekdayPos2);
}
}
Then you can sort listA using your custom Comparator.
Collections.sort(listA, new ItemWeekdayComparator(weekdayOrder));

Speed improvement on JB Nizet's answer (from the suggestion he made himself). With this method:
Sorting a 1000 items list 100 times improves speed 10 times on my
unit tests.
Sorting a 10000 items list 100 times improves speed 140 times (265 ms for the whole batch instead of 37 seconds) on my
unit tests.
This method will also work when both lists are not identical:
/**
* Sorts list objectsToOrder based on the order of orderedObjects.
*
* Make sure these objects have good equals() and hashCode() methods or
* that they reference the same objects.
*/
public static void sortList(List<?> objectsToOrder, List<?> orderedObjects) {
HashMap<Object, Integer> indexMap = new HashMap<>();
int index = 0;
for (Object object : orderedObjects) {
indexMap.put(object, index);
index++;
}
Collections.sort(objectsToOrder, new Comparator<Object>() {
public int compare(Object left, Object right) {
Integer leftIndex = indexMap.get(left);
Integer rightIndex = indexMap.get(right);
if (leftIndex == null) {
return -1;
}
if (rightIndex == null) {
return 1;
}
return Integer.compare(leftIndex, rightIndex);
}
});
}

Problem : sorting a list of Pojo on the basis of one of the field's all possible values present in another list.
Take a look at this solution, may be this is what you are trying to achieve:
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;
public class Test {
public static void main(String[] args) {
List<Employee> listToSort = new ArrayList<>();
listToSort.add(new Employee("a", "age11"));
listToSort.add(new Employee("c", "age33"));
listToSort.add(new Employee("b", "age22"));
listToSort.add(new Employee("a", "age111"));
listToSort.add(new Employee("c", "age3"));
listToSort.add(new Employee("b", "age2"));
listToSort.add(new Employee("a", "age1"));
List<String> listWithOrder = new ArrayList<>();
listWithOrder.add("a");
listWithOrder.add("b");
listWithOrder.add("c");
Collections.sort(listToSort, Comparator.comparing(item ->
listWithOrder.indexOf(item.getName())));
System.out.println(listToSort);
}
}
class Employee {
String name;
String age;
public Employee(String name, String age) {
super();
this.name = name;
this.age = age;
}
public String getName() {
return name;
}
public String getAge() {
return age;
}
#Override
public String toString() {
return "[name=" + name + ", age=" + age + "]";
}
}
O U T P U T
[[name=a, age=age11], [name=a, age=age111], [name=a, age=age1], [name=b, age=age22], [name=b, age=age2], [name=c, age=age33], [name=c, age=age3]]

Here is a solution that increases the time complexity by 2n, but accomplishes what you want. It also doesn't care if the List R you want to sort contains Comparable elements so long as the other List L you use to sort them by is uniformly Comparable.
public class HeavyPair<L extends Comparable<L>, R> implements Comparable<HeavyPair<L, ?>> {
public final L left;
public final R right;
public HeavyPair(L left, R right) {
this.left = left;
this.right = right;
}
public compareTo(HeavyPair<L, ?> o) {
return this.left.compareTo(o.left);
}
public static <L extends Comparable<L>, R> List<R> sort(List<L> weights, List<R> toSort) {
assert(weights.size() == toSort.size());
List<R> output = new ArrayList<>(toSort.size());
List<HeavyPair<L, R>> workHorse = new ArrayList<>(toSort.size());
for(int i = 0; i < toSort.size(); i++) {
workHorse.add(new HeavyPair(weights.get(i), toSort.get(i)))
}
Collections.sort(workHorse);
for(int i = 0; i < workHorse.size(); i++) {
output.add(workHorse.get(i).right);
}
return output;
}
}
Excuse any terrible practices I used while writing this code, though. I was in a rush.
Just call HeavyPair.sort(listB, listA);
Edit: Fixed this line return this.left.compareTo(o.left);. Now it actually works.

Here is an example of how to sort a list and then make the changes in another list according to the changes exactly made to first array list. This trick will never fails and ensures the mapping between the items in list. The size of both list must be same to use this trick.
ArrayList<String> listA = new ArrayList<String>();
ArrayList<String> listB = new ArrayList<String>();
int j = 0;
// list of returns of the compare method which will be used to manipulate
// the another comparator according to the sorting of previous listA
ArrayList<Integer> sortingMethodReturns = new ArrayList<Integer>();
public void addItemstoLists() {
listA.add("Value of Z");
listA.add("Value of C");
listA.add("Value of F");
listA.add("Value of A");
listA.add("Value of Y");
listB.add("this is the value of Z");
listB.add("this is the value off C");
listB.add("this is the value off F");
listB.add("this is the value off A");
listB.add("this is the value off Y");
Collections.sort(listA, new Comparator<String>() {
#Override
public int compare(String lhs, String rhs) {
// TODO Auto-generated method stub
int returning = lhs.compareTo(rhs);
sortingMethodReturns.add(returning);
return returning;
}
});
// now sort the list B according to the changes made with the order of
// items in listA
Collections.sort(listB, new Comparator<String>() {
#Override
public int compare(String lhs, String rhs) {
// TODO Auto-generated method stub
// comparator method will sort the second list also according to
// the changes made with list a
int returning = sortingMethodReturns.get(j);
j++;
return returning;
}
});
}

try this for java 8:
listB.sort((left, right) -> Integer.compare(list.indexOf(left), list.indexOf(right)));
or
listB.sort(Comparator.comparingInt(item -> list.indexOf(item)));

import java.util.Comparator;
import java.util.List;
public class ListComparator implements Comparator<String> {
private final List<String> orderedList;
private boolean appendFirst;
public ListComparator(List<String> orderedList, boolean appendFirst) {
this.orderedList = orderedList;
this.appendFirst = appendFirst;
}
#Override
public int compare(String o1, String o2) {
if (orderedList.contains(o1) && orderedList.contains(o2))
return orderedList.indexOf(o1) - orderedList.indexOf(o2);
else if (orderedList.contains(o1))
return (appendFirst) ? 1 : -1;
else if (orderedList.contains(o2))
return (appendFirst) ? -1 : 1;
return 0;
}
}
You can use this generic comparator to sort list based on the the other list.
For example, when appendFirst is false below will be the output.
Ordered list: [a, b]
Un-ordered List: [d, a, b, c, e]
Output:
[a, b, d, c, e]

One way of doing this is looping through listB and adding the items to a temporary list if listA contains them:
List<?> tempList = new ArrayList<?>();
for(Object o : listB) {
if(listA.contains(o)) {
tempList.add(o);
}
}
listA.removeAll(listB);
tempList.addAll(listA);
return tempList;

Not completely clear what you want, but if this is the situation:
A:[c,b,a]
B:[2,1,0]
And you want to load them both and then produce:
C:[a,b,c]
Then maybe this?
List c = new ArrayList(b.size());
for(int i=0;i<b.size();i++) {
c.set(b.get(i),a.get(i));
}
that requires an extra copy, but I think to to it in place is a lot less efficient, and all kinds of not clear:
for(int i=0;i<b.size();i++){
int from = b.get(i);
if(from == i) continue;
T tmp = a.get(i);
a.set(i,a.get(from));
a.set(from,tmp);
b.set(b.lastIndexOf(i),from);
}
Note I didn't test either, maybe got a sign flipped.

Another solution that may work depending on your setting is not storing instances in listB but instead indices from listA. This could be done by wrapping listA inside a custom sorted list like so:
public static class SortedDependingList<E> extends AbstractList<E> implements List<E>{
private final List<E> dependingList;
private final List<Integer> indices;
public SortedDependingList(List<E> dependingList) {
super();
this.dependingList = dependingList;
indices = new ArrayList<>();
}
#Override
public boolean add(E e) {
int index = dependingList.indexOf(e);
if (index != -1) {
return addSorted(index);
}
return false;
}
/**
* Adds to this list the element of the depending list at the given
* original index.
* #param index The index of the element to add.
*
*/
public boolean addByIndex(int index){
if (index < 0 || index >= this.dependingList.size()) {
throw new IllegalArgumentException();
}
return addSorted(index);
}
/**
* Returns true if this list contains the element at the
* index of the depending list.
*/
public boolean containsIndex(int index){
int i = Collections.binarySearch(indices, index);
return i >= 0;
}
private boolean addSorted(int index){
int insertIndex = Collections.binarySearch(indices, index);
if (insertIndex < 0){
insertIndex = -insertIndex-1;
this.indices.add(insertIndex, index);
return true;
}
return false;
}
#Override
public E get(int index) {
return dependingList.get(indices.get(index));
}
#Override
public int size() {
return indices.size();
}
}
Then you can use this custom list as follows:
public static void main(String[] args) {
class SomeClass{
int index;
public SomeClass(int index) {
super();
this.index = index;
}
#Override
public String toString() {
return ""+index;
}
}
List<SomeClass> listA = new ArrayList<>();
for (int i = 0; i < 100; i++) {
listA.add(new SomeClass(i));
}
SortedDependingList<SomeClass> listB = new SortedDependingList<>(listA);
Random rand = new Random();
// add elements by index:
for (int i = 0; i < 5; i++) {
int index = rand.nextInt(listA.size());
listB.addByIndex(index);
}
System.out.println(listB);
// add elements by identity:
for (int i = 0; i < 5; i++) {
int index = rand.nextInt(listA.size());
SomeClass o = listA.get(index);
listB.add(o);
}
System.out.println(listB);
}
Of course, this custom list will only be valid as long as the elements in the original list do not change. If changes are possible, you would need to somehow listen for changes to the original list and update the indices inside the custom list.
Note also, that the SortedDependingList does currently not allow to add an element from listA a second time - in this respect it actually works like a set of elements from listA because this is usually what you want in such a setting.
The preferred way to add something to SortedDependingList is by already knowing the index of an element and adding it by calling sortedList.addByIndex(index);

If the two lists are guaranteed to contain the same elements, just in a different order, you can use List<T> listA = new ArrayList<>(listB) and this will be O(n) time complexity. Otherwise, I see a lot of answers here using Collections.sort(), however there is an alternative method which is guaranteed O(2n) runtime, which should theoretically be faster than sort's worst time complexity of O(nlog(n)), at the cost of 2n storage
Set<T> validItems = new HashSet<>(listB);
listA.clear();
listB.forEach(item -> {
if(validItems.contains(item)) {
listA.add(item);
}
});

List<String> listA;
Comparator<B> comparator = Comparator.comparing(e -> listA.indexOf(e.getValue()));
//call your comparator inside your list to be sorted
listB.stream().sorted(comparator)..

Like Tim Herold wrote, if the object references should be the same, you can just copy listB to listA, either:
listA = new ArrayList(listB);
Or this if you don't want to change the List that listA refers to:
listA.clear();
listA.addAll(listB);
If the references are not the same but there is some equivalence relationship between objects in listA and listB, you could sort listA using a custom Comparator that finds the object in listB and uses its index in listB as the sort key. The naive implementation that brute force searches listB would not be the best performance-wise, but would be functionally sufficient.

IMO, you need to persist something else. May be not the full listB, but something. May be just the indexes of the items that the user changed.

Try this. The code below is general purpose for a scenario where listA is a list of Objects since you did not indicate a particular type.
Object[] orderedArray = new Object[listA.size()];
for(int index = 0; index < listB.size(); index ++){
int position = listB.get(index); //this may have to be cast as an int
orderedArray[position] = listA.get(index);
}
//if you receive UnsupportedOperationException when running listA.clear()
//you should replace the line with listA = new List<Object>()
//using your actual implementation of the List interface
listA.clear();
listA.addAll(orderedArray);

Just encountered the same problem.
I have a list of ordered keys, and I need to order the objects in a list according to the order of the keys.
My lists are long enough to make the solutions with time complexity of N^2 unusable.
My solution:
<K, T> List<T> sortByOrder(List<K> orderedKeys, List<T> objectsToOrder, Function<T, K> keyExtractor) {
AtomicInteger ind = new AtomicInteger(0);
Map<K, Integer> keyToIndex = orderedKeys.stream().collect(Collectors.toMap(k -> k, k -> ind.getAndIncrement(), (oldK, newK) -> oldK));
SortedMap<Integer, T> indexToObj = new TreeMap<>();
objectsToOrder.forEach(obj -> indexToObj.put(keyToIndex.get(keyExtractor.apply(obj)), obj));
return new ArrayList<>(indexToObj.values());
}
The time complexity is O(N * Log(N)).
The solution assumes that all the objects in the list to sort have distinct keys. If not then just replace SortedMap<Integer, T> indexToObj by SortedMap<Integer, List<T>> indexToObjList.

To avoid having a very inefficient look up, you should index the items in listB and then sort listA based on it.
Map<Item, Integer> index = IntStream.range(0, listB.size()).boxed()
.collect(Collectors.toMap(listB::get, x -> x));
listA.sort((e1, e2) -> Integer.compare(index.get(c1), index.get(c2));

So for me the requirement was to sort originalList with orderedList. originalList always contains all element from orderedList, but not vice versa. No new elements.
fun <T> List<T>.sort(orderedList: List<T>): List<T> {
return if (size == orderedList.size) {
orderedList
} else {
var keepIndexCount = 0
mapIndexed { index, item ->
if (orderedList.contains(item)) {
orderedList[index - keepIndexCount]
} else {
keepIndexCount++
item
}
}
}}
P.S. my case was that I have list that user can sort by drag and drop, but some items might be filtered out, so we preserve hidden items position.

If you want to do it manually. Solution based on bubble sort (same length required):
public void sortAbasedOnB(String[] listA, double[] listB) {
for (int i = 0; i < listB.length - 1; i++) {
for (int j = listB.length - 1; j > i; j--) {
if (listB[j] < listB[j - 1]){
double tempD = listB[j - 1];
listB[j - 1] = listB[j];
listB[j] = tempD;
String tempS = listA[j - 1];
listA[j - 1] = listA[j];
listA[j] = tempS;
}
}
}
}

If the object references should be the same, you can initialize listA new.
listA = new ArrayList(listB)

In Java there are set of classes which can be useful to sort lists or arrays. Most of the following examples will use lists but the same concept can be applied for arrays. A example will show this.
We can use this by creating a list of Integers and sort these using the Collections.sort(). The Collections (Java Doc) class (part of the Java Collection Framework) provides a list of static methods which we can use when working with collections such as list, set and the like. So in a nutshell, we can sort a list by simply calling: java.util.Collections.sort(the list) as shown in the following example:
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
public class example {
public static void main(String[] args) {
List<Integer> ints = new ArrayList<Integer>();
ints.add(4);
ints.add(3);
ints.add(7);
ints.add(5);
Collections.sort(ints);
System.out.println(ints);
}
}
The above class creates a list of four integers and, using the collection sort method, sorts this list (in one line of code) without us having to worry about the sorting algorithm.

How to interleave (merge) two Java 8 Streams?

Stream<String> a = Stream.of("one", "three", "five");
Stream<String> b = Stream.of("two", "four", "six");
What do I need to do for the output to be the below?
// one
// two
// three
// four
// five
// six
I looked into concat but as the javadoc explains, it just appends one after the other, it does not interleave / intersperse.
Stream<String> out = Stream.concat(a, b);
out.forEach(System.out::println);
Creates a lazily concatenated stream whose elements are all the
elements of the first stream followed by all the elements of the
second stream.
Wrongly gives
// one
// three
// five
// two
// four
// six
Could do it if I collected them and iterated, but was hoping for something more Java8-y, Streamy :-)
Note
I don't want to zip the streams
“zip” operation will take an element from each collection and combine them.
the result of a zip operation would be something like this: (unwanted)
// onetwo
// threefour
// fivesix

I’d use something like this:
public static <T> Stream<T> interleave(Stream<? extends T> a, Stream<? extends T> b) {
Spliterator<? extends T> spA = a.spliterator(), spB = b.spliterator();
long s = spA.estimateSize() + spB.estimateSize();
if(s < 0) s = Long.MAX_VALUE;
int ch = spA.characteristics() & spB.characteristics()
& (Spliterator.NONNULL|Spliterator.SIZED);
ch |= Spliterator.ORDERED;
return StreamSupport.stream(new Spliterators.AbstractSpliterator<T>(s, ch) {
Spliterator<? extends T> sp1 = spA, sp2 = spB;
#Override
public boolean tryAdvance(Consumer<? super T> action) {
Spliterator<? extends T> sp = sp1;
if(sp.tryAdvance(action)) {
sp1 = sp2;
sp2 = sp;
return true;
}
return sp2.tryAdvance(action);
}
}, false);
}
It retains the characteristics of the input streams as far as possible, which allows certain optimizations (e.g. for count()and toArray()). Further, it adds the ORDERED even when the input streams might be unordered, to reflect the interleaving.
When one stream has more elements than the other, the remaining elements will appear at the end.

A much dumber solution than Holger did, but may be it would fit your requirements:
private static <T> Stream<T> interleave(Stream<T> left, Stream<T> right) {
Spliterator<T> splLeft = left.spliterator();
Spliterator<T> splRight = right.spliterator();
T[] single = (T[]) new Object[1];
Stream.Builder<T> builder = Stream.builder();
while (splRight.tryAdvance(x -> single[0] = x) && splLeft.tryAdvance(builder)) {
builder.add(single[0]);
}
return builder.build();
}

As you can see from the question comments, I gave this a go using zip:
Stream<String> a = Stream.of("one", "three", "five");
Stream<String> b = Stream.of("two", "four", "six");
Stream<String> out = interleave(a, b);
public static <T> Stream<T> interleave(Stream<T> streamA, Stream<T> streamB) {
return zip(streamA, streamB, (o1, o2) -> Stream.of(o1, o2)).flatMap(s -> s);
}
/**
* https://stackoverflow.com/questions/17640754/zipping-streams-using-jdk8-with-lambda-java-util-stream-streams-zip
**/
private static <A, B, C> Stream<C> zip(Stream<A> streamA, Stream<B> streamB, BiFunction<A, B, C> zipper) {
final Iterator<A> iteratorA = streamA.iterator();
final Iterator<B> iteratorB = streamB.iterator();
final Iterator<C> iteratorC = new Iterator<C>() {
#Override
public boolean hasNext() {
return iteratorA.hasNext() && iteratorB.hasNext();
}
#Override
public C next() {
return zipper.apply(iteratorA.next(), iteratorB.next());
}
};
final boolean parallel = streamA.isParallel() || streamB.isParallel();
return iteratorToFiniteStream(iteratorC, parallel);
}
private static <T> Stream<T> iteratorToFiniteStream(Iterator<T> iterator, boolean parallel) {
final Iterable<T> iterable = () -> iterator;
return StreamSupport.stream(iterable.spliterator(), parallel);
}

This may not be a good answer because
(1) it collects to map, which you don't want to do I guess and
(2) it is not completely stateless as it uses AtomicIntegers.
Still adding it because
(1) it is readable and
(2) community can get an idea from this and try to improve it.
Stream<String> a = Stream.of("one", "three", "five");
Stream<String> b = Stream.of("two", "four", "six");
AtomicInteger i = new AtomicInteger(0);
AtomicInteger j = new AtomicInteger(1);
Stream.of(a.collect(Collectors.toMap(o -> i.addAndGet(2), Function.identity())),
b.collect(Collectors.toMap(o -> j.addAndGet(2), Function.identity())))
.flatMap(m -> m.entrySet().stream())
.sorted(Comparator.comparing(Map.Entry::getKey))
.forEach(e -> System.out.println(e.getValue())); // or collect
Output
one
two
three
four
five
six
#Holger's edit
Stream.concat(a.map(o -> new AbstractMap.SimpleEntry<>(i.addAndGet(2), o)),
b.map(o -> new AbstractMap.SimpleEntry<>(j.addAndGet(2), o)))
.sorted(Map.Entry.comparingByKey())
.forEach(e -> System.out.println(e.getValue())); // or collect

without any external lib (using jdk11)
import java.util.ArrayList;
import java.util.List;
import java.util.stream.Collectors;
import java.util.stream.IntStream;
import java.util.stream.Stream;
public class MergeUtil {
private static <T> Stream<T> zipped(List<T> lista, List<T> listb) {
int maxSize = Math.max(lista.size(), listb.size());
final var listStream = IntStream
.range(0, maxSize)
.mapToObj(i -> {
List<T> result = new ArrayList<>(2);
if (i < lista.size()) result.add(lista.get(i));
if (i < listb.size()) result.add(listb.get(i));
return result;
});
return listStream.flatMap(List::stream);
}
public static void main(String[] args) {
var l1 = List.of(1, 2, 3);
var l2 = List.of(4, 5, 6, 7, 8, 9);
final var zip = zipped(l1, l2);
System.out.println(zip.collect(Collectors.toList()));
}
}
listStream is a Stream<List<A>> that flatted in return.
The result is:
[1, 4, 2, 5, 3, 6, 7, 8, 9]

One solution with Iterator
final Iterator<String> iterA = a.iterator();
final Iterator<String> iterB = b.iterator();
final Iterator<String> iter = new Iterator<String>() {
private final AtomicInteger idx = new AtomicInteger();
#Override
public boolean hasNext() {
return iterA.hasNext() || iterB.hasNext();
}
#Override
public String next() {
return idx.getAndIncrement() % 2 == 0 && iterA.hasNext() ? iterA.next() : iterB.next();
}
};
// Create target Stream with StreamEx from: https://github.com/amaembo/streamex
StreamEx.of(iter).forEach(System.out::println);
// Or Streams from Google Guava
Streams.stream(iter).forEach(System.out::println);
Or simply by the solution in abacus-common provided by me:
AtomicInteger idx = new AtomicInteger();
StreamEx.merge(a, b, (s1, s2) -> idx.getAndIncrement() % 2 == 0 ? Nth.FIRST : Nth.SECOND).forEach(Fn.println());

Using Guava's Streams.zip and Stream.flatMap:
Stream<String> interleaved = Streams
.zip(a, b, (x, y) -> Stream.of(x, y))
.flatMap(Function.identity());
interleaved.forEach(System.out::println);
Prints:
one
two
three
four
five
six

Java 8 Stream API - Select the lowest key after group by

I have a stream of Foo objects.
class Foo {
private int variableCount;
public Foo(int vars) {
this.variableCount = vars;
}
public Integer getVariableCount() {
return variableCount;
}
}
I want a list of Foo's that all have the lowest variableCount.
For example
new Foo(3), new Foo(3), new Foo(2), new Foo(1), new Foo(1)
I only want the stream to return the last 2 Foos, since they have the lowest value.
I've tried doing a collect with grouping by
.collect(Collectors.groupingBy((Foo foo) -> {
return foo.getVariableCount();
})
And that returns a Map<Integer, List<Foo>> and I'm not sure how to transform that into what I want.
Thanks in advance

You can use a sorted map for grouping and then just get the first entry.
Something along the lines:
Collectors.groupingBy(
Foo::getVariableCount,
TreeMap::new,
Collectors.toList())
.firstEntry()
.getValue()

Here is a solution that:
Only streams the list once.
Doesn't build a map or other structure that contains all of the input items (unless the variable counts are all the same), only keeping those that are currently the minimum.
Is O(n) time, O(n) space. It's entirely possible that all Foos have the same variable count, in which case this solution would store all items like other solutions. But in practice, with different, varied values and higher cardinality, the number of items in the list is likely to be much lower.
Edited
I've improved my solution according to the suggestions in the comments.
I implemented an accumulator object, which supplies functions to the Collector for this.
/**
* Accumulator object to hold the current min
* and the list of Foos that are the min.
*/
class Accumulator {
Integer min;
List<Foo> foos;
Accumulator() {
min = Integer.MAX_VALUE;
foos = new ArrayList<>();
}
void accumulate(Foo f) {
if (f.getVariableCount() != null) {
if (f.getVariableCount() < min) {
min = f.getVariableCount();
foos.clear();
foos.add(f);
} else if (f.getVariableCount() == min) {
foos.add(f);
}
}
}
Accumulator combine(Accumulator other) {
if (min < other.min) {
return this;
}
else if (min > other.min) {
return other;
}
else {
foos.addAll(other.foos);
return this;
}
}
List<Foo> getFoos() { return foos; }
}
Then all we have to do is collect, referencing the accumulator's methods for its functions.
List<Foo> mins = foos.stream().collect(Collector.of(
Accumulator::new,
Accumulator::accumulate,
Accumulator::combine,
Accumulator::getFoos
)
);
Testing with
List<Foo> foos = Arrays.asList(new Foo(3), new Foo(3), new Foo(2), new Foo(1), new Foo(1), new Foo(4));
The output is (with a suitable toString defined on Foo):
[Foo{1}, Foo{1}]

IF you are OK streaming (iterating) twice:
private static List<Foo> mins(List<Foo> foos) {
return foos.stream()
.map(Foo::getVariableCount)
.min(Comparator.naturalOrder())
.map(x -> foos.stream()
.filter(y -> y.getVariableCount() == x)
.collect(Collectors.toList()))
.orElse(Collections.emptyList());
}

To avoid creating the entire map and also avoiding streaming twice, I copied a custom collector from here https://stackoverflow.com/a/30497254/1264846 and modified it to work with min instead of max. I didn't even know custom collectors were possible so I thank #lexicore for pointing me in that direction.
This is the resulting function minAll
public static <T, A, D> Collector<T, ?, D> minAll(Comparator<? super T> comparator,
Collector<? super T, A, D> downstream) {
Supplier<A> downstreamSupplier = downstream.supplier();
BiConsumer<A, ? super T> downstreamAccumulator = downstream.accumulator();
BinaryOperator<A> downstreamCombiner = downstream.combiner();
class Container {
A acc;
T obj;
boolean hasAny;
Container(A acc) {
this.acc = acc;
}
}
Supplier<Container> supplier = () -> new Container(downstreamSupplier.get());
BiConsumer<Container, T> accumulator = (acc, t) -> {
if(!acc.hasAny) {
downstreamAccumulator.accept(acc.acc, t);
acc.obj = t;
acc.hasAny = true;
} else {
int cmp = comparator.compare(t, acc.obj);
if (cmp < 0) {
acc.acc = downstreamSupplier.get();
acc.obj = t;
}
if (cmp <= 0)
downstreamAccumulator.accept(acc.acc, t);
}
};
BinaryOperator<Container> combiner = (acc1, acc2) -> {
if (!acc2.hasAny) {
return acc1;
}
if (!acc1.hasAny) {
return acc2;
}
int cmp = comparator.compare(acc1.obj, acc2.obj);
if (cmp < 0) {
return acc1;
}
if (cmp > 0) {
return acc2;
}
acc1.acc = downstreamCombiner.apply(acc1.acc, acc2.acc);
return acc1;
};
Function<Container, D> finisher = acc -> downstream.finisher().apply(acc.acc);
return Collector.of(supplier, accumulator, combiner, finisher);
}

You could use collect wisely on the sorted list and in accumulator add the logic to add only either first element to empty list or add any other Foo having variable count same as of the first element of the list.
A complete working example below:-
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.List;
class Foo {
private int variableCount;
public Foo(int vars) {
this.variableCount = vars;
}
public Integer getVariableCount() {
return variableCount;
}
public static void main(String[] args) {
List<Foo> list = Arrays.asList(
new Foo(2),
new Foo(2),
new Foo(3),
new Foo(3),
new Foo(1),
new Foo(1)
);
System.out.println(list.stream()
.sorted(Comparator.comparing(Foo::getVariableCount))
.collect(() -> new ArrayList<Foo>(),
(ArrayList<Foo> arrayList, Foo e) -> {
if (arrayList.isEmpty()
|| arrayList.get(0).getVariableCount() == e.getVariableCount()) {
arrayList.add(e);
}
},
(ArrayList<Foo> foos, ArrayList<Foo> foo) -> foos.addAll(foo)
)
);
}
#Override
public String toString() {
return "Foo{" +
"variableCount=" + variableCount +
'}';
}
}
Also, you could first find the minimum variableCount in one stream and use that inside filter of another stream.
list.sort(Comparator.comparing(Foo::getVariableCount));
int min = list.get(0).getVariableCount();
list.stream().filter(foo -> foo.getVariableCount() == min)
.collect(Collectors.toList());
I think in any case either sorting is required or a way to find the minimum number which later can be used inside the predicate. Even if you are using the map to group the values.
Cheers!

Here is alternative with one stream and custom reducer. The idea is to first sort and then collect only elements with first min value:
List<Foo> newlist = list.stream()
.sorted( Comparator.comparing(Foo::getVariableCount) )
.reduce( new ArrayList<>(),
(l, f) -> {
if ( l.isEmpty() || l.get(0).getVariableCount() == f.getVariableCount() ) l.add(f);
return l;
},
(l1, l2) -> {
l1.addAll(l2);
return l1;
}
);
Or using collect is even more compact:
List<Foo> newlist = list.stream()
.sorted( Comparator.comparing(Foo::getVariableCount) )
.collect( ArrayList::new,
(l, f) -> if ( l.isEmpty() || l.get(0).getVariableCount() == f.getVariableCount() ) l.add(f),
List::addAll
);

To avoid creating the map you could use two streams :
the first finds the minimum value.
the second filters elements with this value.
It could give :
List<Foo> foos = ...;
int min = foos.stream()
.mapToInt(Foo::getVariableCount)
.min()
.orElseThrow(RuntimeException::new); // technical error
List<Foo> minFoos = foos.stream()
.filter(f -> f.getVariableCount() == min)
.collect(Collectors.toList());

How to get a random element from a list with stream api?

What is the most effective way to get a random element from a list with Java8 stream api?
Arrays.asList(new Obj1(), new Obj2(), new Obj3());
Thanks.

Why with streams? You just have to get a random number from 0 to the size of the list and then call get on this index:
Random r = new Random();
ElementType e = list.get(r.nextInt(list.size()));
Stream will give you nothing interesting here, but you can try with:
Random r = new Random();
ElementType e = list.stream().skip(r.nextInt(list.size())).findFirst().get();
Idea is to skip an arbitrary number of elements (but not the last one!), then get the first element if it exists. As a result you will have an Optional<ElementType> which will be non empty and then extract its value with get. You have a lot of options here after having skip.
Using streams here is highly inefficient...
Note: that none of these solutions take in account empty lists, but the problem is defined on non-empty lists.

There are much more efficient ways to do it, but if this has to be Stream the easiest way is to create your own Comparator, which returns random result (-1, 0, 1) and sort your stream:
List<String> strings = Arrays.asList("a", "b", "c", "d", "e", "f");
String randomString = strings
.stream()
.sorted((o1, o2) -> ThreadLocalRandom.current().nextInt(-1, 2))
.findAny()
.get();
ThreadLocalRandom has ready "out of the box" method to get random number in your required range for comparator.

While all the given answers work, there is a simple one-liner that does the trick without having to check if the list is empty first:
List<String> list = List.of("a", "b", "c");
list.stream().skip((int) (list.size() * Math.random())).findAny();
For an empty list this will return an Optional.empty.

In the last time I needed to do something like that I did that:
List<String> list = Arrays.asList("a", "b", "c");
Collections.shuffle(list);
String letter = list.stream().findAny().orElse(null);
System.out.println(letter);

If you HAVE to use streams, I wrote an elegant, albeit very inefficient collector that does the job:
/**
* Returns a random item from the stream (or null in case of an empty stream).
* This operation can't be lazy and is inefficient, and therefore shouldn't
* be used on streams with a large number or items or in performance critical sections.
* #return a random item from the stream or null if the stream is empty.
*/
public static <T> Collector<T, List<T>, T> randomItem() {
final Random RANDOM = new Random();
return Collector.of(() -> (List<T>) new ArrayList<T>(),
(acc, elem) -> acc.add(elem),
(list1, list2) -> ListUtils.union(list1, list2), // Using a 3rd party for list union, could be done "purely"
list -> list.isEmpty() ? null : list.get(RANDOM.nextInt(list.size())));
}
Usage:
#Test
public void standardRandomTest() {
assertThat(Stream.of(1, 2, 3, 4).collect(randomItem())).isBetween(1, 4);
}

Another idea would be to implement your own Spliterator and then use it as a source for Stream:
import java.util.List;
import java.util.Random;
import java.util.Spliterator;
import java.util.function.Consumer;
import java.util.function.Supplier;
public class ImprovedRandomSpliterator<T> implements Spliterator<T> {
private final Random random;
private final T[] source;
private int size;
ImprovedRandomSpliterator(List<T> source, Supplier<? extends Random> random) {
if (source.isEmpty()) {
throw new IllegalArgumentException("RandomSpliterator can't be initialized with an empty collection");
}
this.source = (T[]) source.toArray();
this.random = random.get();
this.size = this.source.length;
}
#Override
public boolean tryAdvance(Consumer<? super T> action) {
if (size > 0) {
int nextIdx = random.nextInt(size);
int lastIdx = size - 1;
action.accept(source[nextIdx]);
source[nextIdx] = source[lastIdx];
source[lastIdx] = null; // let object be GCed
size--;
return true;
} else {
return false;
}
}
#Override
public Spliterator<T> trySplit() {
return null;
}
#Override
public long estimateSize() {
return source.length;
}
#Override
public int characteristics() {
return SIZED;
}
}
public static <T> Collector<T, ?, Stream<T>> toShuffledStream() {
return Collectors.collectingAndThen(
toCollection(ArrayList::new),
list -> !list.isEmpty()
? StreamSupport.stream(new ImprovedRandomSpliterator<>(list, Random::new), false)
: Stream.empty());
}
and then simply:
list.stream()
.collect(toShuffledStream())
.findAny();
Details can be found here.
...but it's definitely an overkill, so if you're looking for a pragmatic approach. Definitely go for Jean's solution.

If you don't know in advance the size of the your list, you could do something like that :
yourStream.collect(new RandomListCollector<>(randomSetSize));
I guess that you will have to write your own Collector implementation like this one to have an homogeneous randomization :
public class RandomListCollector<T> implements Collector<T, RandomListCollector.ListAccumulator<T>, List<T>> {
private final Random rand;
private final int size;
public RandomListCollector(Random random , int size) {
super();
this.rand = random;
this.size = size;
}
public RandomListCollector(int size) {
this(new Random(System.nanoTime()), size);
}
#Override
public Supplier<ListAccumulator<T>> supplier() {
return () -> new ListAccumulator<T>();
}
#Override
public BiConsumer<ListAccumulator<T>, T> accumulator() {
return (l, t) -> {
if (l.size() < size) {
l.add(t);
} else if (rand.nextDouble() <= ((double) size) / (l.gSize() + 1)) {
l.add(t);
l.remove(rand.nextInt(size));
} else {
// in any case gSize needs to be incremented
l.gSizeInc();
}
};
}
#Override
public BinaryOperator<ListAccumulator<T>> combiner() {
return (l1, l2) -> {
int lgSize = l1.gSize() + l2.gSize();
ListAccumulator<T> l = new ListAccumulator<>();
if (l1.size() + l2.size()<size) {
l.addAll(l1);
l.addAll(l2);
} else {
while (l.size() < size) {
if (l1.size()==0 || l2.size()>0 && rand.nextDouble() < (double) l2.gSize() / (l1.gSize() + l2.gSize())) {
l.add(l2.remove(rand.nextInt(l2.size()), true));
} else {
l.add(l1.remove(rand.nextInt(l1.size()), true));
}
}
}
// set the gSize of l :
l.gSize(lgSize);
return l;
};
}
#Override
public Function<ListAccumulator<T>, List<T>> finisher() {
return (la) -> la.list;
}
#Override
public Set<Characteristics> characteristics() {
return Collections.singleton(Characteristics.CONCURRENT);
}
static class ListAccumulator<T> implements Iterable<T> {
List<T> list;
volatile int gSize;
public ListAccumulator() {
list = new ArrayList<>();
gSize = 0;
}
public void addAll(ListAccumulator<T> l) {
list.addAll(l.list);
gSize += l.gSize;
}
public T remove(int index) {
return remove(index, false);
}
public T remove(int index, boolean global) {
T t = list.remove(index);
if (t != null && global)
gSize--;
return t;
}
public void add(T t) {
list.add(t);
gSize++;
}
public int gSize() {
return gSize;
}
public void gSize(int gSize) {
this.gSize = gSize;
}
public void gSizeInc() {
gSize++;
}
public int size() {
return list.size();
}
#Override
public Iterator<T> iterator() {
return list.iterator();
}
}
}
If you want something easier and still don't want to load all your list in memory:
public <T> Stream<T> getRandomStreamSubset(Stream<T> stream, int subsetSize) {
int cnt = 0;
Random r = new Random(System.nanoTime());
Object[] tArr = new Object[subsetSize];
Iterator<T> iter = stream.iterator();
while (iter.hasNext() && cnt <subsetSize) {
tArr[cnt++] = iter.next();
}
while (iter.hasNext()) {
cnt++;
T t = iter.next();
if (r.nextDouble() <= (double) subsetSize / cnt) {
tArr[r.nextInt(subsetSize)] = t;
}
}
return Arrays.stream(tArr).map(o -> (T)o );
}
but you are then away from the stream api and could do the same with a basic iterator

The selected answer has errors in its stream solution...
You cannot use Random#nextInt with a non-positive long, "0" in this case.
The stream solution will also never choose the last in the list
Example:
List<Integer> intList = Arrays.asList(0, 1, 2, 3, 4);
// #nextInt is exclusive, so here it means a returned value of 0-3
// if you have a list of size = 1, #next Int will throw an IllegalArgumentException (bound must be positive)
int skipIndex = new Random().nextInt(intList.size()-1);
// randomInt will only ever be 0, 1, 2, or 3. Never 4
int randomInt = intList.stream()
.skip(skipIndex) // max skip of list#size - 2
.findFirst()
.get();
My recommendation would be to go with the non-stream approach that Jean-Baptiste Yunès put forth, but if you must do a stream approach, you could do something like this (but it's a little ugly):
list.stream()
.skip(list.isEmpty ? 0 : new Random().nextInt(list.size()))
.findFirst();

Sometimes you may want to get a random item somewhere in the stream. If you want to get random items even after filtering your list, this code snippet will work for you:
List<String> items = Arrays.asList("A", "B", "C", "D", "E");
List<String> shuffledAndFilteredItems = items.stream()
.filter(value -> value.equals("A") || value.equals("B"))
//filter, map...
.collect(Collectors.collectingAndThen(
Collectors.toCollection(ArrayList::new),
list -> {
Collections.shuffle(list);
return list;
}));
String randomItem = shuffledAndFilteredItems
.stream()
.findFirst()
.orElse(null);
Of course there may be faster / optimized ways, but it allows you to do it all at once.

Is there a Java equivalent of Python's 'enumerate' function?

In Python, the enumerate function allows you to iterate over a sequence of (index, value) pairs. For example:
>>> numbers = ["zero", "one", "two"]
>>> for i, s in enumerate(numbers):
... print i, s
...
0 zero
1 one
2 two
Is there any way of doing this in Java?

For collections that implement the List interface, you can call the listIterator() method to get a ListIterator. The iterator has (amongst others) two methods - nextIndex(), to get the index; and next(), to get the value (like other iterators).
So a Java equivalent of the Python above might be:
import java.util.ListIterator;
import java.util.List;
List<String> numbers = Arrays.asList("zero", "one", "two");
ListIterator<String> it = numbers.listIterator();
while (it.hasNext()) {
System.out.println(it.nextIndex() + " " + it.next());
}
which, like the Python, outputs:
0 zero
1 one
2 two

I find this to be the most similar to the python approach.
Usage
public static void main(String [] args) {
List<String> strings = Arrays.asList("zero", "one", "two");
for(EnumeratedItem<String> stringItem : ListUtils.enumerate(strings)) {
System.out.println(stringItem.index + " " + stringItem.item);
}
System.out.println();
for(EnumeratedItem<String> stringItem : ListUtils.enumerate(strings, 3)) {
System.out.println(stringItem.index + " " + stringItem.item);
}
}
Output
0 zero
1 one
2 two
3 zero
4 one
5 two
Features
Works on any iterable
Does not create an in-memory list copy (suitable for large lists)
Supports native for each syntax
Accepts a start parameter which can be added to the index
Implementation
import java.util.Iterator;
public class ListUtils {
public static class EnumeratedItem<T> {
public T item;
public int index;
private EnumeratedItem(T item, int index) {
this.item = item;
this.index = index;
}
}
private static class ListEnumerator<T> implements Iterable<EnumeratedItem<T>> {
private Iterable<T> target;
private int start;
public ListEnumerator(Iterable<T> target, int start) {
this.target = target;
this.start = start;
}
#Override
public Iterator<EnumeratedItem<T>> iterator() {
final Iterator<T> targetIterator = target.iterator();
return new Iterator<EnumeratedItem<T>>() {
int index = start;
#Override
public boolean hasNext() {
return targetIterator.hasNext();
}
#Override
public EnumeratedItem<T> next() {
EnumeratedItem<T> nextIndexedItem = new EnumeratedItem<T>(targetIterator.next(), index);
index++;
return nextIndexedItem;
}
};
}
}
public static <T> Iterable<EnumeratedItem<T>> enumerate(Iterable<T> iterable, int start) {
return new ListEnumerator<T>(iterable, start);
}
public static <T> Iterable<EnumeratedItem<T>> enumerate(Iterable<T> iterable) {
return enumerate(iterable, 0);
}
}

Strictly speaking, no, as the enumerate() function in Python returns a list of tuples, and tuples do not exist in Java.
If however, all you're interested in is printing out an index and a value, then you can follow the suggestion from Richard Fearn & use nextIndex() and next() on an iterator.
Note as well that enumerate() can be defined using the more general zip() function (using Python syntax):
mylist = list("abcd")
zip(range(len(mylist)), mylist)
gives [(0, 'a'), (1, 'b'), (2, 'c'), (3, 'd')]
If you define your own Tuple class (see Using Pairs or 2-tuples in Java as a starting point), then you could certainly easily write your own zip() function in Java to make use of it (using the Tuple class defined in the link):
public static <X,Y> List<Tuple<X,Y>> zip(List<X> list_a, List<Y> list_b) {
Iterator<X> xiter = list_a.iterator();
Iterator<Y> yiter = list_b.iterator();
List<Tuple<X,Y>> result = new LinkedList<Tuple<X,Y>>();
while (xiter.hasNext() && yiter.hasNext()) {
result.add(new Tuple<X,Y>(xiter.next(), yiter.next()));
}
return result;
}
And once you have zip(), implementing enumerate() is trivial.
Edit: slow day at work, so to finish it off:
public static <X> List<Tuple<Integer,X>> enumerate (List<X> list_in) {
List<Integer> nums = new ArrayList<Integer>(list_in.size());
for (int x = 0; x < list_in.size(); x++) {
nums.add(Integer.valueOf(x));
}
return zip (nums, list_in);
}
Edit 2: as pointed out in the comments to this question, this is not entirely equivalent. While it produces the same values as Python's enumerate, it doesn't do so in the same generative fashion that Python's enumerate does. Thus for large collections this approach could be quite prohibitive.

Simple and straightforward
public static <T> void enumerate(Iterable<T> iterable, java.util.function.ObjIntConsumer<T> consumer) {
int i = 0;
for(T object : iterable) {
consumer.accept(object, i);
i++;
}
}
Sample usage:
void testEnumerate() {
List<String> strings = Arrays.asList("foo", "bar", "baz");
enumerate(strings, (str, i) -> {
System.out.println(String.format("Index:%d String:%s", i, str));
});
}

According to the Python docs (here), this is the closest you can get with Java, and it's no more verbose:
String[] numbers = {"zero", "one", "two"}
for (int i = 0; i < numbers.length; i++) // Note that length is a property of an array, not a function (hence the lack of () )
System.out.println(i + " " + numbers[i]);
}
If you need to use the List class...
List<String> numbers = Arrays.asList("zero", "one", "two");
for (int i = 0; i < numbers.size(); i++) {
System.out.println(i + " " + numbers.get(i));
}
*NOTE: if you need to modify the list as you're traversing it, you'll need to use the Iterator object, as it has the ability to modify the list without raising a ConcurrentModificationException.

Now with Java 8s Stream API together with the small ProtonPack library providing StreamUtils it can be achieved easily.
The first example uses the same for-each notation as in the question:
Stream<String> numbers = Arrays.stream("zero one two".split(" "));
List<Indexed<String>> indexedNumbers = StreamUtils.zipWithIndex(numbers)
.collect(Collectors.toList());
for (Indexed<String> indexed : indexedNumbers) {
System.out.println(indexed.getIndex() + " " + indexed.getValue());
}
Above although does not provide the lazy evaluation as in Python.
For that you must use the forEach() Stream API method:
Stream<String> numbers = Arrays.stream("zero one two".split(" "));
StreamUtils.zipWithIndex(numbers)
.forEach(n -> System.out.println(n.getIndex() + " " + n.getValue()));
The lazy evaluation can be verified with the following infinite stream:
Stream<Integer> infStream = Stream.iterate(0, i -> i++);
StreamUtils.zipWithIndex(infStream)
.limit(196)
.forEach(n -> System.out.println(n.getIndex() + " " + n.getValue()));

No. Maybe there are some libraries for supporting such a functionality. But if you resort to the standard libraries it is your job to count.

List<String> list = { "foo", "bar", "foobar"};
int i = 0;
for (String str : list){
System.out.println(i++ + str );
}

I think this should be the java functionality that resemble the python "enumerate" most, though it is quite complicated and inefficent. Basically, just map the list's indices to its elements, using ListIterator or Collector:
List<String> list = new LinkedList<>(Arrays.asList("one", "two", "three", "four"));
Map<Integer, String> enumeration = new Map<>();
ListIterator iter = list.listIterator();
while(iter.hasNext){
map.put(iter.nextIndex(), iter.next());
}
or using lambda expression:
Set<Integer, String> enumeration = IntStream.range(0, list.size()).boxed.collect(Collectors.toMap(index -> index, index -> list.get(index)));
then you can use it with an enhanced for loop:
for (Map.Entry<Integer, String> entry : enumeration.entrySet){
System.out.println(entry.getKey() + "\t" + entry.getValue());
}

By combining generics with anonymous interfaces, you can essentially create a factory method for handing enumeration. The Enumerator callback hides the messiness of the iterator underneath.
import java.util.Arrays;
import java.util.List;
import java.util.ListIterator;
public class ListUtils2 {
public static interface Enumerator<T> {
void execute(int index, T value);
};
public static final <T> void enumerate(final List<T> list,
final Enumerator<T> enumerator) {
for (ListIterator<T> it = list.listIterator(); it.hasNext();) {
enumerator.execute(it.nextIndex(), it.next());
}
}
public static final void enumerate(final String[] arr,
final Enumerator<String> enumerator) {
enumerate(Arrays.asList(arr), enumerator);
}
public static void main(String[] args) {
String[] names = { "John", "Paul", "George", "Ringo" };
enumerate(names, new Enumerator<String>() {
#Override
public void execute(int index, String value) {
System.out.printf("[%d] %s%n", index, value);
}
});
}
}
Result
[0] John
[1] Paul
[2] George
[3] Ringo
Extended Thoughts
Map, Reduce, Filter
I have taken this a step further and created map, reduce, and filter functions based on this concept.
Both Google's Guava and Apache common-collections dependencies include similar functionality. You can check them out as you wish.
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.ListIterator;
public class ListUtils {
// =========================================================================
// Enumerate
// =========================================================================
public static abstract interface Enumerator<T> {
void execute(int index, T value, List<T> list);
};
public static final <T> void enumerate(final List<T> list,
final Enumerator<T> enumerator) {
for (ListIterator<T> it = list.listIterator(); it.hasNext();) {
enumerator.execute(it.nextIndex(), it.next(), list);
}
}
// =========================================================================
// Map
// =========================================================================
public static interface Transformer<T, U> {
U execute(int index, T value, List<T> list);
};
public static final <T, U> List<U> transform(final List<T> list,
final Transformer<T, U> transformer) {
List<U> result = new ArrayList<U>();
for (ListIterator<T> it = list.listIterator(); it.hasNext();) {
result.add(transformer.execute(it.nextIndex(), it.next(), list));
}
return result;
}
// =========================================================================
// Reduce
// =========================================================================
public static interface Reducer<T, U> {
U execute(int index, T value, U result, List<T> list);
};
public static final <T, U> U reduce(final List<T> list,
final Reducer<T, U> enumerator, U result) {
for (ListIterator<T> it = list.listIterator(); it.hasNext();) {
result = enumerator.execute(it.nextIndex(), it.next(), result, list);
}
return result;
}
// =========================================================================
// Filter
// =========================================================================
public static interface Predicate<T> {
boolean execute(int index, T value, List<T> list);
};
public static final <T> List<T> filter(final List<T> list,
final Predicate<T> predicate) {
List<T> result = new ArrayList<T>();
for (ListIterator<T> it = list.listIterator(); it.hasNext();) {
int index = it.nextIndex();
T value = it.next();
if (predicate.execute(index, value, list)) {
result.add(value);
}
}
return result;
}
// =========================================================================
// Predefined Methods
// =========================================================================
// Enumerate
public static <T> String printTuples(List<T> list) {
StringBuffer buff = new StringBuffer();
enumerate(list, new Enumerator<T>() {
#Override
public void execute(int index, T value, List<T> list) {
buff.append('(').append(index).append(", ")
.append(value).append(')');
if (index < list.size() - 1) {
buff.append(", ");
}
}
});
return buff.toString();
}
// Map
public static List<String> intToHex(List<Integer> list) {
return transform(list, new Transformer<Integer, String>() {
#Override
public String execute(int index, Integer value, List<Integer> list) {
return String.format("0x%02X", value);
}
});
}
// Reduce
public static Integer sum(List<Integer> list) {
return reduce(list, new Reducer<Integer, Integer>() {
#Override
public Integer execute(int index, Integer value, Integer result,
List<Integer> list) {
return result + value;
}
}, 0);
}
// Filter
public static List<Integer> evenNumbers(List<Integer> list) {
return filter(list, new Predicate<Integer>() {
#Override
public boolean execute(int index, Integer value, List<Integer> list) {
return value % 2 == 0;
}
});
}
// =========================================================================
// Driver
// =========================================================================
public static void main(String[] args) {
List<Integer> numbers = Arrays.asList(8, 6, 7, 5, 3, 0, 9);
// Enumerate
System.out.printf("%-10s: %s%n", "Enumerate", printTuples(numbers));
// Map
System.out.printf("%-10s: %s%n", "Map", intToHex(numbers));
// Reduce
System.out.printf("%-10s: %d%n", "Reduce", sum(numbers));
// Filter
System.out.printf("%-10s: %s%n", "Filter", evenNumbers(numbers));
}
}

Pretty much the same syntax using Java8 Streams
ArrayList<String> numbers = new ArrayList<String>();
numbers.add("one");
numbers.add("two");
numbers.add("three");
numbers.stream().forEach(num ->
{
System.out.println(numbers.indexOf(num) + " " + num);
});