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.
I am trying to use a comparator to sort my List 2 based on List 1.
So the two lists are:
ListA = [2,3,4]
ListB = [8,2,4]
I need to sort list 2 based on list 1.
Expected output:
List1: [2,3,4]
List2: [2,8,4]
Here is the code I am trying
Collections.sort(list1);
Collections.sort(list2,new Comparator<Integer>(){
public int compare(Integer left,Integer right){
return Integer.compare(list1.indexOf(left),list1.indexOf(right));
}
})
Here sorting will be based on index of List1 elements. The comparator above is not working for me Please help?
It's not exactly clear to me from your example what you're trying to do, so this answer might not actually address your question. If what you're trying to do is to apply the sort-order of ListA to both ListA and ListB then you can do it as follows.
Create an auxiliary class that allows you to pair up the two lists with respect to the same index:
private class IntPair implements Comparable<IntPair> {
int i1;
int i2;
public IntPair(int i1, int i2) {
this.i1 = i1;
this.i2 = i2;
}
public int compareTo(IntPair other) {
return Integer.compare(i1, other.i1);
}
}
Then create a new list of IntPair instances from your original lists and sort it.
int size = list1.size();
List<IntPair> aux = new ArrayList(size);
for (int i = 0; i < size; i++) {
aux.add(new IntPair(list1.get(i), list2.get(i)))
}
Collections.sort(aux);
Finally, copy back the values of the resulting pair list onto your original lists.
for (int i = 0; i < size; i++) {
IntPair ip = aux.get(i);
list1.set(i, ip.i1);
list2.set(i, ip.i2);
}
Note that in terms of algorithmic complexity, this approach still is O(nlogn).
This might be a little heavy-weight, but it gets the job done and it's generic.
Essentially, you can zip-up two lists, sort/compare them based off an index, unzip them, and return the result.
This does not modify the list in-place. I tried to give the variables meaningful, but discreet names. Please acknowledge me for clarification.
import java.util.*;
public class SortUtil {
public static void main(String[] args) {
List<Integer> indicies = Arrays.asList(3, 2, 4);
List<Integer> values = Arrays.asList(8, 2, 4);
List<Integer> sorted = doSort(indicies, values, new TupleComparator<Integer>(0, 1) {
#Override
public int doCompare(Integer valueA, Integer valueB) {
return Integer.compare(valueA, valueB);
}
});
System.out.println(sorted); // [2, 8, 4]
}
public static List<Integer> doSort(List<Integer> listA, List<Integer> listB, TupleComparator<Integer> comparator) {
List<Tuple<Integer>> tuples = zip(listA, listB);
Collections.sort(tuples, comparator);
return unzip(tuples, comparator.getValuePos());
}
private static <E> List<E> unzip(List<Tuple<E>> tuples, int index) {
List<E> list = new ArrayList<E>();
for (Tuple<E> tuple : tuples) {
list.add(tuple.getData().get(index));
}
return list;
}
private static <E> List<Tuple<E>> zip(List<E> listA, List<E> listB) {
List<Tuple<E>> listC = new ArrayList<Tuple<E>>();
for (int i = 0; i < listA.size(); i++) {
listC.add(new Tuple<E>(Arrays.asList(listA.get(i), listB.get(i))));
}
return listC;
}
private static abstract class TupleComparator<E> implements Comparator<Tuple<E>> {
private int indexPos;
private int valuePos;
public TupleComparator(int indexPos, int valuePos) {
this.indexPos = indexPos;
this.valuePos = valuePos;
}
public int compare(Tuple<E> left, Tuple<E> right) {
E indexA = left.getData().get(this.getIndexPos());
E indexB = right.getData().get(this.getIndexPos());
return doCompare(indexA, indexB);
}
public int getIndexPos() { return indexPos; }
public int getValuePos() { return valuePos; }
public abstract int doCompare(E valueA, E valueB);
}
private static class Tuple<T> {
private List<T> data;
public Tuple(List<T> data) { this.data = data; }
public List<T> getData() {return data; }
}
}
I have the following map: Map<Integer,String[]> map = new HashMap<Integer,String[]>();
The keys are integers and the values are arrays (could also be replaced by lists).
Now, I would like to get all possible combinations of the values among the keys. For example, let's say the map contains the following entries:
key 1: "test1", "stackoverflow"
key 2: "test2", "wow"
key 3: "new"
The combinations consists of
("test1","test2","new")
("test1","wow","new")
("stackoverflow", "test2", "new")
("stackoverflow", "wow", "new")
For this I imagine a method boolean hasNext() which returns true if there is a next pair and a second method which just returns the next set of values (if any).
How can this be done? The map could also be replaced by an other data structure.
The algorithm is essentially almost the same as the increment algorithm for decimal numbers ("x -> x+1").
Here the iterator class:
import java.util.Iterator;
import java.util.Map;
import java.util.NoSuchElementException;
import java.util.TreeSet;
public class CombinationsIterator implements Iterator<String[]> {
// Immutable fields
private final int combinationLength;
private final String[][] values;
private final int[] maxIndexes;
// Mutable fields
private final int[] currentIndexes;
private boolean hasNext;
public CombinationsIterator(final Map<Integer,String[]> map) {
combinationLength = map.size();
values = new String[combinationLength][];
maxIndexes = new int[combinationLength];
currentIndexes = new int[combinationLength];
if (combinationLength == 0) {
hasNext = false;
return;
}
hasNext = true;
// Reorganize the map to array.
// Map is not actually needed and would unnecessarily complicate the algorithm.
int valuesIndex = 0;
for (final int key : new TreeSet<>(map.keySet())) {
values[valuesIndex++] = map.get(key);
}
// Fill in the arrays of max indexes and current indexes.
for (int i = 0; i < combinationLength; ++i) {
if (values[i].length == 0) {
// Set hasNext to false if at least one of the value-arrays is empty.
// Stop the loop as the behavior of the iterator is already defined in this case:
// the iterator will just return no combinations.
hasNext = false;
return;
}
maxIndexes[i] = values[i].length - 1;
currentIndexes[i] = 0;
}
}
#Override
public boolean hasNext() {
return hasNext;
}
#Override
public String[] next() {
if (!hasNext) {
throw new NoSuchElementException("No more combinations are available");
}
final String[] combination = getCombinationByCurrentIndexes();
nextIndexesCombination();
return combination;
}
private String[] getCombinationByCurrentIndexes() {
final String[] combination = new String[combinationLength];
for (int i = 0; i < combinationLength; ++i) {
combination[i] = values[i][currentIndexes[i]];
}
return combination;
}
private void nextIndexesCombination() {
// A slightly modified "increment number by one" algorithm.
// This loop seems more natural, but it would return combinations in a different order than in your example:
// for (int i = 0; i < combinationLength; ++i) {
// This loop returns combinations in the order which matches your example:
for (int i = combinationLength - 1; i >= 0; --i) {
if (currentIndexes[i] < maxIndexes[i]) {
// Increment the current index
++currentIndexes[i];
return;
} else {
// Current index at max:
// reset it to zero and "carry" to the next index
currentIndexes[i] = 0;
}
}
// If we are here, then all current indexes are at max, and there are no more combinations
hasNext = false;
}
#Override
public void remove() {
throw new UnsupportedOperationException("Remove operation is not supported");
}
}
Here the sample usage:
final Map<Integer,String[]> map = new HashMap<Integer,String[]>();
map.put(1, new String[]{"test1", "stackoverflow"});
map.put(2, new String[]{"test2", "wow"});
map.put(3, new String[]{"new"});
final CombinationsIterator iterator = new CombinationsIterator(map);
while (iterator.hasNext()) {
System.out.println(
org.apache.commons.lang3.ArrayUtils.toString(iterator.next())
);
}
It prints exactly what's specified in your example.
P.S. The map is actually not needed; it could be replaced by a simple array of arrays (or list of lists). The constructor would then get a bit simpler:
public CombinationsIterator(final String[][] array) {
combinationLength = array.length;
values = array;
// ...
// Reorganize the map to array - THIS CAN BE REMOVED.
I took this as a challenge to see whether the new Java 8 APIs help with these kind of problems. So here's my solution for the problem:
public class CombinatorIterator implements Iterator<Collection<String>> {
private final String[][] arrays;
private final int[] indices;
private final int total;
private int counter;
public CombinatorIterator(Collection<String[]> input) {
arrays = input.toArray(new String[input.size()][]);
indices = new int[arrays.length];
total = Arrays.stream(arrays).mapToInt(arr -> arr.length)
.reduce((x, y) -> x * y).orElse(0);
counter = 0;
}
#Override
public boolean hasNext() {
return counter < total;
}
#Override
public Collection<String> next() {
List<String> nextValue = IntStream.range(0, arrays.length)
.mapToObj(i -> arrays[i][indices[i]]).collect(Collectors.toList());
//rolling carry over the indices
for (int j = 0;
j < arrays.length && ++indices[j] == arrays[j].length; j++) {
indices[j] = 0;
}
counter++;
return nextValue;
}
}
Note that I don't use a map as an input as the map keys actually don't play any role here. You can use map.values() though to pass in the input for the iterator. With the following test code:
List<String[]> input = Arrays.asList(
new String[] {"such", "nice", "question"},
new String[] {"much", "iterator"},
new String[] {"very", "wow"}
);
Iterator<Collection<String>> it = new CombinatorIterator(input);
it.forEachRemaining(System.out::println);
the output will be:
[such, much, very]
[nice, much, very]
[question, much, very]
[such, iterator, very]
[nice, iterator, very]
[question, iterator, very]
[such, much, wow]
[nice, much, wow]
[question, much, wow]
[such, iterator, wow]
[nice, iterator, wow]
[question, iterator, wow]
I have following data in string(comma format) (Name,Mark)
A,20,B,10,C,30
I want to convert into Descending order like :
C,30,A,29,B,10
Please help me how can i implement in android ?
Here is code what i have prepared....
public class custom_sort {
public String name;
public int mark;
public custom_sort(String a, int b) {
// TODO Auto-generated constructor stub
name = a;
mark = b;
}
void setname(String s)
{
name=s;
}
void setmark(int s)
{
mark = s;
}
String getname()
{
return(name);
}
int getmark()
{
return(mark);
}
}
Thanks in Advance,
There are many ways of doing this. I see that you have already made a class: custom_sort. We can use this to sort it, if we just make it comparable. We do this by implementing the Comparable interface.
public class custom_sort implements Comparable{
Then all you need to do is implement the one required method:
#Override
public int compareTo(custom_sort cs) {
/*
This method should return 0 if the two objects are equal,
1 if this is biggest
and -1 if cs is biggest */
}
Then you can put all the custom_sorts in an List and just do Arrays.sort(yourArray).
You could also check out this post Android sort array
Use ArrayList rather than array of string.
This might not be the best solution but it works.
Initialize the ArrayList
ArrayList<String> names = new ArrayList<>();
ArrayList<Integer> scores = new ArrayList<>();
Fill data inside the ArrayList
names.add("A");
scores.add(20);
....
Now sort them
private void sortScoreAndName() {
for (int i = 0; i < scores.size(); i++) {
for (int j = 0; j < i; j++) {
if (scores.get(i) > scores.get(j))
swap(i, j);
}
}
}
private void swap(int i, int j) {
int tempSco = scores.get(i);
String tempName = names.get(i);
scores.remove(i);
names.remove(i);
scores.add(i, scores.get(j));
names.add(i, names.get(j));
scores.remove(j);
names.remove(j);
scores.add(j, tempSco);
names.add(j, tempName);
}
Now your ArrayList is in descending order. You can get the corresponding data using.
names.get(poition);
scores.get(position);
try this:
import android.support.v4.util.Pair;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;
String[] values = {"A", "20", "B", "10", "C", "30"};
List<Pair> pairs = new ArrayList<>();
for (int i = 0; i < values.length; i += 2) {
pairs.add(new Pair<>(values[i], Integer.parseInt(values[i + 1])));
}
Collections.sort(pairs, new Comparator<Pair>() {
#Override
public int compare(Pair lhs, Pair rhs) {
return ((Integer) rhs.second).compareTo((Integer) lhs.second);
}
});
The List pairs is now sorted like you wish. Check out the Pair class i used http://developer.android.com/reference/android/util/Pair.html
And if you want to convert it back following code will help you:
String[] newValues = new String[values.length];
int i = 0;
for (Pair pair : pairs) {
newValues[i] = (String) pair.first;
newValues[i + 1] = Integer.toString((Integer) pair.second);
i += 2;
}
Is there an easy way of finding the MAX number from the list where number is stored in x.y.z format? e.g. To manage some system versions.
I have tried Collection.max(list) and that does not work.
Sample Code:
public static void main(String args[])
{
List<String> list = new ArrayList<String>();
list.add("1.0.0");
list.add("1.1.0");
list.add("1.9.0");
list.add("1.10.0");
System.out.println(Collections.max(list));
}
Expected: 1.10.0
Result: 1.9
Thanks for your time.
Try to use this one :
Collections.max(myList, new Comparator<String>() {
#Override
public int compare(String lhs, String rhs) {
String[] first = lhs.split("\\.");
String[] second = rhs.split("\\.");
for (int i = 0; i < first.length; i++) {
if(Integer.valueOf(first[i]) > Integer.valueOf(second[i])) {
return 1;
}
if(Integer.valueOf(first[i]) < Integer.valueOf(second[i])) {
return -1;
}
}
return 0;
}
});
Well for one thing, you need to ensure that Java knows they are numbers - at the moment they're just Strings, and strings sort lexigraphically (i.e. in "alphabetical order").
My approach to this would be to create a small class that implements Comparable, which will then work automatically with sorting and comparison logic. Something like this perhaps:
public class VersionNumber implements Comparable<VersionNumber> {
public final int major;
public final int minor;
public final int patch;
// Constructor etc. elided
public int compareTo(VersionNumber other) {
if (other.major != major) return major - other.major;
if (other.minor != minor) return minor - other.minor;
return patch - other.patch;
}
}
Parsing the string to create instances of this class is left as an exercise to the reader!
You may have to write a custom Comparator for comparing version number strings:
public class VersionComparator extends Comparator<String> {
#Override
public int compare(String o1, String o2) {
// Get major/minor/revison numbers by splitting strings at dots
String[] p1 = o1.split("\\.");
String[] p2 = o2.split("\\.");
// Compare major versions then minor then revision until a difference found
for(int i = 0; i < (p1.length < p2.length) ? p1.length : p2.length; i++) {
int result = Integer.valueOf(p1[i]).compareTo(Integer.valueOf(p2[i]));
if(result != 0) return result;
}
// Return zero if they're identical
return 0;
}
}
The you can use this comparator with the Collections.max function:
Collections.max(list, new VarsionComparator());
You can use version of max with the specified comparator:
System.out.println(Collections.max(list, new Comparator<String>() {
public int compare(String s1, String s2)
{
StringTokenizer st1 = new StringTokenizer(s1,".");
StringTokenizer st2 = new StringTokenizer(s2,".");
int res = 0;
String t1, t2;
while(st1.hasMoreTokens() && st2.hasMoreTokens())
{
t1 = st1.nextToken();
t2 = st2.nextToken();
res = Integer.valueOf(t1).compareTo(Integer.valueOf(t2));
}
if(res == 0)
{
res = st1.hasMoreTokens() ? 1 : (st2.hasMoreTokens() ? -1 : 0);
}
return res;
}
public boolean equals(Object obj) { return false; }
}));
This will give you 1.9 because it will not consider second number to be 10, it will treat it as 1 first and then 9
Edit
If you want to do it manually, then
Split your number on basis of "."
Check manually which number is greater.