I need a Collection that sorts the element, but does not removes the duplicates.
I have gone for a TreeSet, since TreeSet actually adds the values to a backed TreeMap:
public boolean add(E e) {
return m.put(e, PRESENT)==null;
}
And the TreeMap removes the duplicates using the Comparators compare logic
I have written a Comparator that returns 1 instead of 0 in case of equal elements. Hence in the case of equal elements the TreeSet with this Comparator will not overwrite the duplicate and will just sort it.
I have tested it for simple String objects, but I need a Set of Custom objects.
public static void main(String[] args)
{
List<String> strList = Arrays.asList( new String[]{"d","b","c","z","s","b","d","a"} );
Set<String> strSet = new TreeSet<String>(new StringComparator());
strSet.addAll(strList);
System.out.println(strSet);
}
class StringComparator implements Comparator<String>
{
#Override
public int compare(String s1, String s2)
{
if(s1.compareTo(s2) == 0){
return 1;
}
else{
return s1.compareTo(s2);
}
}
}
Is this approach fine or is there a better way to achieve this?
EDIT
Actually I am having a ArrayList of the following class:
class Fund
{
String fundCode;
BigDecimal fundValue;
.....
public boolean equals(Object obj) {
// uses fundCode for equality
}
}
I need all the fundCode with highest fundValue
You can use a PriorityQueue.
PriorityQueue<Integer> pQueue = new PriorityQueue<Integer>();
PriorityQueue(): Creates a PriorityQueue with the default initial capacity (11) that orders its elements according to their natural ordering.
This is a link to doc: https://docs.oracle.com/javase/8/docs/api/java/util/PriorityQueue.html
I need all the fundCode with highest fundValue
If that's the only reason why you want to sort I would recommend not to sort at all. Sorting comes mostly with a complexity of O(n log(n)). Finding the maximum has only a complexity of O(n) and is implemented in a simple iteration over your list:
List<Fund> maxFunds = new ArrayList<Fund>();
int max = 0;
for (Fund fund : funds) {
if (fund.getFundValue() > max) {
maxFunds.clear();
max = fund.getFundValue();
}
if (fund.getFundValue() == max) {
maxFunds.add(fund);
}
}
You can avoid that code by using a third level library like Guava. See: How to get max() element from List in Guava
you can sort a List using Collections.sort.
given your Fund:
List<Fund> sortMe = new ArrayList(...);
Collections.sort(sortMe, new Comparator<Fund>() {
#Override
public int compare(Fund left, Fund right) {
return left.fundValue.compareTo(right.fundValue);
}
});
// sortMe is now sorted
In case of TreeSet either Comparator or Comparable is used to compare and store objects . Equals are not called and that is why it does not recognize the duplicate one
Instead of the TreeSet we can use List and implement the Comparable interface.
public class Fund implements Comparable<Fund> {
String fundCode;
int fundValue;
public Fund(String fundCode, int fundValue) {
super();
this.fundCode = fundCode;
this.fundValue = fundValue;
}
public String getFundCode() {
return fundCode;
}
public void setFundCode(String fundCode) {
this.fundCode = fundCode;
}
public int getFundValue() {
return fundValue;
}
public void setFundValue(int fundValue) {
this.fundValue = fundValue;
}
public int compareTo(Fund compareFund) {
int compare = ((Fund) compareFund).getFundValue();
return compare - this.fundValue;
}
public static void main(String args[]){
List<Fund> funds = new ArrayList<Fund>();
Fund fund1 = new Fund("a",100);
Fund fund2 = new Fund("b",20);
Fund fund3 = new Fund("c",70);
Fund fund4 = new Fund("a",100);
funds.add(fund1);
funds.add(fund2);
funds.add(fund3);
funds.add(fund4);
Collections.sort(funds);
for(Fund fund : funds){
System.out.println("Fund code: " + fund.getFundCode() + " Fund value : " + fund.getFundValue());
}
}
}
Add the elements to the arraylist and then sort the elements using utility Collections.sort,. then implement comparable and write your own compareTo method according to your key.
wont remove duplicates as well, can be sorted also:
List<Integer> list = new ArrayList<>();
Collections.sort(list,new Comparator<Integer>()
{
#Override
public int compare(Objectleft, Object right) {
**your logic**
return '';
}
}
)
;
I found a way to get TreeSet to store duplicate keys.
The problem originated when I wrote some code in python using SortedContainers. I have a spatial index of objects where I want to find all objects between a start/end longitude.
The longitudes could be duplicates but I still need the ability to efficiently add/remove specific objects from the index. Unfortunately I could not find the Java equivalent of the Python SortedKeyList that separates the sort key from the type being stored.
To illustrate this consider that we have a large list of retail purchases and we want to get all purchases where the cost is in a specific range.
// We are using TreeSet as a SortedList
TreeSet _index = new TreeSet<PriceBase>()
// populate the index with the purchases.
// Note that 2 of these have the same cost
_index.add(new Purchase("candy", 1.03));
Purchase _bananas = new Purchase("bananas", 1.45);
_index.add(new Purchase(_bananas);
_index.add(new Purchase("celery", 1.45));
_index.add(new Purchase("chicken", 4.99));
// Range scan. This iterator should return "candy", "bananas", "celery"
NavigableSet<PriceBase> _iterator = _index.subset(
new PriceKey(0.99), new PriceKey(3.99));
// we can also remove specific items from the list and
// it finds the specific object even through the sort
// key is the same
_index.remove(_bananas);
There are 3 classes created for the list
PriceBase: Base class that returns the sort key (the price).
Purchase: subclass that contains transaction data.
PriceKey: subclass used for the range search.
When I initially implemented this with TreeSet it worked except in the case where the prices are the same. The trick is to define the compareTo() so that it is polymorphic:
If we are comparing Purchase to PriceKey then only compare the price.
If we are comparing Purchase to Purchase then compare the price and the name if the prices are the same.
For example here are the compareTo() functions for the PriceBase and Purchase classes.
// in PriceBase
#Override
public int compareTo(PriceBase _other) {
return Double.compare(this.getPrice(), _other.getPrice());
}
// in Purchase
#Override
public int compareTo(PriceBase _other) {
// compare by price
int _compare = super.compareTo(_other);
if(_compare != 0) {
// prices are not equal
return _compare;
}
if(_other instanceof Purchase == false) {
throw new RuntimeException("Right compare must be a Purchase");
}
// compare by item name
Purchase _otherPurchase = (Purchase)_other;
return this.getName().compareTo(_otherChild.getName());
}
This trick allows the TreeSet to sort the purchases by price but still do a real comparison when one needs to be uniquely identified.
In summary I needed an object index to support a range scan where the key is a continuous value like double and add/remove is efficient.
I understand there are many other ways to solve this problem but I wanted to avoid writing my own tree class. My solution seems like a hack and I am surprised that I can't find anything else. if you know of a better way then please comment.
Related
I have a collection of strings in an array like this:
ArrayList<String> collection = new ArrayList<>();
That stores:
collection: ["(,0,D=1", "(,1,D=2", "),2,D=2", "),3,D=1", "(,4,D=1", "(,5,D=2", "),6,D=2", "),7,D=1"]
I have a lot of d=1 and d=2, as you can see. How do I organize this from 1 first to 2? I tried to use a for loop but the list can contain an infinite number of d=x's. Can you help me organize?
Also, please help me so I don't change the ORDER of any numbers. Example:
collection: ["(,0,D=1", "),3,D=1", "(,4,D=1", "),7,D=1", "(,1,D=2", "),2,D=2", "(,5,D=2", "),6,D=2"]
So like, every parentheses will be aligned.
I should note that collection[0] = "(,0,D=1"
You should use a class for the items, not a string, e.g. Class Item {char c; int i; int depth;} and ArrayList. Then you can easily sort the list with a custom Comparator.
You can implement your own Comparator to do the sorting. A Comparator is a sorting algorithms that you define for your application which written in programming language. Give Collections.sort() a Comparator basically you teach Java how you want to sort the list. And it will sort the list for you.
This implementation is based on the following assumptions:
The comparison will only take effect on the first D=x pattern, subsequent will be ignored.
Element is sorted in ascending order base on x.
Elements do not have D=x will be placed at the back
class DeeEqualComparator implements Comparator<String> {
private static final String REGEX = "D=([0-9])+";
#Override
public int compare(String s1, String s2) {
// find a D=x pattern from the element
Matcher s1Matcher = Pattern.compile(REGEX).matcher(s1);
Matcher s2Matcher = Pattern.compile(REGEX).matcher(s2);
boolean s1Match = s1Matcher.find();
boolean s2Match = s2Matcher.find();
if (s1Match && s2Match) {
// if match is found on s1 and s2, return their integer comparison result
Integer i1 = Integer.parseInt(s1Matcher.group(1));
Integer i2 = Integer.parseInt(s2Matcher.group(1));
return i1.compareTo(i2);
} else if (s1Match) {
// if only s1 found a match
return -1;
} else if (s2Match) {
// if only s2 found a match
return 1;
} else {
// if no match is found on both, return their string comparison result
return s1.compareTo(s2);
}
}
Test run
public static void main(String[] args) {
String[] array = {
// provided example
"(,0,D=1", "(,1,D=2", "),2,D=2", "),3,D=1", "(,4,D=1", "(,5,D=2", "),6,D=2", "),7,D=1"
// extra test case
, "exception-5", "exception-0", "D=68" };
List<String> list = Arrays.asList(array);
Collections.sort(list, new DeeEqualComparator());
System.out.print(list);
}
output
[(,0,D=1, ),3,D=1, (,4,D=1, ),7,D=1, (,1,D=2, ),2,D=2, (,5,D=2, ),6,D=2, D=68, exception-0, exception-5]
So I'm trying to sort this Arraylist of "Movies" type, alphabetically based on the actor's name and containing no duplicates, which means using a treeSet.
So I have implemented the "comparable" interface and then didn't know what to do after for making the list sorted by alphabets
public int compareTo(Movie m){
for(Movie movileTitle: moviesList){
return movileTitle.getActor().compareTo(m.getActor());
}
return 0;
}
//this is where I'm trying add the arrayList of the movies and sort them using a treeSet
public TreeSet<Movie> movieListByOrder(){
TreeSet<Movie> movieTemp = new TreeSet<>();
//Note, "allMoviesList" is the the list that contains all movies(duplicates) that I'm trying
to add to the treeSet, so that it gets rid of duplicates and that is also ordered.
movieTemp.addAll(allMoviesList);
System.out.println(movieTemp);
return movieTemp;
}
Continuing in the direction in which you are already headed, you may try using an inline comparator when creating your TreeSet:
Set<Movie> treeSet = new TreeSet<Movie>(new Comparator<Movie>() {
#Override
public int compare(Movie m1, Movie m2) {
if (m1 == null) {
if (m2 != null) return -1;
else return 0;
}
else if (m2 == null) {
return 1;
}
else {
return m1.getActor().compareTo(m2.getActor());
}
}
});
A TreeSet is a sorted collection, meaning it will maintain the sort order imposed by the above comparator. So if you want to print out the contents of the set in order, you need only iterate through it.
All that looks archaic, sorry.
Try to look into this code and decide if it's more interesting.
movies.stream()
.sorted(comparing(Movie::getActor))
.distinct()
.collect(toList());
PS: play with performance tuning for extra credit
I have an ArrayList of Items and I want to be able remove one Item from the list by entering only one Item attribute, for example its number (int ItemNumber). I also wanna do the same when I check Item quantities.
These are my equals() & contains() methods, do I need to make any changes here?
public boolean contains(T anEntry) {
boolean found = false;
for (int index = 0; !found && (index < numberOfEntries); index++) {
if (anEntry.equals(list[index]))
found = true;
}//end for
return found;
} // end contains
public boolean equals(Object object){
Item item = (Item) object;
if (itemNo == item.itemNo)
return true;
return false;
}
If you change the class Item equals() and compareTo() methods, so that they check only one object field, such as a quantity, it could result in strange behavior in other parts of your application. For example, two items with different itemNo, itemName, and itemPrice, but with the same quantities could be considered equal. Besides, you wouldn't be able to change the comparison attribute without changing the equals() code every time.
Also, creating a custom contains() method makes no sense, since it belongs to the ArrayList class, and not to Item.
If you can use Java 8, a clean way to do it is to use the new Collection's removeIf method:
Suppose you have an Item class with the num and name properties:
class Item {
final int num;
final String name;
Item(int num, String name) {
this.num = num;
this.name = name;
}
}
Given a List<Item> called items and an int variable called number, representing the number of the item you want to remove, you could simply do:
items.removeIf(item -> item.num == number);
If you are unable to use Java 8, you can achieve this by using custom comparators, binary search, and dummy objects.
You can create a custom comparator for each attribute you need to look for. The comparator for num would look like this:
class ItemNumComparator implements Comparator<Item> {
#Override
public int compare(Item a, Item b) {
return (a.num < b.num) ? -1 : ((a.num == b.num) ? 0 : 1);
}
}
Then you can use the comparator to sort and search for the desired elements in your list:
public static void main(String[] args) {
List<Item> items = new ArrayList<>();
items.add(new Item(2, "ball"));
items.add(new Item(5, "cow"));
items.add(new Item(3, "gum"));
Comparator<Item> itemNumComparator = new ItemNumComparator();
Collections.sort(items, itemNumComparator);
// Pass a dummy object containing only the relevant attribute to be searched
int index = Collections.binarySearch(items, new Item(5, ""), itemNumComparator);
Item removedItem = null;
// binarySearch will return -1 if it does not find the element.
if (index > -1) {
// This will remove the element, Item(5, "cow") in this case, from the list
removedItem = items.remove(index);
}
System.out.println(removedItem);
}
To search for another field like name, for example, you would need to create a name comparator and use it to sort and run the binary search on your list.
Note this solution has some drawbacks though. Unless you are completely sure that the list didn't change since the last sort, you must re-sort it before running the binarySearch() method. Otherwise, it may not be able to find the correct element. Sorting complexity is O(nlogn), so running it multiple times can get quite expensive depending on the size of your list.
Do you want to remove an object at a specific index? I'm not entirely sure what you mean by 'number field'. If so, jump to method: remove(int):
http://docs.oracle.com/javase/7/docs/api/java/util/ArrayList.html#remove%28int%29
EDIT: If you want to find/adjust a field of an object in the Array list, you can do this (piece of my own code):
public boolean studentHasISBN(ArrayList<Student> st, String s){
for(Student j : st) {
if(s.equals(j.getRentedBookISBN()))
return true;
}
return false;
}
All you have to do is iterate through the list, and search through the field that you want to find. Then use the remove(int) method.
simply use the remove function of ArrayLists in Java:
theNameOfYourArrayList.remove(ItemNumber);
to remove the element which has the index (int ItemNumber)
to check if the element with item number (int ItemNumber) exists in your ArrayList (hypothetically called theNameOfYourArrayList):
theNameOfYourArrayList.get(ItemNumber);
I'm going to assume that by 'number field' you mean that you invoked ArrayList with the Integer data type. I have a few different solutions to your problem:
ArrayLists, assuming that the ArrayList is ArrayList<Integer> numList = new ArrayList<Integer>(); you can simply write a method that will search 'numList' and delete the index that the number is. The problem is, contains and find in ArrayLists can be slow.
public void deleteNumField(int field) {
// this will stop any error if field isn't actually in numList
// and it will remove the first index of field in the ArrayList
if(numList.contains(field)) numList.remove(numList.find(field));
}
HashSets, HashSets are a handy data type that is like an ArrayList, except, its data is its 'index' (sortof). I won't go in depth about how they work, but I will say that searching in them is considered O(1). This will make your deletion really easy, and fast. Note: the HashSet assumes there are no duplicate numbers, if there are use a HashMap.
HashSet<Integer> numList = new HashSet<Integer>();
public void deleteNumField(int field) {
// this will stop errors from attempting to remove a
// non-existant element, and remove it if it exists.
if(numList.contains(field)) numList.remove(field);
}
For more information on HashMaps, HashSets and ArrayLists, see:
http://docs.oracle.com/javase/8/docs/api/
This is what i have so far, i'm trying to sort a bunch of List<String>'s based on the value of an index.
LinkedHashSet<List<String>> sorted = new LinkedHashSet<List<String>>();
How do i sort the LinkedHashSet in order from Highest to Lowest index 2 value of the List's?
Example input:
List<String> data1 = Database.getData(uuid);
double price = Double.valueOf(data1.get(2))
data1.add("testval");
data1.add("testval");
data1.add("100.00");
sorted.add(data1);
and on another seperate List:
List<String> data2 = Database.getData(uuid);
double price = Double.valueOf(data2.get(2))
data2.add("anotherval");
data2.add("anotherval");
data2.add("50.00");
sorted.add(data2);
Output of the sorted LinkedHashSet in descending order.
testval testval 100.00
anotherval anotherval 50.00
Sorry if this is confusing, im not sure where to go about sorting like this.
Create a new class to represent you complex objects. There is no need to store multiple values in a list when you can do it in objects.
public class ComplexObject {
private String description1;
private String description2;
private Double value;
public ComplexObject(String description1, String description2, Double value) {
this.description1 = description1;
this.description2 = description2;
this.value = value;
}
public void setDescription1(String description1) {
this.description1 = description1;
}
public String getDescription1() {
return description1;
}
public void setDescription2(String description2) {
this.description2 = description2;
}
public String getDescription2() {
return description2;
}
public void setValue(Double value) {
this.value = value;
}
public Double getValue() {
return value;
}
}
Then add elements to the list and sort it using a new, custom, comparator:
public static void main(String[] args) {
List<ComplexObject> complexObjectList = new ArrayList<ComplexObject>();
//add elements to the list
complexObjectList.add(new ComplexObject("testval","testval",100.00d));
complexObjectList.add(new ComplexObject("anotherval","anotherval",50.00d));
//sort the list in descending order based on the value attribute of complexObject
Collections.sort(complexObjectList, new Comparator<ComplexObject>() {
public int compare(ComplexObject obj1, ComplexObject obj2) {
return obj2.getValue().compareTo(obj1.getValue()); //compares 2 Double values, -1 if less , 0 if equal, 1 if greater
}
});
//print objects from sorted list
for(ComplexObject co : complexObjectList){
System.out.println(co.getDescription1()+" "+co.getDescription2()+" "+co.getValue());
}
}
Output:
testval testval 100.0
anotherval anotherval 50.0
Firstly, you shouldn't use a LinkedHashSet but a TreeSet. LinkedHashSet will retain the insertion order without sorting.
Secondly, you need to initialize your TreeSet with a Comparator that compares based on whichever value of your List is required, that is, if you know the index of the String that will represent a double value in advance. Otherwise I would recommend using custom objects instead of List.
If you decide to use custom objects, you don't necessarily need to initialize your TreeSet with a Comparator as second argument.
Instead, you could have your custom objects implement Comparable, and implement a one-time comparation logic there.
It all depends on whether you only need to sort in a particular order.
Finally, custom objects will require you to override equals and hashCode.
First, and extracted from Oracle's Java reference:
This linked list defines the iteration ordering, which is the order in which elements were inserted into the set
So you can't sort your data just inserting it into the LinkedHashSet.
You may be confusing that set implementation with SortedSet. SortedSet allows you to pass a comparator which will determine the elements order in the data structure.
On the other hand, I don't know whether you chose you List<String> arbitrarily but it seems to me a wiser option to aggregate your the 3 strings as a class attributes. The point is that, if your elements are always going to be 3 elements, being the last one a double value: Why do you need a dynamic structure as a List?
EDIT
Here you have a possible better implementation of what you want:
public class Element
{
public Element(String a, String b, double val) {
this.a = a;
this.b = b;
this.val = val;
}
#Override
public String toString() {
return a + "\t" + b + "\t" + val;
}
public String a;
public String b;
public double val;
}
And you can use this class to store your elements. An example of use:
SortedSet<Element> sorted = new TreeSet<>(new Comparator<Element>() {
#Override
public int compare(Element o1, Element o2) {
return (new Double(o1.val)).compareTo(o2.val);
}
});
sorted.add(new Element("testval", "testval", 100.0));
sorted.add(new Element("anotherval", "anotherval", 50.0));
for(Element el: sorted)
{
System.out.println(el);
}
Note that the comparator is given as an instance of an anonympous inner class implementing Java's Comparator interface.
I know that TreeSet stores objects in a sorted manner. But is there a way that i can customize the order?
For example if i have a treeSet:
TreeSet t1 = new TreeSet();
t1.add("c");
t1.add("d");
t1.add("a");
And now if i iterate over it>
Iterator it1 =t1.iterator();
while(it1.hasNext()){
Object o1 = it1.next();
System.out.println(o1);
}
i will always get the order as: a>c>d, however i want it to return the order the same as i added the elements in it i.e c>d>a?
Use LinkedHashSet for it,
TreeSet sorts the element, and for string it sorts based on natural order( this is how its comparator is implemented), If you want to manage the insertion order then you need to user LinkedHashSet
and if you don't need the uniquness (feature of set) then go for List
Since you mention about being bound to use TreeSet, something like this comes to my mind:
Set<String> result = new TreeSet<String>(new Comparator<String>(){
#Override
public int compare(String arg0, String arg1) {
return returnCode(arg0).compareTo(returnCode(arg1));
}
});
where:
private Integer returnCode(String p){
int code = 0;
String id = p.toLowerCase();
if ("a".equalsIgnoreCase(id)) code = 3;
else if ("b".equalsIgnoreCase(id)) code = 2;
else if ("c".equalsIgnoreCase(id)) code = 1;
//etc
return new Integer(code);
}
So basically you are implementing your own comparator which is nothing but assigning certain integer values to the inserted String (which i assume you know already).
NOTE: this solution will not work, incase you do not catch the option in your returnCode() method. I assume you already know the data that is being fed to the TreeSet.
It is much more easy:
TreeSet treeSetObj = new TreeSet(Collections.reverseOrder());
I have got a good answer. It's my first one.
import java.util.*;
class M
{
public static void main(String args[])
{
TreeSet<String> t=new TreeSet<>(new MyComparator());
t.add("c");
t.add("d");
t.add("a");
System.out.println(t);//[c,d,a] not [a,c,d] ....your answer
}
}
class MyComparator implements Comparator
{
public int compare(Object o1,Object o2)
{
String i1=(String)o1;
String i2=(String)o2;
return +1; //Store elements as they come
}
}
What basically +1 does store the elements in the order in which they come.
Always remember
if I write return +1 it means to store the elements in the order in which they come.
if I write return -1 it means to store the elements in reverse order in which they come.