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Sorting an arraylist with integers

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]

How to sort a Map by Key and Value, whereas the Val is a Map/List itself

I am having a hard time understanding the right syntax to sort Maps which values aren't simply one type, but can be nested again.
I'll try to come up with a fitting example here:
Let's make a random class for that first:
class NestedFoo{
int valA;
int valB;
String textA;
public NestedFoo(int a, int b, String t){
this.valA = a;
this.valB = b;
this.textA = t;
}
}
Alright, that is our class.
Here comes the list:
HashMap<Integer, ArrayList<NestedFoo>> sortmePlz = new HashMap<>();
Let's create 3 entries to start with, that should show sorting works already.
ArrayList<NestedFoo> l1 = new ArrayList<>();
n1 = new NestedFoo(3,2,"a");
n2 = new NestedFoo(2,2,"a");
n3 = new NestedFoo(1,4,"c");
l1.add(n1);
l1.add(n2);
l1.add(n3);
ArrayList<NestedFoo> l2 = new ArrayList<>();
n1 = new NestedFoo(3,2,"a");
n2 = new NestedFoo(2,2,"a");
n3 = new NestedFoo(2,2,"b");
n4 = new NestedFoo(1,4,"c");
l2.add(n1);
l2.add(n2);
l2.add(n3);
l2.add(n4);
ArrayList<NestedFoo> l3 = new ArrayList<>();
n1 = new NestedFoo(3,2,"a");
n2 = new NestedFoo(2,3,"b");
n3 = new NestedFoo(2,2,"b");
n4 = new NestedFoo(5,4,"c");
l3.add(n1);
l3.add(n2);
l3.add(n3);
l3.add(n4);
Sweet, now put them in our Map.
sortmePlz.put(5,l1);
sortmePlz.put(2,l2);
sortmePlz.put(1,l3);
What I want now, is to sort the Entire Map first by its Keys, so the order should be l3 l2 l1.
Then, I want the lists inside each key to be sorted by the following Order:
intA,intB,text (all ascending)
I have no idea how to do this. Especially not since Java 8 with all those lambdas, I tried to read on the subject but feel overwhelmed by the code there.
Thanks in advance!
I hope the code has no syntatical errors, I made it up on the go

You can use TreeSet instead of regular HashMap and your values will be automatically sorted by key:
Map<Integer, ArrayList<NestedFoo>> sortmePlz = new TreeMap<>();
Second step I'm a little confused.
to be sorted by the following Order: intA,intB,text (all ascending)
I suppose you want to sort the list by comparing first the intA values, then if they are equal compare by intB and so on. If I understand you correctly you can use Comparator with comparing and thenComparing.
sortmePlz.values().forEach(list -> list
.sort(Comparator.comparing(NestedFoo::getValA)
.thenComparing(NestedFoo::getValB)
.thenComparing(NestedFoo::getTextA)));

I'm sure there are way of doing it with lambda but it is not actually required. See answer from Schidu Luca for a lambda like solution.
Keep reading if you want an 'old school solution'.
You cannot sort a map. It does not make sense because there is no notion of order in a map. Now, there are some map objects that store the key in a sorted way (like the TreeMap).
You can order a list. In your case, makes the class NestedFoo comparable (https://docs.oracle.com/javase/8/docs/api/java/lang/Comparable.html). Then you can invoke the method Collections.sort (https://docs.oracle.com/javase/8/docs/api/java/util/Collections.html#sort-java.util.List-) on your lists.

Use TreeMap instead of HashMap, it solves the 1st problem: ordering entries by key.
After getting the needed list from the Map, you can sort the ArrayList by valA, valB, text:
l1.sort(
Comparator.comparing(NestedFoo::getValA).thenComparing(NestedFoo::getValB).thenComparing(NestedFoo::getTextA)
);
And change your NestedFoo class definition like this:
class NestedFoo {
int valA;
int valB;
String textA;
public NestedFoo(int a, int b, String t) {
this.valA = a;
this.valB = b;
this.textA = t;
}
public int getValA() {
return valA;
}
public void setValA(int valA) {
this.valA = valA;
}
public int getValB() {
return valB;
}
public void setValB(int valB) {
this.valB = valB;
}
public String getTextA() {
return textA;
}
public void setTextA(String textA) {
this.textA = textA;
}
}

When using treemap for sorting keep in mind that treemap uses compareTo instead of equals for sorting and to find duplicity. compareTo should be incosistent with equals and hashcode when implemented for any object which will be used as key. You can look for a detailed example on this link https://codingninjaonline.com/2017/09/29/unexpected-results-for-treemap-with-inconsistent-compareto-and-equals/

Sort List based on index value

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.

A TreeSet or TreeMap that allow duplicates

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.

Comparator based on a configurable order

Is there a way to write custom comparator, following this example:
There are at most 10 items coming in at a random order
i.e.
first item: item_one
second: second_one
third: third_one
I want result them to be sorted like : second_one, third_one, first_one. I'd like to pull this order from configuration file, sort of like template for sorting.
Am I using the wrong data structure, does anyone have experience with this?

Sure. Here is an "OrderedComparator" that compares elements according to a predefined order:
class OrderedComparator implements Comparator<String> {
List<String> predefinedOrder;
public OrderedComparator(String[] predefinedOrder) {
this.predefinedOrder = Arrays.asList(predefinedOrder);
}
#Override
public int compare(String o1, String o2) {
return predefinedOrder.indexOf(o1) - predefinedOrder.indexOf(o2);
}
}
And here is some test code. (I used a List instead of a Set since it 1) seem more natural when talking about the order of the elements and 2) better illustrate what happens with duplicate elements upon sorting using this comparator.)
class Test {
public static void main(String[] args) {
// Order (could be read from config file)
String[] order = { "lorem", "ipsum", "dolor", "sit" };
List<String> someList = new ArrayList<String>();
// Insert elements in random order.
someList.add("sit");
someList.add("ipsum");
someList.add("sit");
someList.add("lorem");
someList.add("dolor");
someList.add("lorem");
someList.add("ipsum");
someList.add("lorem");
System.out.println(someList);
Collections.sort(someList, new OrderedComparator(order));
System.out.println(someList);
}
}
Output:
[sit, ipsum, sit, lorem, dolor, lorem, ipsum, lorem]
[lorem, lorem, lorem, ipsum, ipsum, dolor, sit, sit]

Take a look at TreeSet (http://download.oracle.com/javase/6/docs/api/java/util/TreeSet.html). You can provide a custom Comparator in a constructor. This Comparator will take into account your config. file . The logic of the comparator will not be pretty though since you want arbitrary order. You will most probably end up enumerating all possible comparisons.

A set stores unordered elements. If you want to compare and sort, you should probably go with a list. Here's a quick snippet for you:
List<X> sorted = new ArrayList<X>(myset);
Collections.sort(sorted, new Comparator<X>() {
public int compare(X o1, X o2) {
if (/* o1 < o2 */) {
return -1;
} else if (/* o1 > o2 */) {
return 1;
} else {
return 0;
}
}
});
Now you've got sorted, which has all the same elements of myset, which was unordered by virtue of being a set.
You can also look at TreeSet, which orders its elements, but it's generally not a good idea to rely on a set being ordered.

Use Guava's com.google.common.collect.Ordering:
Ordering.explicit(second_one, third_one, first_one);