As I used Comparator for sorting a library after the author's name, I just coincidentally "found" something, which actually works perfectly, but I don't understand why. Firstly please have a look at my code:
public class Bookshelf{
Collection<Literature> shelf = new ArrayList<Literature>();
ArrayList<Literature> unsorted = (ArrayList<Literature>)shelf;
public void printShelf() {
Comparator<Literature> compareBySurname= new Comparator<Literature>() {
#Override
public int compare(Literature o1, Literature o2) {
return o1.author.surname.compareTo(o2.author.surname);
}
};
unsorted.sort(compareBySurname);
for (Literature c : shelf)
System.out.println(c);
}
}
As you can see, I am sorting the ArrayList "unsorted". But after I sort it, I am iterating through the Collection "shelf" and printing the elements of the Collection "shelf".And the output is a list of sorted elements by surname.
To achive my intention, I actually should iterate through the ArrayList "unsorted" and print the elements (of course this option works too). So my question is, why the first methode actually works too? :D So I am not sorting the Collection "shelf" directly, but I get a sorted list.
Thanks in advance!
ArrayList<Literature> unsorted = (ArrayList<Literature>)shelf; does not create a new ArrayList. It simply makes unsorted refer to the same ArrayList as shelf. They are not different objects. You want something like
ArrayList<Literature> unsorted = new ArrayList<>(shelf); // <-- a different List.
Because both lists share the same memory reference when you assign the list with the "=" operator. To have a new list with another reference, you must use the key name "new".
For lists, we use the Collections.sort(List) method. What if we want to sort a HashSet?
A HashSet does not guarantee any order of its elements. If you need this guarantee, consider using a TreeSet to hold your elements.
However if you just need your elements sorted for this one occurrence, then just temporarily create a List and sort that:
Set<?> yourHashSet = new HashSet<>();
...
List<?> sortedList = new ArrayList<>(yourHashSet);
Collections.sort(sortedList);
Add all your objects to the TreeSet, you will get a sorted Set. Below is a raw example.
HashSet myHashSet = new HashSet();
myHashSet.add(1);
myHashSet.add(23);
myHashSet.add(45);
myHashSet.add(12);
TreeSet myTreeSet = new TreeSet();
myTreeSet.addAll(myHashSet);
System.out.println(myTreeSet); // Prints [1, 12, 23, 45]
Update
You can also use TreeSet's constructor that takes a HashSet as a parameter.
HashSet myHashSet = new HashSet();
myHashSet.add(1);
myHashSet.add(23);
myHashSet.add(45);
myHashSet.add(12);
TreeSet myTreeSet = new TreeSet(myHashSet);
System.out.println(myTreeSet); // Prints [1, 12, 23, 45]
Thanks #mounika for the update.
Java 8 way to sort it would be:
fooHashSet.stream()
.sorted(Comparator.comparing(Foo::getSize)) //comparator - how you want to sort it
.collect(Collectors.toList()); //collector - what you want to collect it to
*Foo::getSize it's an example how to sort the HashSet of YourItem's naturally by size.
*Collectors.toList() is going to collect the result of sorting into a List the you will need to capture it with List<Foo> sortedListOfFoo =
You can use a TreeSet instead.
Use java.util.TreeSet as the actual object. When you iterate over this collection, the values come back in a well-defined order.
If you use java.util.HashSet then the order depends on an internal hash function which is almost certainly not lexicographic (based on content).
Just in-case you don't wanna use a TreeSet you could try this using java stream for concise code.
set = set.stream().sorted().collect(Collectors.toCollection(LinkedHashSet::new));
You can use Java 8 collectors and TreeSet
list.stream().collect(Collectors.toCollection(TreeSet::new))
Based on the answer given by #LazerBanana i will put my own example of a Set sorted by the Id of the Object:
Set<Clazz> yourSet = [...];
yourSet.stream().sorted(new Comparator<Clazz>() {
#Override
public int compare(Clazz o1, Clazz o2) {
return o1.getId().compareTo(o2.getId());
}
}).collect(Collectors.toList()); // Returns the sorted List (using toSet() wont work)
Elements in HashSet can't be sorted. Whenever you put elements into HashSet, it can mess up the ordering of the whole set. It is deliberately designed like that for performance. When you don't care about the order, HashSet will be the most efficient set for frequent insertions and queries.
TreeSet is the alternative that you can use. When you iterate on the tree set, you will get sorted elements automatically.
But it will adjust the tree to try to remain sorted every time you insert an element.
Perhaps, what you are trying to do is to sort just once. In that case, TreeSet is not the most efficient option because it needs to determine the placing of newly added elements all the time. Use TreeSet only when you want to sort often.
If you only need to sort once, use ArrayList. Create a new list and add all the elements then sort it once. If you want to retain only unique elements (remove all duplicates), then put the list into a LinkedHashSet, it will retain the order you have already sorted.
List<Integer> list = new ArrayList<>();
list.add(6);
list.add(4);
list.add(4);
list.add(5);
Collections.sort(list);
Set<Integer> unique = new LinkedHashSet<>(list); // 4 5 6
Now, you've gotten a sorted set if you want it in a list form then convert it into list.
You can use TreeSet as mentioned in other answers.
Here's a little more elaboration on how to use it:
TreeSet<String> ts = new TreeSet<String>();
ts.add("b1");
ts.add("b3");
ts.add("b2");
ts.add("a1");
ts.add("a2");
System.out.println(ts);
for (String s: ts)
System.out.println(s);
Output:
[a1, a2, a3, a4, a5]
a1
a2
b1
b2
b3
In my humble opinion , LazerBanana's answer should be the top rated answer & accepted because all the other answers pointing to java.util.TreeSet ( or first convert to list then call Collections.sort(...) on the converted list ) didn't bothered to ask OP as what kind of objects your HashSet has i.e. if those elements have a predefined natural ordering or not & that is not optional question but a mandatory question.
You just can't go in & start putting your HashSet elements into a TreeSet if element type doesn't already implement Comparable interface or if you are not explicitly passing Comparator to TreeSet constructor.
From TreeSet JavaDoc ,
Constructs a new, empty tree set, sorted according to the natural
ordering of its elements. All elements inserted into the set must
implement the Comparable interface. Furthermore, all such elements
must be mutually comparable: e1.compareTo(e2) must not throw a
ClassCastException for any elements e1 and e2 in the set. If the user
attempts to add an element to the set that violates this constraint
(for example, the user attempts to add a string element to a set whose
elements are integers), the add call will throw a ClassCastException.
That is why only all Java8 stream based answers - where you define your comparator on the spot - only make sense because implementing comparable in POJO becomes optional. Programmer defines comparator as and when needed. Trying to collect into TreeSet without asking this fundamental question is also incorrect ( Ninja's answer). Assuming object types to be String or Integer is also incorrect.
Having said that, other concerns like ,
Sorting Performance
Memory Foot Print ( retaining original set and creating new sorted sets each time sorting is done or wish to sort the set in - place etc etc )
should be the other relevant points too. Just pointing to API shouldn't be only intention.
Since Original set already contains only unique elements & that constraint is also maintained by sorted set so original set needs to be cleared from memory since data is duplicated.
1. Add all set element in list -> al.addAll(s);
2. Sort all the elements in list using -> Collections.sort(al);
public class SortSetProblem {
public static void main(String[] args) {
ArrayList<String> al = new ArrayList();
Set<String> s = new HashSet<>();
s.add("ved");
s.add("prakash");
s.add("sharma");
s.add("apple");
s.add("ved");
s.add("banana");
System.out.println("Before Sorting");
for (String s1 : s) {
System.out.print(" " + s1);
}
System.out.println("After Sorting");
al.addAll(s);
Collections.sort(al);
for (String set : al) {
System.out.print(" " + set);
}
}
}
input - ved prakash sharma apple ved banana
Output - apple banana prakash sharma ved
If you want want the end Collection to be in the form of Set and if you want to define your own natural order rather than that of TreeSet then -
Convert the HashSet into List
Custom sort the List using Comparator
Convert back the List into LinkedHashSet to maintain order
Display the LinkedHashSet
Sample program -
package demo31;
import java.util.*;
public class App26 {
public static void main(String[] args) {
Set<String> set = new HashSet<>();
addElements(set);
List<String> list = new LinkedList<>();
list = convertToList(set);
Collections.sort(list, new Comparator<String>() {
#Override
public int compare(String s1, String s2) {
int flag = s2.length() - s1.length();
if(flag != 0) {
return flag;
} else {
return -s1.compareTo(s2);
}
}
});
Set<String> set2 = new LinkedHashSet<>();
set2 = convertToSet(list);
displayElements(set2);
}
public static void addElements(Set<String> set) {
set.add("Hippopotamus");
set.add("Rhinocerous");
set.add("Zebra");
set.add("Tiger");
set.add("Giraffe");
set.add("Cheetah");
set.add("Wolf");
set.add("Fox");
set.add("Dog");
set.add("Cat");
}
public static List<String> convertToList(Set<String> set) {
List<String> list = new LinkedList<>();
for(String element: set) {
list.add(element);
}
return list;
}
public static Set<String> convertToSet(List<String> list) {
Set<String> set = new LinkedHashSet<>();
for(String element: list) {
set.add(element);
}
return set;
}
public static void displayElements(Set<String> set) {
System.out.println(set);
}
}
Output -
[Hippopotamus, Rhinocerous, Giraffe, Cheetah, Zebra, Tiger, Wolf, Fox, Dog, Cat]
Here the collection has been sorted as -
First - Descending order of String length
Second - Descending order of String alphabetical hierarchy
you can do this in the following ways:
Method 1:
Create a list and store all the hashset values into it
sort the list using Collections.sort()
Store the list back into LinkedHashSet as it preserves the insertion order
Method 2:
Create a treeSet and store all the values into it.
Method 2 is more preferable because the other method consumes lot of time to transfer data back and forth between hashset and list.
We can not decide that the elements of a HashSet would be sorted automatically. But we can sort them by converting into TreeSet or any List like ArrayList or LinkedList etc.
// Create a TreeSet object of class E
TreeSet<E> ts = new TreeSet<E> ();
// Convert your HashSet into TreeSet
ts.addAll(yourHashSet);
System.out.println(ts.toString() + "\t Sorted Automatically");
You can use guava library for the same
Set<String> sortedSet = FluentIterable.from(myHashSet).toSortedSet(new Comparator<String>() {
#Override
public int compare(String s1, String s2) {
// descending order of relevance
//required code
}
});
SortedSet has been added Since java 7
https://docs.oracle.com/javase/8/docs/api/java/util/SortedSet.html
You can wrap it in a TreeSet like this:
Set mySet = new HashSet();
mySet.add(4);
mySet.add(5);
mySet.add(3);
mySet.add(1);
System.out.println("mySet items "+ mySet);
TreeSet treeSet = new TreeSet(mySet);
System.out.println("treeSet items "+ treeSet);
output :
mySet items [1, 3, 4, 5]
treeSet items [1, 3, 4, 5]
Set mySet = new HashSet();
mySet.add("five");
mySet.add("elf");
mySet.add("four");
mySet.add("six");
mySet.add("two");
System.out.println("mySet items "+ mySet);
TreeSet treeSet = new TreeSet(mySet);
System.out.println("treeSet items "+ treeSet);
output:
mySet items [six, four, five, two, elf]
treeSet items [elf, five, four, six, two]
requirement for this method is that the objects of the set/list should be comparable (implement the Comparable interface)
The below is my sample code and its already answered by pointing the code in comments , am still sharing because it contains the complete code
package Collections;
import java.util.*;
public class TestSet {
public static void main(String[] args) {
Set<String> objset = new HashSet<>();
objset.add("test");
objset.add("abc");
objset.add("abc");
objset.add("mas");
objset.add("vas");
Iterator itset = objset.iterator();
while(itset.hasNext())
{
System.out.println(itset.next());
}
TreeSet<String> treeobj = new TreeSet(objset);
System.out.println(treeobj);
}
}
TreeSet treeobj = new TreeSet(objset); here we are invoking the treeset constructor which will call the addAll method to add the objects .
See this below code from the TreeSet class how its mentioned ,
public TreeSet(Collection<? extends E> c) {
this();
addAll(c);
}
Convert HashSet to List then sort it using Collection.sort()
List<String> list = new ArrayList<String>(hset);
Collections.sort(List)
This simple command did the trick for me:
myHashSet.toList.sorted
I used this within a print statement, so if you need to actually persist the ordering, you may need to use TreeSets or other structures proposed on this thread.
I have a List<MyClass> and there is an attribute named attribute1 in MyClass.
Now the question is, how i can get the attribute1 values from List<MyClass> as an array without looping through the list in traditional way?
You can use Guava's FluentIterable to collect your elements, here's an example assuming attribute1 is an Integer
//populated
List<MyClass> yourList;
List<Integer> listbyAttribute = FluentIterable.from(yourList)
.transform(new Function<MyClass, Integer>() {
public Integer apply(MyClass f) {
return f.getAttribute1();
}
}).toList();
More fun with this Guava class: here
You can create your own class which implements LIST interface. YOu should basically do the implementation which is exactly the same as one of the other implementation of list except;
1)you in your add method, every time you add a new element, append it in a string array which is a variable of your class.
2)and add an extra method, lets say giveMyAttribute1List and return the variable list I mentioned earlier.
you you basically have your answer.
List<MyClass> a = new myListIMpl<MyClass>();
a.giveMyAttribute1List();
If this is some kind of tricky question where you (yourself) are not allowed to use a for-each, an iterator nor an old fashioned for-loop (like being asked in a job interview), I would suggest using the following:
Collections.sort(myList, myComparator)
and create the comparator
public class MyClassComparator implements Comparator<MyClass> {
#Override
public int compare(final MyClass o1, final MyClass o2) {
// implement
}
}
and implement the compare method in such a way, that the attribute1 is either in front (or back, or somewhere you know). Then you can easily fetch the element from the list without manually looping through it.
But this answer applies only if this is some kind of "job interview question", otherwise looping through the list is most likely the only option you have.
If you want to get the value of an attribute of each element in a List, you have to visit each element of that List. In other words, you have to iterate over each element. Whether you visit the elements in order with a loop or randomly in some ridiculous way, you're looping/iterating.
What you are asking for is not possible.
I'm having array of list like this
array list={Rank,Shyam,"Sores","Hintus",Ander}
I'm using Collections.sort(arraylist);
I'm getting like this:
"Hintus"
"Sores"
Ander
Rank
I'm getting strings with special characters before any other string in my list.
But I want to get results like this:
Ander
Rank
"Hintus"
"Sores"
How do I get desired order?
Specify a Comparator that ignores them... note that this may be costly.
Collectons.sort(list, new Comparator<String>() {
public int compare(final String a, final String b) {
return a.replace("\"", "").compareTo(b.replace("\"", ""));
}
});
you can create a comparator by your own, and in comparator, write your sorting criteria. and then sort the arraylist using this comparator.
I have a nested list like below (but it has 1,000's of the holder lists within the one main list). Say I need to sort the main list listEmailData by the value for each of its holder lists on the holder.get(2) index. I can't seem to figure out how to do this any advice is appreciated.
ArrayList listEmailData;
ArrayList holder = new ArrayList();
listEmailData.add(3)
listEmailData.add(323)
listEmailData.add(2342)
listEmailData.add(holder)
EDIT: To clarify, I have a list where each list entry contains a sub-list, within this sub-list a specific index contains a value that is a ranking. I need to sort the main list based on this ranking value within each sub-list.
2ND EDIT: Thanks for the help on this, got it working but its seems that its putting larger numbers first and large numbers later, I was hoping to reverse this so it goes from largest to smallest as I am
You should implement Comparator<T> to compare lists, then call
Collections.sort(listEmailData, comparator);
Your comparator would have to compare any two "sublists" - e.g. by fetching a particular value. For example:
public class ListComparator implements Comparator<List<Integer>>
{
private final int indexToCompare;
public ListComparator(int indexToCompare)
{
this.indexToCompare = indexToCompare;
}
public int compare(List<Integer> first, List<Integer> second)
{
// TODO: null checking
Integer firstValue = first.get(indexToCompare);
Integer secondValue = second.get(indexToCompare);
return firstValue.compareTo(secondValue);
}
}
Note that this is using generics - hopefully your real code is too.