Develop Reference

Reduce for loops in the code below

I am trying to get friends recommendations for a user based on their direct friends and rank them on their frequency(i.e number of times the recommended friend appeared on user's direct friend lists). Below is the working code to solve the problem.
public List<String> getFriendsRecommendations(String user)
{
Recommendations rd = new Recommendations();
List<String> result= new ArrayList<String>();
List<String> drfriend = rd.getdirectfriends(user); //return list of direct friend for the user.
List<ArrayList<String>> allfriends = new ArrayList<ArrayList<String>>();
Map<String, Integer> mapfriend = new HashMap<String, Integer>();
List<String> userfriend = rd.getfriends(user); //returns the list of all the friends for a given user.
int counter =0;
for(String s: drfriend)
{
allfriends.add(new ArrayList<String>(rd.getfriends(s)));
rd.intersection(userfriend, allfriends.get(counter), mapfriend);
counter++;
}
result.addAll(mapfriend.keySet());
//Sorting based on the value of hashmap. friend with highest value will be recommended first
Collections.sort(result, new Comparator<String>(){
public int compare(String s1, String s2)
{
if(mapfriend.get(s1) > mapfriend.get(s2))
return -1;
else if(mapfriend.get(s1) < mapfriend.get(s2))
return 1;
else if(mapfriend.get(s1) == mapfriend.get(s2))
{
return s1.compareTo(s2);
}
return 0;
}
});
return result;
}
public void intersection(List<String> lt1, ArrayList<String> lt2, Map<String, Integer> ranked)
{
lt2.removeAll(lt1); // ignoring the friends that user is already connected to
for(String st: lt2)
{
boolean val = ranked.containsKey(st);
if(val)
{
int getval = ranked.get(st);
ranked.put(st, getval+1); //friend name as a key and the value would be the count.
}
else
{
ranked.put(st, 1);
}
}
}
I would like to know if there is more efficient way to solve the above problem instead of using 2 for loops?

Quick tip for your Comparator: Get the two values you are interested in comparing at the start and store them in variables, that way you only do a maximimum of 2 get calls instead of 6 in your current worst case scenario (each get call will hash the String, so less is better).
As for simplifying the for loops, could you just get a list of friends of friends and count the occurrences of each friend in that list? Then afterwards, remove any friends you are already friends with.

List of string with occurrences count and sort

I'm developing a Java Application that reads a lot of strings data likes this:
1 cat (first read)
2 dog
3 fish
4 dog
5 fish
6 dog
7 dog
8 cat
9 horse
...(last read)
I need a way to keep all couple [string, occurrences] in order from last read to first read.
string occurrences
horse 1 (first print)
cat 2
dog 4
fish 2 (last print)
Actually i use two list:
1) List<string> input; where i add all data
In my example:
input.add("cat");
input.add("dog");
input.add("fish");
...
2)List<string> possibilities; where I insert the strings once in this way:
if(possibilities.contains("cat")){
possibilities.remove("cat");
}
possibilities.add("cat");
In this way I've got a sorted list where all possibilities.
I use it like that:
int occurrence;
for(String possible:possibilities){
occurrence = Collections.frequency(input, possible);
System.out.println(possible + " " + occurrence);
}
That trick works good but it's too slow(i've got millions of input)... any help?
(English isn’t my first language, so please excuse any mistakes.)

Use a Map<String, Integer>, as #radoslaw pointed, to keep the insertion sorting use LinkedHashMap and not a TreeMap as described here:
LinkedHashMap keeps the keys in the order they were inserted, while a TreeMap is kept sorted via a Comparator or the natural Comparable ordering of the elements.
Imagine you have all the strings in some array, call it listOfAllStrings, iterate over this array and use the string as key in your map, if it does not exists, put in the map, if it exists, sum 1 to actual result...
Map<String, Integer> results = new LinkedHashMap<String, Integer>();
for (String s : listOfAllStrings) {
if (results.get(s) != null) {
results.put(s, results.get(s) + 1);
} else {
results.put(s, 1);
}
}

Make use of a TreeMap, which will keep ordering on the keys as specified by the compare of your MyStringComparator class handling MyString class which wraps String adding insertion indexes, like this:
// this better be immutable
class MyString {
private MyString() {}
public static MyString valueOf(String s, Long l) { ... }
private String string;
private Long index;
public hashcode(){ return string.hashcode(); }
public boolean equals() { // return rely on string.equals() }
}
class MyStringComparator implements Comparator<MyString> {
public int compare(MyString s1, MyString s2) {
return -s1.getIndex().compareTo(s2.gtIndex());
}
}
Pass the comparator while constructing the map:
Map<MyString,Integer> map = new TreeMap<>(new MyStringComparator());
Then, while parsing your input, do
Long counter = 0;
while (...) {
MyString item = MyString.valueOf(readString, counter++);
if (map.contains(item)) {
map.put(map.get(item)+1);
} else {
map.put(item,1);
}
}
There will be a lot of instantiation because of the immutable class, and the comparator will not be consistent with equals, but it should work.
Disclaimer: this is untested code just to show what I'd do, I'll come back and recheck it when I get my hands on a compiler.

Here is the complete solution for your problem,
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
public class DataDto implements Comparable<DataDto>{
public int count = 0;
public String string;
public long lastSeenTime;
public DataDto(String string) {
this.string = string;
this.lastSeenTime = System.currentTimeMillis();
}
public boolean equals(Object object) {
if(object != null && object instanceof DataDto) {
DataDto temp = (DataDto) object;
if(temp.string != null && temp.string.equals(this.string)) {
return true;
}
}
return false;
}
public int hashcode() {
return string.hashCode();
}
public int compareTo(DataDto o) {
if(o != null) {
return o.lastSeenTime < this.lastSeenTime ? -1 : 1;
}
return 0;
}
public String toString() {
return this.string + " : " + this.count;
}
public static final void main(String[] args) {
String[] listOfAllStrings = {"horse", "cat", "dog", "fish", "cat", "fish", "dog", "cat", "horse", "fish"};
Map<String, DataDto> results = new HashMap<String, DataDto>();
for (String s : listOfAllStrings) {
DataDto dataDto = results.get(s);
if(dataDto != null) {
dataDto.count = dataDto.count + 1;
dataDto.lastSeenTime = System.nanoTime();
} else {
dataDto = new DataDto(s);
results.put(s, dataDto);
}
}
List<DataDto> finalResults = new ArrayList<DataDto>(results.values());
System.out.println(finalResults);
Collections.sort(finalResults);
System.out.println(finalResults);
}
}
Ans
[horse : 1, cat : 2, fish : 2, dog : 1]
[fish : 2, horse : 1, cat : 2, dog : 1]
I think this solution will be suitable for your requirement.

If you know that your data is not going to exceed your memory capacity when you read it all into memory, then the solution is simple - using a LinkedList or a and a LinkedHashMap.
For example, if you use a Linked list:
LinkedList<String> input = new LinkedList();
You then proceed to use input.add() as you did originally. But when the input list is full, you basically use Jordi Castilla's solution - but put the entries in the linked list in reverse order. To do that, you do:
Iterator<String> iter = list.descendingIterator();
LinkedHashMap<String,Integer> map = new LinkedHashMap<>();
while (iter.hasNext()) {
String s = iter.next();
if ( map.containsKey(s)) {
map.put( s, map.get(s) + 1);
} else {
map.put(s, 1);
}
}
Now, the only real difference between his solution and mine is that I'm using list.descendingIterator() which is a method in LinkedList that gives you the entries in backwards order, from "horse" to "cat".
The LinkedHashMap will keep the proper order - whatever was entered first will be printed first, and because we entered things in reverse order, then whatever was read last will be printed first. So if you print your map the result will be:
{horse=1, cat=2, dog=4, fish=2}
If you have a very long file, and you can't load the entire list of strings into memory, you had better keep just the map of frequencies. In this case, in order to keep the order of entry, we'll use an object such as this:
private static class Entry implements Comparable<Entry> {
private static long nextOrder = Long.MIN_VALUE;
private String str;
private int frequency = 1;
private long order = nextOrder++;
public Entry(String str) {
this.str = str;
}
public String getString() {
return str;
}
public int getFrequency() {
return frequency;
}
public void updateEntry() {
frequency++;
order = nextOrder++;
}
#Override
public int compareTo(Entry e) {
if ( order > e.order )
return -1;
if ( order < e.order )
return 1;
return 0;
}
#Override
public String toString() {
return String.format( "%s: %d", str, frequency );
}
}
The trick here is that every time you update the entry (add one to the frequency), it also updates the order. But the compareTo() method orders Entry objects from high order (updated/inserted later) to low order (updated/inserted earlier).
Now you can use a simple HashMap<String,Entry> to store the information as you read it (I'm assuming you are reading from some sort of scanner):
Map<String,Entry> m = new HashMap<>();
while ( scanner.hasNextLine() ) {
String str = scanner.nextLine();
Entry entry = m.get(str);
if ( entry == null ) {
entry = new Entry(str);
m.put(str, entry);
} else {
entry.updateEntry();
}
}
Scanner.close();
Now you can sort the values of the entries:
List<Entry> orderedList = new ArrayList<Entry>(m.values());
m = null;
Collections.sort(orderedList);
Running System.out.println(orderedList) will give you:
[horse: 1, cat: 2, dog: 4, fish: 2]
In principle, you could use a TreeMap whose keys contained the "order" stuff, rather than a plain HashMap like this followed by sorting, but I prefer not having either mutable keys in a map, nor changing the keys constantly. Here we are only changing the values as we fill the map, and each key is inserted into the map only once.

What you could do:
Reverse the order of the list using
Collections.reverse(input). This runs in linear time - O(n);
Create a Set from the input list. A Set garantees uniqueness.
To preserve insertion order, you'll need a LinkedHashSet;
Iterate over this set, just as you did above.
Code:
/* I don't know what logic you use to create the input list,
* so I'm using your input example. */
List<String> input = Arrays.asList("cat", "dog", "fish", "dog",
"fish", "dog", "dog", "cat", "horse");
/* by the way, this changes the input list!
* Copy it in case you need to preserve the original input. */
Collections.reverse(input);
Set<String> possibilities = new LinkedHashSet<String>(strings);
for (String s : possibilities) {
System.out.println(s + " " + Collections.frequency(strings, s));
}
Output:
horse 1
cat 2
dog 4
fish 2

Java - How to use a for each loop to check the different occurrences of a value in a list of objects

Sorry if the title isn't clear, I wasn't sure how to word it. I have an arraylist of objects and within each of these objects I store an integer value referring to a category and one referring to an ID.
I want to find the number of unique combinations of category and IDs that there are.
So at the moment I have
for(Object object: listofObjects){
//For each unique type of object.getID
//For each unique type of object.getCategory
//Add 1 to counter
}
I can't figure out how to do this. Doing things like for(int cat: object.getCategory()) brings up an error.
I can add the values to a new list within the initial for each loop like so,
ArrayList<Integer> aList= new ArrayList<Integer>();
for (Object object : spriteExplore) {
aList.add(object.getCategory());
}
for (int cat : aList) {
testCounter++;
}
but this obviosuly does not take into account uniqueness and also makes it awkward for factoring in the other variable of ID.
I feel like there is probably some easier work around that I am missing. Any advice?
Thanks in advance.

So you list of UserDefine object in ArrayList and you want to find unique Object.Just create set from list.
For e.g Suppose you have
List<Customer> list=new ArrayList<Custeomer>();
list.add(new Customer("A",12));
list.add(new Customer("B",13));
list.add(new Customer("A",12));
now
create set From this list
Set<Customer> set = new HashSet<Customer>(list);
this will have unique Customer
IMP : dont forget to override equals and hashcode method for Customer

Your best approach would be storing the data correctly.
It's possible that you still need to store non-unique items, if that's so - continue using an ArrayList, but in addition, use the following:
Override the hashcode & equels function as shown in this link:
What issues should be considered when overriding equals and hashCode in Java?
Then, use a Set (HashSet would probably be enough for you) to store all your objects. This data structure will disregard elements which are not unique to elements already inside the set.
Then, all you need to do is query the size of the set, and that gives you the amount of unique elements in the list.

I don't know any library that does this automatically, but you can do it manually using sets. Sets will retain only unique object so if you try to add the same value twice it will only keep one reference.
Set<Integer> categories = new HashSet<Integer>();
Set<Integer> ids= new HashSet<Integer>();
for (Object object : listofObjects) {
categories.add(object.getCategory());
ids.add(object.getID());
}
Then you get the number of unique categories / ids by doing
categories.size()
ids.size()
And all your unique values are stored in the sets if you want to use them.

I would look into using a (Hash)Map<Integer, Integer>. Then just have 1 foreach loop, checking to see if the value of Map<object.getId(), object.getCategory()> is null by checking if map.get(object.getId()) is null - if it is, then this pair does not exist yet, so add this pair into the map by using map.put(object.getId(), object.getCategory()). If not, do nothing. Then at the end, to find the number of unique pairs you can just use map.size()
Hope this helps

Map<Integer,List<Integer>> uniqueCombinations = new HashMap<Integer,List<Integer>>();
for (Object object : listofObjects) {
if(uniqueCombinations.get(object.getCategoryId())==null) {
uniqueCombinations.put(object.getCategoryId(), new LinkedList<Integer>);
}
uniqueCombinations.get(object.getCategoryId()).add(object.getId());
}
return uniqueCombinations.size()

I believe you want unique combinations of both category and id, right?
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
public class SO {
class MyObject{
private int id;
private int category;
private String name;
private MyObject(int id, int category,String name) {
super();
this.id = id;
this.category = category;
this.name = name;
}
protected int getId() {
return id;
}
protected int getCategory() {
return category;
}
#Override
public String toString() {
return "MyObject [id=" + id + ", category=" + category + ", name=" + name + "]";
}
}
public static void main(String[] args) {
SO so = new SO();
List<Object> listofObjects = new ArrayList<Object>();
listofObjects.add(so.new MyObject(1,1,"One"));
listofObjects.add(so.new MyObject(1,1,"Two"));
listofObjects.add(so.new MyObject(1,2,"Three"));
Map<String,List<MyObject>> combinations = new HashMap<String,List<MyObject>>();
for(Object object: listofObjects ){
//For each unique type of object.getID
//For each unique type of object.getCategory
//Add 1 to counter
if (object instanceof MyObject){
MyObject obj = (MyObject)object;
String unique = obj.id+"-"+obj.category;
if (combinations.get(unique) == null){
combinations.put(unique, new ArrayList<MyObject>());
}
combinations.get(unique).add(obj);
}
}
System.out.println(combinations);
//counts
for(Entry<String,List<MyObject>> entry:combinations.entrySet()){
System.out.println(entry.getKey()+"="+entry.getValue().size());
}
}
}

Use the Hashmap to save occurence. Dont forget to implement hashcode und equals Methods. You can generate them if you work with Eclipse IDE.
public static void main(String[] args) {
List<MyObject> myObjects = Arrays.asList(new MyObject(1, 2), new MyObject(2, 3), new MyObject(3, 4), new MyObject(3, 4));
Map<MyObject, Integer> map = new HashMap<>();
for (MyObject myObject : myObjects) {
Integer counter = map.get(myObject);
if(counter == null){
counter = 1;
} else {
counter = counter + 1;
}
map.put(myObject, counter);
}
long uniqueness = 0;
for(Integer i : map.values()){
if(i == 1){
++uniqueness;
}
}
System.out.println(uniqueness);
}
The last part can be replaced by this one line expression if you are working with Java 8:
long uniqueness = map.values().stream().filter(i -> i == 1).count();

split an array list into multiple lists in java

I have an array list which when populated has a key and a value I want to know if there is a way of splitting it on repeating keys for example my current data is like this:
[RoleID_123.0, UserHandel_tom, Password_12345.0, prevPassword_null, userCaption_thomas, Email_tom#tom.tom, RoleID_124.0, UserHandel_dave, Password_ghadf, prevPassword_sdfsd, userCaption_david, Email_dave#dave.dave, RoleID_125.0, UserHandel_trevor, Password_tre, prevPassword_null, userCaption_trev, Email_trev#trev.trev]
I want it to come out more like this:
[RoleID_123.0, UserHandel_tom, Password_12345.0, prevPassword_null, userCaption_thomas, Email_tom#tom.tom]
[RoleID_124.0, UserHandel_dave, Password_ghadf, prevPassword_sdfsd, userCaption_david, Email_dave#dave.dave]
[RoleID_125.0, UserHandel_trevor, Password_tre, prevPassword_null, userCaption_trev, Email_trev#trev.trev]
Is there a way to split it on say role id or am I going about this the wrong way?

You can try by using HashMap
private static class MyItemHashMap extends HashMap {
public Item add(Item item) {
get(item).add(item);
return item;
}
public List get(Item key) {
List list = (List) get(createItemKey((Item) key));
return list == null ? createItemEntry((Item) key) : list;
}
private List createItemEntry(Item item) {
List list = new ArrayList();
put(createItemKey(item), list);
return list;
}
private Object createItemKey(Item item) {
return item.getSplitterProperty();
}
}
public static void main(String[] args) {
MyItemHashMap itemMapped = new MyItemHashMap();
List items = Arrays.asList(new Object[]{new Item("A"), new Item("B"),
new Item("C")});
for (Iterator iter = items.iterator(); iter.hasNext();) {
Item item = (Item) iter.next();
itemMapped.add(item);
}
}

If it is an ArrayList, there is no built-in function to split data like this; you will have to do it manually. If you know the number of consecutive fields that make a single structure, this shouldn't be too hard; something like this:
// 6 because there are 6 fields
for (int i = 0; i < arrayList.size(); i = i + 6) {
List thisList = arrayList.subList(i, i + 5);
// ... Now do whatever you want with thisList - it contains one structure.
}
If the number of fields can change then you'll have to do something a little more dynamic and loop through looking for a RoleID field, for example.

I'd use a HashMap to seperate the data instead of one long ArrayList ( you shouldn't have stored the data like this in the first instance )
HashMap<String,ArrayList<String>> hm = new HashMap<String,ArrayList<String>>;
// For each list:
ArrayList<String> arr = new ArrayList<String>;
arr.add("each element");
hm.put("RoleID_123.0", arr);
This way you will end up with a three dimensional structure with a key ( "RoleID..." ) pointing to its child elements.

Try this
String[] str=new String[]{"RoleID_123.0", "UserHandel_tom", "Password_12345.0", "prevPassword_null", "userCaption_thomas", "Email_tom#tom.tom", "RoleID_124.0", "UserHandel_dave", "Password_ghadf", "prevPassword_sdfsd", "userCaption_david", "Email_dave#dave.dave", "RoleID_125.0", "UserHandel_trevor", "Password_tre", "prevPassword_null", "userCaption_trev", "Email_trev#trev.trev"};
List<String> list=new ArrayList<>(Arrays.asList(str));
List<String> subList=list.subList(0,5);
You can try something similar to this

If you feel like taking a Linq-ee Libraried approach, this is about as good as it gets, and it requires use of a couple delegate objects:
import static com.google.common.collect.Collections2.filter;
import static com.google.common.collect.Collections2.transform;
//...
final List<String> yourList = //...
final int RECORD_LENGTH = 6;
Collection<String> roleIdValues = filter(yourList, new Predicate<String>() {
public boolean apply(#Nullable String input) {
return input != null && input.startsWith("RoleID");
}
});
Collection<Collection<String>> splitRecords = transform(roleIdValues, new Function<String, Collection<String>>() {
#Nullable public Collection<String> apply(#Nullable String input) {
return yourList.subList(yourList.indexOf(input), RECORD_LENGTH);
}
});
If Oracle had delivered Java 8 on time you would be able to do this in a way more slick manor. Ironically the reason you cant was provided by the same people providing the guava library

How to get the list elements using list inside another list in java

I have a list like this
List contains set of dtime,uptime values.I want to get the list items i.e., dtime into one and
uptime into another variable.Likewise I want to get all the dtime and uptime pair values seperatly into
the variables using for loop in java.How can I achieve this.Is it possible list or vector?Please help me.
Pseudo code
List.get(0).get(0)-->gives 1st dtime
List.get(0).get(1)-->gives 1st uptime
List.get(1).get(0)-->gives 2nd dtime
List.get(1).get(1)-->gives 2nd uptime
And so on..
How to implement this with for loop I am not getting.I am new to java>please help me..

First Convert That ArrayList into Object[] array then get the value like given below code...driver_ModelsObj is an array convert that into drives object array then get the value from inside the array.
for(int indx=0;indx<driver_ModelsObj.size();indx++){
Object[] drivers=(Object[]) driver_ModelsObj.get(indx);
String Device_ID=drivers[0].toString();
}

If your list is as below
List list = [[1],[2],[3]];
We can retrieve the each value as below.
((List)list.get(0)).get(0); //This will retrieve value 1
((List)list.get(1)).get(0); //This will retrieve value 2

Sounds like you could use a domain object containing uptime and downtime.
For example,
public class Stats {
int dtime;
int uptime;
}
Then you can have a List<Stats> and access it like this:
mylist.get(0).dtime
mylist.get(0).uptime
mylist.get(1).dtime
mylist.get(1).uptime
Part of the (newer) Collcetions framework, List is almost always a better alternative than Vector

List.get(0).get(0)-->gives 1st dtime
List.get(0).get(1)-->gives 1st uptime
Well, what you're doing here, is getting the list at position 0, and getting item 1 from that list. In a for loop we can express this as:
for(int x = 0; x < List.size(); x++)
{
for(int y = 0; y < List.get(x).size(); y++)
{
if(y % 2 == 0)
{
// It's a dtime object.
}
else
{
// It's an uptime object.
}
}
}
Before this, you could declare some lists of your own:
List<DTime> listD = new ArrayList<ATimeObject>();
List<UpTime> listUp = new ArrayList<UpTime>();
Now when you're cycling through, all you need to do is add the relevant object:
if(y % 2 == 0)
{
listD.add(List.get(x).get(y));
}
else
{
listUp.add(List.get(x).get(y));
}

You should create a POJO like
public class TimeData {
double dtime;
Date uptime;
}
Then add each POJO to array list and then iterate it.
List<TimeData> oList = new ArrayList<TimeData>();
int nSize = oList.size();
for(int i=0;i<nSize;i++){
TimeData child = oList.get(i);
// get value using getters
}

You can try this ,Let say you have variables like
double dtime;
Timestamp tp;
And listofresults is coming from query results.
listofresults = results.getResultList();
If list is coming from query then put it in the loop this way in the condition of for loop
for(int i=0;i< listofresults.size() ;i=i+2)
{
dtime=(double) listofresults.get(i);
//cast because the value is of object type
tp=(TimeStamp) listofresults.get(i+1);
//cast because the value is of object type
//do something with these variables
}

I recommend creating a wrapper for it.
public class UpTimeDownTime {
MyTimeDataClass downtime;
MyTimeDataClass uptime;
public UpTimeDownTime(MyTimeDataClass downtime, MyTimeDataClass uptime){
this.downtime = downtime;
this.uptime = uptime;
}
public MyTimeDataClass getDowntime () {
return downtime;
}
public MyTimeDataClass getUptime () {
return uptime;
}
public static void main (String[] args) {
List<List<MyTimeDataClass>> List = ...;
List<UpTimeDownTime> uptimeDowntime = new ArrayList<UpTimeDownTime>();
for(List<MyTimeDataClass> timeList : List){
UpTimeDownTime u = new UpTimeDownTime(timeList.get(0), timeList.get(1));
uptimeDowntime.add(u);
}
}
}