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Same word occurrence in a string. Java [duplicate]

I am writing a very basic java program that calculates frequency of each word in a sentence so far i managed to do this much
import java.io.*;
class Linked {
public static void main(String args[]) throws IOException {
BufferedReader br = new BufferedReader(
new InputStreamReader(System.in));
System.out.println("Enter the sentence");
String st = br.readLine();
st = st + " ";
int a = lengthx(st);
String arr[] = new String[a];
int p = 0;
int c = 0;
for (int j = 0; j < st.length(); j++) {
if (st.charAt(j) == ' ') {
arr[p++] = st.substring(c,j);
c = j + 1;
}
}
}
static int lengthx(String a) {
int p = 0;
for (int j = 0; j < a.length(); j++) {
if (a.charAt(j) == ' ') {
p++;
}
}
return p;
}
}
I have extracted each string and stored it in a array , now problem is actually how to count the no of instances where each 'word' is repeated and how to display so that repeated words not get displayed multiple times , can you help me in this one ?

Use a map with word as a key and count as value, somthing like this
Map<String, Integer> map = new HashMap<>();
for (String w : words) {
Integer n = map.get(w);
n = (n == null) ? 1 : ++n;
map.put(w, n);
}
if you are not allowed to use java.util then you can sort arr using some sorting algoritm and do this
String[] words = new String[arr.length];
int[] counts = new int[arr.length];
words[0] = words[0];
counts[0] = 1;
for (int i = 1, j = 0; i < arr.length; i++) {
if (words[j].equals(arr[i])) {
counts[j]++;
} else {
j++;
words[j] = arr[i];
counts[j] = 1;
}
}
An interesting solution with ConcurrentHashMap since Java 8
ConcurrentMap<String, Integer> m = new ConcurrentHashMap<>();
m.compute("x", (k, v) -> v == null ? 1 : v + 1);

In Java 8, you can write this in two simple lines! In addition you can take advantage of parallel computing.
Here's the most beautiful way to do this:
Stream<String> stream = Stream.of(text.toLowerCase().split("\\W+")).parallel();
Map<String, Long> wordFreq = stream
.collect(Collectors.groupingBy(String::toString,Collectors.counting()));

import java.util.*;
public class WordCounter {
public static void main(String[] args) {
String s = "this is a this is this a this yes this is a this what it may be i do not care about this";
String a[] = s.split(" ");
Map<String, Integer> words = new HashMap<>();
for (String str : a) {
if (words.containsKey(str)) {
words.put(str, 1 + words.get(str));
} else {
words.put(str, 1);
}
}
System.out.println(words);
}
}
Output:
{a=3, be=1, may=1, yes=1, this=7, about=1, i=1, is=3, it=1, do=1, not=1, what=1, care=1}

Try this
public class Main
{
public static void main(String[] args)
{
String text = "the quick brown fox jumps fox fox over the lazy dog brown";
String[] keys = text.split(" ");
String[] uniqueKeys;
int count = 0;
System.out.println(text);
uniqueKeys = getUniqueKeys(keys);
for(String key: uniqueKeys)
{
if(null == key)
{
break;
}
for(String s : keys)
{
if(key.equals(s))
{
count++;
}
}
System.out.println("Count of ["+key+"] is : "+count);
count=0;
}
}
private static String[] getUniqueKeys(String[] keys)
{
String[] uniqueKeys = new String[keys.length];
uniqueKeys[0] = keys[0];
int uniqueKeyIndex = 1;
boolean keyAlreadyExists = false;
for(int i=1; i<keys.length ; i++)
{
for(int j=0; j<=uniqueKeyIndex; j++)
{
if(keys[i].equals(uniqueKeys[j]))
{
keyAlreadyExists = true;
}
}
if(!keyAlreadyExists)
{
uniqueKeys[uniqueKeyIndex] = keys[i];
uniqueKeyIndex++;
}
keyAlreadyExists = false;
}
return uniqueKeys;
}
}
Output:
the quick brown fox jumps fox fox over the lazy dog brown
Count of [the] is : 2
Count of [quick] is : 1
Count of [brown] is : 2
Count of [fox] is : 3
Count of [jumps] is : 1
Count of [over] is : 1
Count of [lazy] is : 1
Count of [dog] is : 1

From Java 10 you can use the following:
import java.util.Arrays;
import java.util.stream.Collectors;
public class StringFrequencyMap {
public static void main(String... args){
String[] wordArray = {"One", "One", "Two","Three", "Two", "two"};
var freq = Arrays.stream(wordArray)
.collect(Collectors.groupingBy(x -> x, Collectors.counting()));
System.out.println(freq);
}
}
Output:
{One=2, two=1, Two=2, Three=1}

You could try this
public static void frequency(String s) {
String trimmed = s.trim().replaceAll(" +", " ");
String[] a = trimmed.split(" ");
ArrayList<Integer> p = new ArrayList<>();
for (int i = 0; i < a.length; i++) {
if (p.contains(i)) {
continue;
}
int d = 1;
for (int j = i+1; j < a.length; j++) {
if (a[i].equals(a[j])) {
d += 1;
p.add(j);
}
}
System.out.println("Count of "+a[i]+" is:"+d);
}
}

package naresh.java;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Set;
public class StringWordDuplicates {
static void duplicate(String inputString){
HashMap<String, Integer> wordCount = new HashMap<String,Integer>();
String[] words = inputString.split(" ");
for(String word : words){
if(wordCount.containsKey(word)){
wordCount.put(word, wordCount.get(word)+1);
}
else{
wordCount.put(word, 1);
}
}
//Extracting of all keys of word count
Set<String> wordsInString = wordCount.keySet();
for(String word : wordsInString){
if(wordCount.get(word)>1){
System.out.println(word+":"+wordCount.get(word));
}
}
}
public static void main(String args[]){
duplicate("I am Java Programmer and IT Server Programmer with Java as Best Java lover");
}
}

class find
{
public static void main(String nm,String w)
{
int l,i;
int c=0;
l=nm.length();String b="";
for(i=0;i<l;i++)
{
char d=nm.charAt(i);
if(d!=' ')
{
b=b+d;
}
if(d==' ')
{
if(b.compareTo(w)==0)
{
c++;
}
b="";
}
}
System.out.println(c);
}
}

public class wordFrequency {
private static Scanner scn;
public static void countwords(String sent) {
sent = sent.toLowerCase().replaceAll("[^a-z ]", "");
ArrayList<String> arr = new ArrayList<String>();
String[] sentarr = sent.split(" ");
Map<String, Integer> a = new HashMap<String, Integer>();
for (String word : sentarr) {
arr.add(word);
}
for (String word : arr) {
int count = Collections.frequency(arr, word);
a.put(word, count);
}
for (String key : a.keySet()) {
System.out.println(key + " = " + a.get(key));
}
}
public static void main(String[] args) {
scn = new Scanner(System.in);
System.out.println("Enter sentence:");
String inp = scn.nextLine();
countwords(inp);
}
}

Determine the frequency of words in a file.
File f = new File(fileName);
Scanner s = new Scanner(f);
Map<String, Integer> counts =
new Map<String, Integer>();
while( s.hasNext() ){
String word = s.next();
if( !counts.containsKey( word ) )
counts.put( word, 1 );
else
counts.put( word,
counts.get(word) + 1 );
}

The following program finds the frequency, sorts it accordingly, and prints it.
Below is the output grouped by frequency:
0-10:
The 2
Is 4
11-20:
Have 13
Done 15
Here is my program:
package com.company;
import java.io.*;
import java.util.*;
import java.lang.*;
/**
* Created by ayush on 12/3/17.
*/
public class Linked {
public static void main(String args[]) throws IOException {
BufferedReader br = new BufferedReader(
new InputStreamReader(System.in));
System.out.println("Enter the sentence");
String st = br.readLine();
st=st.trim();
st = st + " ";
int count = lengthx(st);
System.out.println(count);
String arr[] = new String[count];
int p = 0;
int c = 0;
for (int i = 0; i < st.length(); i++) {
if (st.charAt(i) == ' ') {
arr[p] = st.substring(c,i);
System.out.println(arr[p]);
c = i + 1;
p++;
}
}
Map<String, Integer> map = new HashMap<>();
for (String w : arr) {
Integer n = map.get(w);
n = (n == null) ? 1 : ++n;
map.put(w, n);
}
for (String key : map.keySet()) {
System.out.println(key + " = " + map.get(key));
}
Set<Map.Entry<String, Integer>> entries = map.entrySet();
Comparator<Map.Entry<String, Integer>> valueComparator = new Comparator<Map.Entry<String,Integer>>() {
#Override
public int compare(Map.Entry<String, Integer> e1, Map.Entry<String, Integer> e2) {
Integer v1 = e1.getValue();
Integer v2 = e2.getValue();
return v1.compareTo(v2); }
};
List<Map.Entry<String, Integer>> listOfEntries = new ArrayList<Map.Entry<String, Integer>>(entries);
Collections.sort(listOfEntries, valueComparator);
LinkedHashMap<String, Integer> sortedByValue = new LinkedHashMap<String, Integer>(listOfEntries.size());
for(Map.Entry<String, Integer> entry : listOfEntries){
sortedByValue.put(entry.getKey(), entry.getValue());
}
for(Map.Entry<String, Integer> entry : listOfEntries){
sortedByValue.put(entry.getKey(), entry.getValue());
}
System.out.println("HashMap after sorting entries by values ");
Set<Map.Entry<String, Integer>> entrySetSortedByValue = sortedByValue.entrySet();
for(Map.Entry<String, Integer> mapping : entrySetSortedByValue){
System.out.println(mapping.getKey() + " ==> " + mapping.getValue());
}
}
static int lengthx(String a) {
int count = 0;
for (int j = 0; j < a.length(); j++) {
if (a.charAt(j) == ' ') {
count++;
}
}
return count;
}
}

import java.io.*;
class Linked {
public static void main(String args[]) throws IOException {
BufferedReader br = new BufferedReader(
new InputStreamReader(System.in));
System.out.println("Enter the sentence");
String st = br.readLine();
st = st + " ";
int a = lengthx(st);
String arr[] = new String[a];
int p = 0;
int c = 0;
for (int j = 0; j < st.length(); j++) {
if (st.charAt(j) == ' ') {
arr[p++] = st.substring(c,j);
c = j + 1;
}
}
}
static int lengthx(String a) {
int p = 0;
for (int j = 0; j < a.length(); j++) {
if (a.charAt(j) == ' ') {
p++;
}
}
return p;
}
}

Simply use Java 8 Stream collectors groupby function:
import java.util.function.Function;
import java.util.stream.Collectors;
static String[] COUNTRY_NAMES
= { "China", "Australia", "India", "USA", "USSR", "UK", "China",
"France", "Poland", "Austria", "India", "USA", "Egypt", "China" };
Map<String, Long> result = Stream.of(COUNTRY_NAMES).collect(
Collectors.groupingBy(Function.identity(), Collectors.counting()));

Count frequency of elements of list in java 8
List<Integer> list = new ArrayList<Integer>();
Collections.addAll(list,3,6,3,8,4,9,3,6,9,4,8,3,7,2);
Map<Integer, Long> frequencyMap = list.stream().collect(Collectors.groupingBy(Function.identity(),Collectors.counting()));
System.out.println(frequencyMap);
Note :
For String frequency counting split the string and convert it to list and use streams for count frequency => (Map frequencyMap)*
Check below link

String s[]=st.split(" ");
String sf[]=new String[s.length];
int count[]=new int[s.length];
sf[0]=s[0];
int j=1;
count[0]=1;
for(int i=1;i<s.length;i++)
{
int t=j-1;
while(t>=0)
{
if(s[i].equals(sf[t]))
{
count[t]++;
break;
}
t--;
}
if(t<0)
{
sf[j]=s[i];
count[j]++;
j++;
}
}

Created a simple easy to understand solution for this problem covers all test cases-
import java.util.HashMap;
import java.util.Map;
/*
* Problem Statement - Count Frequency of each word in a given string, ignoring special characters and space
* Input 1 - "To be or Not to be"
* Output 1 - to(2 times), be(2 times), or(1 time), not(1 time)
*
* Input 2 -"Star 123 ### 123 star"
* Output - Star(2 times), 123(2 times)
*/
public class FrequencyofWords {
public static void main(String[] args) {
String s1="To be or not **** to be! is all i ask for";
fnFrequencyofWords(s1);
}
//-------Supporting Function-----------------
static void fnFrequencyofWords(String s1) {
//------- Convert String to proper format----
s1=s1.replaceAll("[^A-Za-z0-9\\s]","");
s1=s1.replaceAll(" +"," ");
s1=s1.toLowerCase();
//-------Create String to an array with words------
String[] s2=s1.split(" ");
System.out.println(s1);
//-------- Create a HashMap to store each word and its count--
Map <String , Integer> map=new HashMap<String, Integer>();
for(int i=0;i<s2.length;i++) {
if(map.containsKey(s2[i])) //---- Verify if Word Already Exits---
{
map.put(s2[i], 1+ map.get(s2[i])); //-- Increment value by 1 if word already exits--
}
else {
map.put(s2[i], 1); // --- Add Word to map and set value as 1 if it does not exist in map--
}
}
System.out.println(map); //--- Print the HashMap with Key, Value Pair-------
}
}

public class WordFrequencyProblem {
public static void main(String args[]){
String s="the quick brown fox jumps fox fox over the lazy dog brown";
String alreadyProcessedWords="";
boolean isCount=false;
String[] splitWord = s.split("\\s|\\.");
for(int i=0;i<splitWord.length;i++){
String word = splitWord[i];
int count = 0;
isCount=false;
if(!alreadyProcessedWords.contains(word)){
for(int j=0;j<splitWord.length;j++){
if(word.equals(splitWord[j])){
count++;
isCount = true;
alreadyProcessedWords=alreadyProcessedWords+word+" ";
}
}
}
if(isCount)
System.out.println(word +"Present "+ count);
}
}
}

public class TestSplit {
public static void main(String[] args) {
String input="Find the repeated word which is repeated in this string";
List<String> output= (List) Arrays.asList(input.split(" "));
for(String str: output) {
int occurrences = Collections.frequency(output, str);
System.out.println("Occurence of " + str+ " is "+occurrences);
}
System.out.println(output);
}
}

Please try these it may be help for you
public static void main(String[] args) {
String str1="I am indian , I am proud to be indian proud.";
Map<String,Integer> map=findFrquenciesInString(str1);
System.out.println(map);
}
private static Map<String,Integer> findFrquenciesInString(String str1) {
String[] strArr=str1.split(" ");
Map<String,Integer> map=new HashMap<>();
for(int i=0;i<strArr.length;i++) {
int count=1;
for(int j=i+1;j<strArr.length;j++) {
if(strArr[i].equals(strArr[j]) && strArr[i]!="-1") {
strArr[j]="-1";
count++;
}
}
if(count>1 && strArr[i]!="-1") {
map.put(strArr[i], count);
strArr[i]="-1";
}
}
return map;
}

try this
public void count()throws IOException
{
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
System.out.println("enetr the strring");
String s = in.readLine();
int l = s.length();
int a=0,b=0,c=0,i,j,y=0;
char d;
String x;
String n[] = new String [50];
int m[] = new int [50];
for (i=0;i<50;i++)
{
m[i]=0;
}
for (i=0;i<l;i++)
{
d = s.charAt(i);
if((d==' ')||(d=='.'))
{
x = s.substring(a,i);
a= i+1;
for(j=0;j<b;j++)
{
if(x.equalsIgnoreCase(n[j]) == true)
{
m[j]++;
c = 1;
}
}
if(c==0)
{
n[b] = x;
m[b] = 1;
b++;
}
}
c=0;
}
for(i=0;i<b;i++)
{
for (j=0;j<b;j++)
{
if(y<m[j])
{
y=m[j];
}
}
if(m[i]==y)
{
System.out.println(n[i] + " : " + m[i]);
m[i]=0;
}
y=0;
}
}

Reversing a HashMap storing words and their line numbers

I have a HashMap
HashMap<String, LinkedList<Integer>> indexMap;
which is storing all words in a file and their corresponding line numbers where they appear.
Example -
This is just an example
to demonstrate what I am saying an is
Would display
This [1]
demonstrate [2]
an [1 2]
is [1 2]
...
....
And so on. I want to reverse this HashMap so that it displays the words stored at each line number.
For the particular example above, it should display
1 [This, an, just, example, is]
2 [demonstrate, what, to, I, am, saying, is, an]
For this particular task, this is what I have done -
import java.io.FileReader;
import java.io.LineNumberReader;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.Map;
public class ReverseIndex {
private static Map<String, LinkedList<Integer>> indexMap = new HashMap<String, LinkedList<Integer>>();
public static LinkedList<Integer> getIndex(String word) {
return indexMap.get(word);
}
public static void main(String[] args) {
try {
LineNumberReader rdr = new LineNumberReader(
new FileReader(
args[0]));
String line = "";
int lineNumber = 0;
//CREATING THE INITIAL HASHMAP WHICH WE WANT TO REVERSE
while ((line = rdr.readLine()) != null) {
lineNumber++;
String[] words = line.split("\\s+");
for (int i = 0; i < words.length; i++) {
LinkedList<Integer> temp = new LinkedList<Integer>();
if (getIndex(words[i]) != null)
temp = getIndex(words[i]);
temp.add(lineNumber);
indexMap.put(words[i], temp);
}
}
//FINISHED CREATION
Map<Integer, LinkedList<String>> myNewHashMap = new HashMap<Integer, LinkedList<String>>();
for(Map.Entry<String, LinkedList<Integer>> entry : indexMap.entrySet()){
LinkedList<Integer> values = entry.getValue();
String key = entry.getKey();
LinkedList<String> temp = new LinkedList<String>();
for(int i = 0; i <= lineNumber; i++) {
if(values.contains(i)) {
if(!temp.contains(key))
temp.add(key);
myNewHashMap.put(i, temp);
}
}
}
for(Map.Entry<Integer, LinkedList<String>> entry : myNewHashMap.entrySet()){
Integer tester = entry.getKey();
LinkedList<String> temp2 = new LinkedList<String>();
temp2 = entry.getValue();
System.out.print(tester + " ");
for(int i = 0; i < temp2.size(); i++) {
System.out.print(temp2.get(i) + " ");
}
System.out.println();
}
rdr.close();
} catch (Exception e) {
e.printStackTrace();
}
}
}
However the problem with this is, for the example that we had above, it would print -
1 example
2 an
How could I reverse it so that it works perfectly with the expected output?

Just replace the first for loop in your main with the below code. I have made some changes to you original code as per convention like moved variable declaration out of loop and changed the logic in a way it checks if the LinkedList<'String'> already exists for the line number if so add it to the list or else create a new LinkedList<'String'> and then add word.
LinkedList<Integer> values = null;
String key = null;
LinkedList<String> temp = null;
for(Map.Entry<String, LinkedList<Integer>> entry : indexMap.entrySet())
{
values = entry.getValue();
key = entry.getKey();
temp = new LinkedList<String>();
for(int value : values)
{
temp = myNewHashMap.get(value);
if(temp == null )
{
temp = new LinkedList<String>();
myNewHashMap.put(value,temp);
}
temp.add(key);
}
}

How to iterate through two dimensional ArrayList using iterator?

I would like to iterate through two dimensional ArrayList which includes String objects using iterator. I also would like to iterate in a way that let me choose whether I want to iterate horizontally(row) first or vertically(column) by using a boolean value. How can I implement this in java?
What I've tried so far.
public class IterateThis implements Iterator<String>{
ArrayList<ArrayList<String>> array;
public IterateThis(){
array = new ArrayList<ArrayList<String>>();
array.add(new ArrayList<String>());
array.add(new ArrayList<String>());
array.add(new ArrayList<String>());
array.get(0).add("1");
array.get(0).add("2");
array.get(0).add("2");
array.get(1).add("4");
array.get(1).add("5");
array.get(1).add("6");
}
Iterator<String> it = array.iterator(); //This gives me an error...why?
I don't know how I can implement the boolean value though.

Maybe you need to implement two versions, with a boolean that decides which loop to use:
public void iterate(boolean horizantalFirst){
if(horizontalFirst){
for(int i=0; i<array.size(); i++){ // first iterate through the "outer list"
for(int j=0; j<array.get(i).size(); j++){ // then iterate through all the "inner lists"
array.get(i).get(j)="1";
}
}
}else{
int j=0; // index to iterate through the "inner lists"
for(; j<array.get(j).size(); j++){ //dangerous, you need to be sure that there is a j-th element in array
for(int i=0; i<array.size(); i++){ // iterate here through the outer list, by always working on the j-th element
array.get(i).get(j)="1";
}
}
}
}

Why not try this:
import java.util.ArrayList;
public class Iteration
{
private ArrayList<ArrayList<String>> array;
public Iteration()
{
array = new ArrayList<>();
array.add(new ArrayList<String>());
array.get(0).add("000");
array.get(0).add("001");
array.get(0).add("010");
array.add(new ArrayList<String>());
array.get(1).add("100");
array.get(1).add("101");
array.get(1).add("110");
array.get(1).add("111");
iterateRowWise();
System.out.println("\n\n");
iterateColumnWise();
}
public void iterateRowWise()
{
// This uses iterator behind the scene.
for (ArrayList<String> row : array)
{
for (String element : row)
{
System.out.print(element + " ");
}
System.out.println();
}
}
public void iterateColumnWise()
{
int arraySize = array.size();
int maxColumns = getMaximumListSize();
for (int c = 0; c < maxColumns; c++)
{
for (int r = 0; r < arraySize; r++)
{
ArrayList<String> rowList = array.get(r);
if (c < rowList.size())
{
System.out.print(rowList.get(c) + " ");
}
}
System.out.println();
}
}
private int getMaximumListSize()
{
int maxListSize = 0;
for (ArrayList<String> rowList : array)
{
if (maxListSize < rowList.size())
maxListSize = rowList.size();
}
return maxListSize;
}
public static void main(String[] args)
{
new Iteration();
}
}
The iterateRowWise() method iterates using the iterator, but it does so behind the scene.
The iterateColumnWise() method doesn't use iterator, but its safe to use.

Row-wise iteration is simple as shown in the #Awfully Awesome answer.
Tried a columnwise iteration with assumption that List will always have m cross n elements where m=n
public static void IterateThis() {
ArrayList<ArrayList<String>> array = new ArrayList<ArrayList<String>>();
array.add(new ArrayList<String>());
array.add(new ArrayList<String>());
array.get(0).add("1");
array.get(0).add("2");
array.get(0).add("2");
array.get(1).add("4");
array.get(1).add("5");
array.get(1).add("6");
Iterator<ArrayList<String>> it = array.iterator();
int topLevelIteratorResetCounter = 0;
int noOfIteratorNextRequired = 1;
int size = array.size();
while (it.hasNext()) {
ArrayList<String> strList = it.next();
if (noOfIteratorNextRequired > strList.size())
break;
Iterator<String> itString = strList.iterator();
int numtimes = 0;
String str = null;
while (numtimes != noOfIteratorNextRequired) {
str = itString.next();
numtimes++;
}
System.out.println(str);
numtimes = 0;
topLevelIteratorResetCounter++;
if (topLevelIteratorResetCounter == size) { //as column count is equal to column size
it = array.iterator(); //reset the iterator
noOfIteratorNextRequired++;
topLevelIteratorResetCounter = 0;
}
}
}
The answer uses Iterator.

Create word count of text using hashmap

I am trying to create a program as a tutorial for myself for hashmaps. I ask the user into text and try to split it into hashmaps and then increase the count if the word repeats. This is my program:
import java.util.*;
import java.lang.*;
import javax.swing.JOptionPane;
import java.io.*;
public class TestingTables
{
public static void main(String args[])
{
{
String s = JOptionPane.showInputDialog("Enter any text.");
String[] splitted = s.split(" ");
HashMap hm = new HashMap();
int x;
for (int i=0; i<splitted.length ; i++) {
hm.put(splitted[i], i);
System.out.println(splitted[i] + " " + i);
if (hm.containsKey(splitted[i])) {
x = ((Integer)hm.get(splitted[i])).intValue();
hm.put(splitted[i], new Integer(x+1)); }
}
}
}
}
When I input "random random random", I get:
random 0
random 1
random 2
What do I need to change so I get:
random 3
Also, do I need to use an iterator to print out the hashmap, or is what I used OK?

Your initialization is wrong hm.put(splitted[i], i).
You should initialize to 0 or to 1 (to count, not to index).
So do this loop first.
for (int i = 0; i < splitted.length; i++) {
if (!hm.containsKey(splitted[i])) {
hm.put(splitted[i], 1);
} else {
hm.put(splitted[i], (Integer) hm.get(splitted[i]) + 1);
}
}
Then just do one more loop (iterate through the keys of the HashMap) and print the counts out.
for (Object word : hm.keySet()){
System.out.println(word + " " + (Integer) hm.get(word));
}

import java.util.*;
import java.lang.*;
import javax.swing.JOptionPane;
import java.io.*;
public class TestingTables
{
public static void main(String args[])
{
{
String s = JOptionPane.showInputDialog("Enter any text.");
String[] splitted = s.split(" ");
Map<String, Integer> hm = new HashMap<String, Integer>();
int x;
for (int i=0; i<splitted.length ; i++) {
if (hm.containsKey(splitter[i])) {
int cont = hm.get(splitter[i]);
hm.put(splitter[i], cont + 1)
} else {
hm.put(splitted[i], 1);
}
}
}
}
Your Map declaration is wrong, remember the correct way to implement a Map.

This should work, its a pretty simple implementation..
Map<String, Integer> hm = new HashMap<String, Integer>();
int x;
for (int i = 0; i < splitted.length; i++) {
if (hm.containsKey(splitted[i])) {
x = hm.get(splitted[i]);
hm.put(splitted[i], x + 1);
} else {
hm.put(splitted[i], 1);
}
}
for (String key : hm.keySet()) {
System.out.println(key + " " + hm.get(key));
}

Java count occurrence of each item in an array

Is there any method for counting the occurrence of each item on an array?
Lets say I have:
String[] array = {"name1","name2","name3","name4", "name5"};
Here the output will be:
name1 1
name2 1
name3 1
name4 1
name5 1
and if I have:
String[] array = {"name1","name1","name2","name2", "name2"};
The output would be:
name1 2
name2 3
The output here is just to demonstrate the expected result.

List asList = Arrays.asList(array);
Set<String> mySet = new HashSet<String>(asList);
for(String s: mySet){
System.out.println(s + " " + Collections.frequency(asList,s));
}

With java-8, you can do it like this:
String[] array = {"name1","name2","name3","name4", "name5", "name2"};
Arrays.stream(array)
.collect(Collectors.groupingBy(s -> s))
.forEach((k, v) -> System.out.println(k+" "+v.size()));
Output:
name5 1
name4 1
name3 1
name2 2
name1 1
What it does is:
Create a Stream<String> from the original array
Group each element by identity, resulting in a Map<String, List<String>>
For each key value pair, print the key and the size of the list
If you want to get a Map that contains the number of occurences for each word, it can be done doing:
Map<String, Long> map = Arrays.stream(array)
.collect(Collectors.groupingBy(s -> s, Collectors.counting()));
For more informations:
Stream
Collectors
Hope it helps! :)

You could use a MultiSet from Google Collections/Guava or a Bag from Apache Commons.
If you have a collection instead of an array, you can use addAll() to add the entire contents to the above data structure, and then apply the count() method to each value. A SortedMultiSet or SortedBag would give you the items in a defined order.
Google Collections actually has very convenient ways of going from arrays to a SortedMultiset.

I wrote a solution for this to practice myself. It doesn't seem nearly as awesome as the other answers posted, but I'm going to post it anyway, and then learn how to do this using the other methods as well. Enjoy:
public static Integer[] countItems(String[] arr)
{
List<Integer> itemCount = new ArrayList<Integer>();
Integer counter = 0;
String lastItem = arr[0];
for(int i = 0; i < arr.length; i++)
{
if(arr[i].equals(lastItem))
{
counter++;
}
else
{
itemCount.add(counter);
counter = 1;
}
lastItem = arr[i];
}
itemCount.add(counter);
return itemCount.toArray(new Integer[itemCount.size()]);
}
public static void main(String[] args)
{
String[] array = {"name1","name1","name2","name2", "name2", "name3",
"name1","name1","name2","name2", "name2", "name3"};
Arrays.sort(array);
Integer[] cArr = countItems(array);
int num = 0;
for(int i = 0; i < cArr.length; i++)
{
num += cArr[i]-1;
System.out.println(array[num] + ": " + cArr[i].toString());
}
}

Using HashMap it is walk in the park.
main(){
String[] array ={"a","ab","a","abc","abc","a","ab","ab","a"};
Map<String,Integer> hm = new HashMap();
for(String x:array){
if(!hm.containsKey(x)){
hm.put(x,1);
}else{
hm.put(x, hm.get(x)+1);
}
}
System.out.println(hm);
}

It can be done in a very simple way using collections
please find the code below
String[] array = {"name1","name1","name2","name2", "name2"};
List<String> sampleList=(List<String>) Arrays.asList(array);
for(String inpt:array){
int frequency=Collections.frequency(sampleList,inpt);
System.out.println(inpt+" "+frequency);
}
Here the output will be like
name1 2
name1 2
name2 3
name2 3
name2 3
To avoid printing redundant keys use HashMap and get your desired output

I would use a hashtable with in key takes the element of the array (here string) and in value an Integer.
then go through the list doing something like this :
for(String s:array){
if(hash.containsKey(s)){
Integer i = hash.get(s);
i++;
}else{
hash.put(s, new Interger(1));
}

Count String occurence using hashmap, streams & collections
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Set;
import java.util.stream.Collectors;
public class StringOccurence {
public static void main(String args[]) {
String[] stringArray = { "name1", "name1", "name2", "name2", "name2" };
countStringOccurence(stringArray);
countStringOccurenceUsingStream(stringArray);
countStringOccurenceUsingCollections(stringArray);
}
private static void countStringOccurenceUsingCollections(String[] stringArray) {
// TODO Auto-generated method stub
List<String> asList = Arrays.asList(stringArray);
Set<String> set = new HashSet<String>(asList);
for (String string : set) {
System.out.println(string + " --> " + Collections.frequency(asList, string));
}
}
private static void countStringOccurenceUsingStream(String[] stringArray) {
// TODO Auto-generated method stub
Arrays.stream(stringArray).collect(Collectors.groupingBy(s -> s))
.forEach((k, v) -> System.out.println(k + " --> " + v.size()));
}
private static void countStringOccurence(String[] stringArray) {
// TODO Auto-generated method stub
Map<String, Integer> map = new HashMap<String, Integer>();
for (String s : stringArray) {
if (map.containsKey(s)) {
map.put(s, map.get(s) + 1);
} else {
map.put(s, 1);
}
}
for (Map.Entry<String, Integer> entry : map.entrySet()) {
System.out.println(entry.getKey() + " --> " + entry.getValue());
}
}
}

Here is my solution -
The method takes an array of integers(assuming the range between 0 to 100) as input and returns the number of occurrences of each element.
let's say the input is [21,34,43,21,21,21,45,65,65,76,76,76].
So the output would be in a map and it is: {34=1, 21=4, 65=2, 76=3, 43=1, 45=1}
public Map<Integer, Integer> countOccurrence(int[] numbersToProcess) {
int[] possibleNumbers = new int[100];
Map<Integer, Integer> result = new HashMap<Integer, Integer>();
for (int i = 0; i < numbersToProcess.length; ++i) {
possibleNumbers[numbersToProcess[i]] = possibleNumbers[numbersToProcess[i]] + 1;
result.put(numbersToProcess[i], possibleNumbers[numbersToProcess[i]]);
}
return result;
}

There are several methods which can help, but this is one is using for loop.
import java.util.Arrays;
public class one_dimensional_for {
private static void count(int[] arr) {
Arrays.sort(arr);
int sum = 0, counter = 0;
for (int i = 0; i < arr.length; i++) {
if (arr[0] == arr[arr.length - 1]) {
System.out.println(arr[0] + ": " + counter + " times");
break;
} else {
if (i == (arr.length - 1)) {
sum += arr[arr.length - 1];
counter++;
System.out.println((sum / counter) + " : " + counter
+ " times");
break;
} else {
if (arr[i] == arr[i + 1]) {
sum += arr[i];
counter++;
} else if (arr[i] != arr[i + 1]) {
sum += arr[i];
counter++;
System.out.println((sum / counter) + " : " + counter
+ " times");
sum = 0;
counter = 0;
}
}
}
}
}
public static void main(String[] args) {
int nums[] = { 1, 1, 1, 1, 2, 2, 2, 3, 3, 4, 5, 5, 5, 5, 6 };
count(nums);
}
}

You can do it by using Arrays.sort and Recursion. The same wine but in a different bottle....
import java.util.Arrays;
public class ArrayTest {
public static int mainCount=0;
public static void main(String[] args) {
String prevItem = "";
String[] array = {"name1","name1","name2","name2", "name2"};
Arrays.sort(array);
for(String item:array){
if(! prevItem.equals(item)){
mainCount = 0;
countArray(array, 0, item);
prevItem = item;
}
}
}
private static void countArray(String[] arr, int currentPos, String item) {
if(currentPos == arr.length){
System.out.println(item + " " + mainCount);
return;
}
else{
if(arr[currentPos].toString().equals(item)){
mainCount += 1;
}
countArray(arr, currentPos+1, item);
}
}
}

You can use Hash Map as given in the example below:
import java.util.HashMap;
import java.util.Set;
/**
*
* #author Abdul Rab Khan
*
*/
public class CounterExample {
public static void main(String[] args) {
String[] array = { "name1", "name1", "name2", "name2", "name2" };
countStringOccurences(array);
}
/**
* This method process the string array to find the number of occurrences of
* each string element
*
* #param strArray
* array containing string elements
*/
private static void countStringOccurences(String[] strArray) {
HashMap<String, Integer> countMap = new HashMap<String, Integer>();
for (String string : strArray) {
if (!countMap.containsKey(string)) {
countMap.put(string, 1);
} else {
Integer count = countMap.get(string);
count = count + 1;
countMap.put(string, count);
}
}
printCount(countMap);
}
/**
* This method will print the occurrence of each element
*
* #param countMap
* map containg string as a key, and its count as the value
*/
private static void printCount(HashMap<String, Integer> countMap) {
Set<String> keySet = countMap.keySet();
for (String string : keySet) {
System.out.println(string + " : " + countMap.get(string));
}
}
}

This is a simple script I used in Python but it can be easily adapted. Nothing fancy though.
def occurance(arr):
results = []
for n in arr:
data = {}
data["point"] = n
data["count"] = 0
for i in range(0, len(arr)):
if n == arr[i]:
data["count"] += 1
results.append(data)
return results

you can find using HashMap with simple technic
public class HashMapExample {
public static void main(String[] args) {
stringArray();
}
public static void stringArray()
{
String[] a = {"name1","name2","name3","name4", "name5"};
Map<String, String> hm = new HashMap<String, String>();
for(int i=0;i<a.length;i++)
{
String bl=(String)hm.get(a[i]);
if(bl==null)
{
hm.put(a[i],String.valueOf(1));
}else
{
String k=hm.get(a[i]);
int j=Integer.valueOf(k);
hm.put(a[i],String.valueOf(j+1));
}
}
//hm.entrySet();
System.out.println("map elements are "+hm.toString());
}
}

You could use
for (String x : array){
System.out.println(Collections.frequency(array,x));
}

You can use HashMap, where Key is your string and value - count.

// An Answer w/o using Hashset or map or Arraylist
public class Count {
static String names[] = {"name1","name1","name2","name2", "name2"};
public static void main(String args[]) {
printCount(names);
}
public static void printCount(String[] names){
java.util.Arrays.sort(names);
int n = names.length, c;
for(int i=0;i<n;i++){
System.out.print(names[i]+" ");
}
System.out.println();
int result[] = new int[n];
for(int i=0;i<n;i++){
result[i] = 0;
}
for(int i =0;i<n;i++){
if (i != n-1){
for(int j=0;j<n;j++){
if(names[i] == names[j] )
result[i]++;
}
}
else if (names[n-2] == names[n-1]){
result[i] = result[i-1];
}
else result[i] = 1;
}
int max = 0,index = 0;
for(int i=0;i<n;i++){
System.out.print(result[i]+" ");
if (result[i] >= max){
max = result[i];
index = i;
}
}
}
}

public class Main
{
public static void main(String[] args) {
String[] a ={"name1","name1","name2","name2", "name2"};
for (int i=0;i<a.length ;i++ )
{
int count =0;
int count1=0;
for(int j=0;j<a.length;j++)
{
if(a[i]==a[j])
{
count++;
}
}
for(int j=i-1;j>=0 ;j--)
{
if(a[i]==a[j])
{
count1++;
}
}
if(count1 ==0)
{
System.out.println(a[i]+" occurs :"+count);
}
}
}
}

import java.util.HashMap;
import java.util.Map;
public class FrequencyUsingMap {
public static void main(String[] args) {
int a[] = {1,1,1,1,2,2,3,4,1};
int num = 0;
int maxfreq = 0;
int maxnum =0;
Map<Integer, Integer> map = new HashMap<>();
for(int i = 0; i<a.length; i++){
num = a[i];
if(map.containsKey(num)){
int frq = map.get(num);
frq++;
map.put(num, frq);
if(frq>maxfreq){
maxfreq = frq;
maxnum = num;
}
}else{
map.put(num, 1);
}
}
System.out.println(map);
System.out.println("number "+ maxnum + " having max frequency " +maxfreq);
}
}

I wrote an easy solution for this, have a look:
public class Plus_Minus {
public static void main(String[] args) {
double [] x = { -4, 3, -9, -5, 4, 1 };
double p = 0;
double n = 0;
int z = 0;
for (int i = 0; i < x.length; i++) {
if (x[i] > 0) {
p += 1;
}
if (x[i] < 0) {
n += 1;
}
if (x[i] == 0) {
z += 1;
}
}
double ppi = p / x.length;
double pni = n / x.length;
int pzi = z / x.length;
System.out.println(ppi);
System.out.println(pni);
System.out.println(pzi);
}
}

public class test {
static String uniq[];
public static String[] convertWordArray(String str) {
str = str.toLowerCase();
String test[] = str.split(" ");
return test;
}
public static void findRepetitiveWordsinString(String str) {
String[] test =convertWordArray(str);
int len = test.length;
int count;
List<Integer> l = new ArrayList<>();
for (int i = 0; i < len; i++) {
count = 1;
for (int j = i + 1; j < len; j++) {
if (test[i].equals(test[j])) {
count++;
test[j] = "0";
}
}
if (count > 1 && test[i] != "0") {
System.out.println(test[i]);
l.add(i);
}
}
System.out.println("Repetitive words at index :" +l);
uniq = new String[l.size()];
for (int i = 0; i < l.size(); i++) {
uniq[i] = test[l.get(i)];
}
System.out.println("Number of words that are repeated: " + uniq.length);
}
public static void countMatches(String a[], String b[]) {
int count;
for (int i = 0; i < a.length; i++) {
count = 0;
for (int j = 0; j < b.length; j++) {
if (a[i].equals(b[j]))
count++;
}
if (count > 1) {
System.out.println("Repeating word is: " + a[i] + " and the repeating count is " + count);
}
}
}
public static void main(String[] args) {
String str;
Scanner scanner = new Scanner(System.in);
str = scanner.nextLine();
findRepetitiveWordsinString(str);
countMatches(uniq, convertWordArray(str));
}
}

import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.List;
public class MultiString {
public HashMap<String, Integer> countIntem( String[] array ) {
Arrays.sort(array);
HashMap<String, Integer> map = new HashMap<String, Integer>();
Integer count = 0;
String first = array[0];
for( int counter = 0; counter < array.length; counter++ ) {
if(first.hashCode() == array[counter].hashCode()) {
count = count + 1;
} else {
map.put(first, count);
count = 1;
}
first = array[counter];
map.put(first, count);
}
return map;
}
/**
* #param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
String[] array = { "name1", "name1", "name2", "name2", "name2",
"name3", "name1", "name1", "name2", "name2", "name2", "name3" };
HashMap<String, Integer> countMap = new MultiString().countIntem(array);
System.out.println(countMap);
}
}
Gives you O(n) complexity.