count occurence of word in a line of text

[sinyi88]

which method should i use to count occurence of words in a line of text if i just learn till the string and text i/o this topic.Beside the hashmap,got another method to use?i m not really undestand about hashmap...

[CJSLMAN]

Not sure if I understnd your question, but one way is that you could use the string split() method to separte the words in an array and then cycle through the array looking (by comparing) for the string ocurrences.

String API link:
String (Java Platform SE 6)

Luck,
CJSL

[sinyi88]

can i do like this...it seems no errors but no counting occur at all...why?i dont know how to compare the words inside a line between each other...


import java.util.*;

public class Question3{
public static void main(String[]args){

Scanner scan=new Scanner(System.in);

System.out.println("Enter a string:");
String s=scan.nextLine();

String[] words = s.split("[ ':.!?]");


String a=words[0];

int count=0;

for (int i=1;i<words.length;i++){

if (words[i].equals(a)){

count++;

a= words[i];
}

}

for (int i=1;i<words.length;i++){
System.out.print(words[i]+" "+count);
System.out.println();
}

}
}

[CJSLMAN]

I think what you need is two loops... the first loop (outer) would cycle through the words in the array and the second (inner) loop would compare the words from the first loop to all the words in the array. I think it would be something like...

String word = "";
for (int i=0;i<words.length;i++){
   word = words[i];
   for (int j=0;i<words.length;j++){
   if (words[j+1].equals(word)){
.
.
.
}
}

I would also to rename some of your variables... for example change variable "a" to "word", "words" to "wordArray", etc. give some meaning to the variables name.

Luck,
CJSL

[JVposter]

which method should i use to count occurence of words in a line of text if i just learn till the string and text i/o this topic.Beside the hashmap,got another method to use?i m not really undestand about hashmap...

StringUtils.countMatches("arthur", "r")

You do not used hash map for this :
please try read tutorials before puting questions

[CJSLMAN]

Sorry JVPoster, but countMatches() is not a Java API method. It maybe a third party API, but it is not part of the Java API specification (at least I couldn't find it).

CJSL

[CJSLMAN]

OK... so I'm curious about JVp's comment about using "StringUtils.countMatches()" and I decided to see where is it used. I opened Google and typed "java countmatch" and the first match that showed up was this:
String count matches : String Utils*«*Apache Common*«*Java
and guess what? it's extacly the same example that JVp posted. Now, to be able to use that method, you have to have had the following import statement:

import org.apache.commons.lang.StringUtils;

So... JVp's suggestion is worthless and he just posted the first thing that he found on the web that looked similiar... and from his other posts (equally worthless), he's either a troll or trying to bump his post count up to 20.

CJSL

[sinyi88]

Enter a string:
how are you
Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 3
at Question3.main(Question3.java:24)
Press any key to continue . . .

how to initialise the count array?

import java.util.*;

public class Question3{
public static void main(String[]args){

Scanner scan=new Scanner(System.in);

System.out.println("Enter a string:");
String s=scan.nextLine();

String[] words = s.split("[ ':.!?]");




String word = "";
int [] count=new int[100];
for (int i=0;i<words.length;i++){

word = words[i];

for (int j=0;j<words.length;j++){

if (words[j+1].equals(word)){
count[j+1]++;
System.out.println(words[i]+" "+count[j+1]);
}
}
}

}
}

[CJSLMAN]

OK, I had an error... the array was running out of bounds becasue of the [j+1]... simple to fix...

for (int j=1;j<words.length;j++){

and change all [j+1] to [J]

Luck,
CJSL

[sinyi88]

Enter a string:
how do you do
how 0
do 2
you 1
do 2
Press any key to continue . . .


why the word "do" print 2 times?and the word "how" not counting...



import java.util.*;

public class Question3{
public static void main(String[]args){

Scanner scan=new Scanner(System.in);

System.out.println("Enter a string:");
String s=scan.nextLine();

String[] words = s.split("[ ':.!?]");




String word = "";
int [] count=new int[100];
for (int i=0;i<words.length;i++){

word = words[i];

for (int j=1;j<words.length;j++){

if (words[j].equals(word))
count[j]++;

}
}

for (int x=0;x<words.length;x++){
System.out.println(words[x]+"\t"+count[x]);
}


}
}

[pbrockway2]

why the word "do" print 2 times?

Umm... because there's two of them? ;-)

and the word "how" not counting...

The loop where you count the number of occurrences starts with j=1 so the first word is never compared with itself to get a count of 1.

[pbrockway2]

Thinking about this, if you really don't want "do" to be counted and reported twice then a map of some sort really is the easiest way to go.

[pbrockway2]

Or something like this in the loop that does the counting:

FOR j=0 j<words.length j++
    IF words[j].equals(word)
        increment count[i] (not j)
        IF i!=j
            make count[j]=0

That way only the last of the counts for a given word will be nonzero.

[sinyi88]

is it like this?i dont understand what u mean...sorry...still got errors although can run



import java.util.*;

public class Question3{
public static void main(String[]args){

Scanner scan=new Scanner(System.in);

System.out.println("Enter a string:");
String s=scan.nextLine();

String[] words = s.split("[ ':.!?]");




String word = "";
int [] count=new int[20];
for (int i=0;i<words.length;i++){

word = words[i].toLowerCase();

for (int j=0;j<words.length;j++){

if (words[j].equals(word)){

count[i]++;
}
if (i != j){
count[j]=0;

}
}
}


for (int y=0;y<words.length;y++){

System.out.println(words[y]+"\t"+count[y]);
}





}
}

[pbrockway2]

I have no idea what the errors might be unless you say.

You have misplaced the if statement within the second for loop. To see how to correct it, think for a bit about what the algorithm is trying to do.

Your original version updated the count every time it found two words the same. What I am suggesting is that it continues to do this - but that it should also "reset" the counts for all the other equal words.

For instance when you count the occurrences in "How do you do" the counting would go like (each row represents one time round the outer for loop)

How do you do
1   0  0    0 
How do you do
1   2  0    0 
How do you do
1   2  1    0 
How do you do
1   0  1    2

The idea is that when you are counting all the words that are equal to the second "do" you reset the count for the first one back to zero.

Then when you output the counts you can just ignore words with a zero count. (You are not doing this at the moment. You will have to change the second for loop so that the first occurrence of "do" - the one with count 0 - is not printed.)

One small point - you should decide whether your string comparisons should be case sensitive or not and then stick to that. At the moment you are initialising word as the lower case version of the word in your original string, but then comparing it to the original within the inner for loop.

[sinyi88]

That's all i can do...but the print out got problems...i dont know how to avoid to print words that already printed...the 'do' still print 2 times...:(

Enter a string:
how do you do
how 1
do 2
you 1
do 1
Press any key to continue . . .


import java.util.*;

public class Question3{
public static void main(String[]args){

Scanner scan=new Scanner(System.in);

System.out.println("Enter a string:");
String s=scan.nextLine();

String[] words = s.split("[ ':.!?]");




String word = "";
int [] count=new int[20];
for (int i=0;i<words.length;i++){

word = words[i].toLowerCase();

for (int j=1;j<words.length;j++){
if (i != j){
if (words[j].equals(word)){

count[i]++;

}
}


if (i == j){
if (words[j].equals(word)){

count[i]=1;

}
}

if (!words[j].equals(words[0])){
count[0]=1;
}
}
}

for (int y=0;y<words.length;y++){
System.out.println(words[y]+"\t"+count[y]);
}





}
}

[sinyi88]

can somebody help me?reply me...:(

[pbrockway2]

When you post code, highlight it and use the code button (lebelled #).

import java.util.*;

public class Question3 {
    public static void main(String[]args){

        Scanner scan=new Scanner(System.in);

        System.out.println("Enter a string:");
        String s=scan.nextLine();

        String[] words = s.split("[ ':.!?]");

        String word = "";
        int [] count=new int[20];
        for (int i=0;i<words.length;i++){

            word = words[i].toLowerCase();

            for (int j=0;j<words.length;j++){
                    // first increment i's count for every
                    // equal word
                if (words[j].toLowerCase().equals(word)){
                    count[i]++;
                        // then make sure the other instances of this
                        // word get a zero count
                    if (i != j){
                        count[j] = 0;
                    }
                }
            }
        }
            // correct this loop so that it doesn't
            // print the zero counts
        for (int y=0;y<words.length;y++){
            System.out.println(words[y]+"\t"+count[y]);
        }
    }
}

I've given up and put the if statement where it should be, corrected the string comparison and changed the error reintroduced into the inner loop (j=0;...). But I've left you the output loop!

Once you understand what is happening you might try to make it more efficient. (Would it help to change the words array so that it contained only lower case strings? Do we really need to check every word with every other - or is it enough to check every word with all those with a lower index?)

[sinyi88]

i correct it already...it can be run now...thank you...but can u tell me how to sort the words alphabetically when printed out?

[pbrockway2]

can u tell me how to sort the words alphabetically when printed out?

(Just an aside: using a Map - which your OP said you didn't want to do - makes this trivial. A TreeMap for instance will store the counts for each word so that they can be accessed alphabetically without any extra code at all.)

The alternative is to write a sorting algorithm that sorts both arrays alphabetically on the words array.