Thread: Vector of array [logic error]

General explaination:

Hi, I want to fill a vector of an array of strings tokenized that I pass in the arguments, then. I must fill an array that contain only once each word that appear in all strings, even if there are more than twice, because next step i must do, is to count the times each word is appearing.
That's why I use an string array inside vector, because I need method "split" and Class Vector hasn't any "splitter".

My problem:
Fail is in the condition if(!v.contains(vFrases.elementAt(i)[j])){ that is always true and it should be true when it contains twice a word

Maybe I missunderstood how to fill the array index in a vector
Maybe can I obtain same result I want, with a simple Vector? instead vector of arrays?

 
import java.util.Vector;
 
public class Ejercicio {
 
        /** MAIN */
	public static void main(String[] args) {
 
		if(args.length<1){
			System.err.println("Usage: java Ejercicio <string> <string> ");
			return;
		}
 
		frecuenciasLexicas(args);
	}
 
        /** METHOD */
	static void frecuenciasLexicas (String[] s){
 
		Vector<String> v = new Vector<String>();
		Vector<String[]> vFrases = new Vector<String[]>();
 
                //I fill the vector of arrays with the strings I pass in the args "tokenized" with "split"
		for(int i=0; i<s.length ;i++)
			vFrases.addElement(s[i].split("\\s"));
 
 
		for(int i=0; i<vFrases.size() ;i++){
			for(int j=0; j<vFrases.elementAt(i).length ;j++){
 
                                /** HERE is the PROBLEM, this condition is always false */
                                /** The vector "v" should contain only once each word that vector "vFrase" has inside, so if v doesn't contain the string, he add the string, but next time will not, because only must have it once each word*/
				if(!v.contains(vFrases.elementAt(i)[j])){
					v.addElement(vFrases.elementAt(i)[j]+"|");
				}
			}
			for(int k=0; k<v.size();k++)
				System.out.println(v.elementAt(k));
		}
	}
}

ARG EXAMPLE:
"first string" "second string" "third string"
DESIRED OUTPUT (INSIDE V):

v.elementAt(0) -->first|
v.elementAt(1) -->string|
v.elementAt(2) -->second|
v.elementAt(3) -->third|


WHAT I GET:
v.elementAt(0) --> first|
v.elementAt(1) --> string|
v.elementAt(2) --> second|
v.elementAt(3) --> string|
v.elementAt(4) --> third|
v.elementAt(5) --> string|


Sorry for my english XD,
Thank you.

Why not use a Map<String, Integer> such as a HashMap<String, Integer> to help you do the counting? The keys for the map would be your String tokens, and the values would be the number of times that the token is present.

Originally Posted by curmudgeon

Why not use a Map<String, Integer> such as a HashMap<String, Integer> to help you do the counting? The keys for the map would be your String tokens, and the values would be the number of times that the token is present.

I am java begginer, i don't know what is a map /hasmap i am training with vectors and solving problems with them

--- Update ---

Originally Posted by Norm

I added this line first thing in the main() method to get the input you described:

      args = new String[] {"first string", "second string", "third string"}; //<<<<<<<<<<

and get this error:
ArrayIndexOutOfBoundsException: 2 >= 2

Can you show the contents of the console for when you execute the program?

On windows: To copy the contents of the command prompt window:
Click on Icon in upper left corner
Select Edit
Select 'Select All' - The selection will show
Click in upper left again
Select Edit and click 'Copy'

Paste here.

sorry i edited first post with right code. i forgot a testcode inside the code. fixed
now works

Originally Posted by Norm

Do some debugging by adding a println statement to print out the contents of v and the elementAt()'s value so you can see what is in v and what the contains() method is looking for.

i was many hours debugging, my last step is posting here (i deleted the maaaany comments of printlns and fors that i have for you dont go crazy )

i even drawed it in a paper many times but dont undrstand why i get that output, maybe i dont use good the method "contains"

do you mean this?

for(int k=0; k<v.size();k++)
	System.out.println(v.elementAt(k));

the API says :
contains(Object o)
Returns true if this vector contains the specified element.

Isn't a string an object? or what's wrong, don't get it

or it is the fact it contains a string that it make it true?

for(int i=0; i<vFrases.size() ;i++){
			for(int j=0; j<vFrases.elementAt(i).length ;j++){
				System.out.println("vFrases.elementat("+i+")["+j+"] "+vFrases.elementAt(i)[j]);
				System.out.println(!v.contains(vFrases.elementAt(i)[j]));
				if(!v.contains(vFrases.elementAt(i)[j]))
					v.addElement(vFrases.elementAt(i)[j]+"|");
			}
			for(int k=0; k<v.size();k++)
				System.out.println("v.element("+k+"):\t"+v.elementAt(k));
		}

It prints:

vFrases.elementat(0)[0]: First
true
vFrases.elementat(0)[1]: string
--> HERE it catch the word "string" for first time, into "v" vector <--
true
v.element(0): First|
v.element(1): string|
vFrases.elementat(1)[0]: Second
true
vFrases.elementat(1)[1]: string
---->HERE SHOULD RETURN FALSE, because it contains already the word "string"<----
true
v.element(0): First|
v.element(1): string|
v.element(2): Second|
v.element(3): string|
vFrases.elementat(2)[0]: Thrid
true
vFrases.elementat(2)[1]: string
---->HERE SHOULD RETURN FALSE TOO,
true
v.element(0): First|
v.element(1): string|
v.element(2): Second|
v.element(3): string|
v.element(4): Thrid|
v.element(5): string|

Originally Posted by Norm

Where do you see "string" in v?
v.element(1) is close: "string|" but it is different.
contains() would not find it as a match!

OH MY GOD !

I would delete all my code ant start by zero using only arrays if you don't tell me this.
thank you very much
thank you for your time

FIXED:

if(!v.contains(vFrases.elementAt(i)[j])+"|")    /**  same string that I add if true.....*/
	v.addElement(vFrases.elementAt(i)[j]+"|");

I'll try to examine strings more carefully hehe. (I had problem added that I may not understood how Vector Methods works. I only know a few methods at the moment).
Have a good day