comparing string arrays in java

Question

I am working on a simple Java program where we have sorted string array named arr I am trying to compare two adjacent string and calculate frequency for each string present in that array

for(j1=0;j1<arr.length;j1++){
    if(j1+1 < arr.length){ // To prevent arrayOutofBoundsException
         if(arr[j1].equals(arr[j1+1])){
              counter++;
         }
         else {
              System.out.println(arr[j1]+" "+counter);
              counter=1;
         }

But it's not working right , what's wrong ? edit:problem is not in comparing , it's not calculating frequency as desired

What do you mean by "It's not working right"? And have you read [How do I compare strings in Java?](http://stackoverflow.com/questions/513832/how-do-i-compare-strings-in-java) — PakkuDon, Feb 14 '14 at 13:15
Fixed that problem but the thing is it's not giving desired output — Sigma, Feb 14 '14 at 13:18
@Sigma Use your debugger and try to see what's wrong with your logic. — Alexis C., Feb 14 '14 at 13:19
@ZouZou I have been banging my head since yesterday but still i was't able to solve so asked you — Sigma, Feb 14 '14 at 13:20
@Sigma: Could you provide us with some sample input, output and expected output? — PakkuDon, Feb 14 '14 at 13:23
I enter "how are you how do you do" and i expect output to be how 2 are 1 do 2 you 1 — Sigma, Feb 14 '14 at 13:25
I can't see where do you sort the array. Are you sure you did it? — gaborsch, Feb 14 '14 at 13:26
@GaborSch yes i did it before trying to compare , i just posted here comparing — Sigma, Feb 14 '14 at 13:27
@AJ. Stack Overflow... Best place to get your homework done. ;)(pun intended) — Zeeshan, Feb 14 '14 at 13:27

gaborsch · Answer 1 · 2014-02-14T19:52:19.840

2

OK, besides the equals fix, you want to keep the original order of words:

String orig = "hellow hello hello how how he ho" ;
//String orig = "how are you how do you do";

String[] arr = orig.split(" ");

//Arrays.sort(arr);

for(int j1 = 0; j1 < arr.length; j1++){
    if (arr[j1] != null) {
        int counter = 1;
        for(int j2 = j1+1; j2 < arr.length; j2++) {
            if(arr[j2] != null && arr[j1].equals(arr[j2])){
                counter++;
                arr[j2] = null;
            }
        }
        System.out.println(arr[j1]+" "+counter);
    }
}

The trick is that I run through the array, count all occurrences, null the occurrences, so they don't count again, and print the count. No need to sort the array.

edited Feb 14 '14 at 19:52

answered Feb 14 '14 at 13:16

gaborsch

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please check again , I am sorry for that mistake – Sigma Feb 14 '14 at 13:19
@Sigma I found one bug in your code, please recheck. – gaborsch Feb 14 '14 at 13:32
Please provide some sample input and output. – gaborsch Feb 14 '14 at 13:36
@Sigma Are you splitting the input by whitespace? – Alexis C. Feb 14 '14 at 13:43
@GaborSch any heads on this one? – Sigma Feb 14 '14 at 13:49
OK, that is the required one. But what do you *actually* get? – gaborsch Feb 14 '14 at 14:34
OK, I understand our problem. You want to *keep the oroginal order* while counting (in "how are you do" order, not "are do how you"). Please check the latest code. – gaborsch Feb 14 '14 at 19:47

score 0 · Answer 2 · answered Feb 14 '14 at 13:16

0

== compares object identity in terms of memory address - to compare objects in terms of equality, use the equals-method.

answered Feb 14 '14 at 13:16

Smutje

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score 0 · Answer 3 · answered Feb 14 '14 at 13:36

This should work:

public static void main(String[] args)
    {       
        String text = "how are you how do you do";
        String[] keys = {"how", "are", "you", "how", "do", "you", "do"};
         Arrays.sort(keys);
        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:

how are you how do you do
Count of [how] is : 2
Count of [are] is : 1
Count of [you] is : 2
Count of [do] is : 2

score -1 · Answer 4 · answered Feb 14 '14 at 13:38

Are you looking some sort of this

public static void main(String[] args){
        String[] arr = new String[5];
        arr[0] =  "One";
        arr[1] =  "Two";
        arr[2] =  "One";
        arr[3] =  "Three";
        arr[4] =  "Two";

        List<String> lstString = Arrays.asList(arr);
        Collections.sort(lstString);
        for(String eachString : arr){
            System.out.println("Frequency of " + eachString + " is " + getFrequency(eachString,lstString));
        }
    }

    private static int getFrequency(String word, List lstOfString){
        int frequency = 1;
        if(lstOfString != null && lstOfString.size() > 0){
            int firstIndex = lstOfString.indexOf(word);
            int lastIndex = lstOfString.lastIndexOf(word);
            frequency += lastIndex - firstIndex;
        }
        return frequency;
    }

Result : Frequency of One is 2 Frequency of One is 2 Frequency of Three is 1 Frequency of Two is 2 Frequency of Two is 2

Add the reason or explain properly what is your desire. – Kumar Feb 14 '14 at 13:59 — Kumar, Feb 14 '14 at 13:59

comparing string arrays in java

4 Answers4