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Thread: Array Histogram

  1. #1
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    Default Array Histogram

    Hello, I'm having trouble coding a program which reads in strings (search terms) from a file, orders them, and then creates a histogram showing the occurrence of each string length.

    For example, if the search terms were: frog, cat, dog; the program would print the following output:

    Search Terms
    -------------
    frog
    cat
    dog

    Term Length ... Count
    ------------ ... ------
    3 .................... 2
    4 .................... 1

    I have the program reading in terms and sorting them, but I can't get the histogram to display the number of occurrences for each term.

    I get output like this:
    3 .......... 1
    3 .......... 1
    4 .......... 1

    I know it's a logic error, but what should I be doing?

    package nicepackage;
     
    import java.io.*;
    import java.util.Scanner;
     
    public class Search {
     
    	/**
    	 * @param args
    	 */
    	public static void main(String[] args) throws Exception {
    		System.out.println("\nSearch Terms");
    		System.out.print("------------");
     
    		FileReader fr1 = new FileReader("C:\\users\\T\\inSearchterms.txt");
    		Scanner inFr1 = new Scanner (fr1);
     
    		int n = inFr1.nextInt()+1;
    		String[] a = new String[n];
    		for (int i=0; i<n; i++)
    		{
    			a[i]= inFr1.nextLine();
    			//System.out.println(a[i]);
    		}
    		selectionSort(a);
    		System.out.println("\nTerm Length	Count");
    		System.out.println("-----------	-----");
     
    		int[] histogram = new int[n];
    		int[] ct = new int[n];
     
    		for (int b=1; b <= a.length-1; b++)
    		{
    			for (int d = 30; d >= 0; d--)
    			{
    				if (d == a[b].length())
    				{
    					histogram[b] = a[b].length();
    					ct[b]++;
    				}
    			}
    			System.out.println(histogram[b] + "		" + ct[b]);
    		}
    	}
     
    	public static void selectionSort (String [] a)
    	{
    		for (int i=a.length-1; i >= 1; i--)
    		{
    			String currentMax = a[0];
    			int currentMaxIndex = 0;
    			for (int j=1; j <= i; j++)
    			{
    				if (currentMax.compareToIgnoreCase(a[j])<0)
    				{
    					currentMax = a[j];
    					currentMaxIndex = j;
    				}
    			}
    			if (currentMaxIndex != i)
    			{
    				a[currentMaxIndex] = a[i];
    				a[i] = currentMax;
    			}
    			//System.out.println("/"+a[i]);
    		}
     
    	}
    }

    What am I doing wrong? Any help is greatly appreciated, thanks!


  2. #2
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    Default Re: Array Histogram

    Do you know how to use a Map? It would be useful: key=length of String, value=count

    Otherwise use an array. What is the max String length? Have an array with a slot for each length and increment the slots as per String length.
    If you don't understand my answer, don't ignore it, ask a question.

  3. #3
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    Default Re: Array Histogram

    Unfortunately, I don't know how to use a Map.

    I need the program to determine the max String length, can I do that using the process you described? If I'm understanding correctly, I think that's what I'm trying to do. I set up two int arrays, one called histogram and one called ct, each with as many slots as there are terms in the file (or at least that's what I intended to do).

  4. #4
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    Default Re: Array Histogram

    Read all the strings and remember the length of the longest one, create the array then read them again and count the lengths.
    If you don't understand my answer, don't ignore it, ask a question.

  5. #5
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    Default Re: Array Histogram

    I know that the longest String in the file I'm using is 26 characters, but I can't just create the array specific to the Strings in this file because I need it to work with any file.

  6. #6
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    Default Re: Array Histogram

    One way would be to make two passes. The first to find the longest, the second to count.
    Another way would be to use parallel arrays, have one array contain the length, and the other the count.


    Using the new collection classes would make this easier.
    If you don't understand my answer, don't ignore it, ask a question.

  7. #7
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    Default Re: Array Histogram

    Thanks for your help so far.

    I think I'm trying to use parallel arrays. Am I initializing them correctly in my code? I believe that's where I'm getting errors.

  8. #8
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    Default Re: Array Histogram

    java.lang.ArrayIndexOutOfBoundsException: 16
    The index is past the end of the array. Remember array indexes go from 0 to the length-1
    If you don't understand my answer, don't ignore it, ask a question.

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    Default h

    Can somebody take another look at this, please? I'm losing my mind trying to figure this out.

  10. #10
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    Default Re: Array Histogram

    Please post the program's output and explain what is wrong with it and show what it should be.
    If you don't understand my answer, don't ignore it, ask a question.

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