Solution for Programmming Exercise 7.6
This page contains a sample solution to one of the exercises from Introduction to Programming Using Java.
Exercise 7.6:
Write a program that will read a text file selected by the user, and will make an alphabetical list of all the different words in that file. All words should be converted to lower case, and duplicates should be eliminated from the list. The list should be written to an output file selected by the user. As discussed in Subsection 2.4.5, you can use TextIO to read and write files. Use a variable of type ArrayList<String> to store the words. (See Subsection 7.3.4.) It is not easy to separate a file into words as you are reading it. You can use the following method:
/**
* Read the next word from TextIO, if there is one. First, skip past
* any non-letters in the input. If an end-of-file is encountered before
* a word is found, return null. Otherwise, read and return the word.
* A word is defined as a sequence of letters. Also, a word can include
* an apostrophe if the apostrophe is surrounded by letters on each side.
* @return the next word from TextIO, or null if an end-of-file is
* encountered
*/
private static String readNextWord() {
char ch = TextIO.peek(); // Look at next character in input.
while (ch != TextIO.EOF && ! Character.isLetter(ch)) {
TextIO.getAnyChar(); // Read the character.
ch = TextIO.peek(); // Look at the next character.
}
if (ch == TextIO.EOF) // Encountered end-of-file
return null;
// At this point, we know that the next character, so read a word.
String word = ""; // This will be the word that is read.
while (true) {
word += TextIO.getAnyChar(); // Append the letter onto word.
ch = TextIO.peek(); // Look at next character.
if ( ch == '\'' ) {
// The next character is an apostrophe. Read it, and
// if the following character is a letter, add both the
// apostrophe and the letter onto the word and continue
// reading the word. If the character after the apostrophe
// is not a letter, the word is done, so break out of the loop.
TextIO.getAnyChar(); // Read the apostrophe.
ch = TextIO.peek(); // Look at char that follows apostrophe.
if (Character.isLetter(ch)) {
word += "\'" + TextIO.getAnyChar();
ch = TextIO.peek(); // Look at next char.
}
else
break;
}
if ( ! Character.isLetter(ch) ) {
// If the next character is not a letter, the word is
// finished, so bread out of the loop.
break;
}
// If we haven't broken out of the loop, next char is a letter.
}
return word; // Return the word that has been read.
}
Note that this method will return null when the file has been entirely read. You can use this as a signal to stop processing the input file.
This is actually not a very difficult program to write. The main point of the exercise is to get you to use a list of strings and to do something with files.
There are several possible approaches to this problem. One approach is to simply dump all the words from the file into a list, without worrying about eliminating duplicates or keeping the list in order. After the file has been read, the list can be sorted and printed. Although the list can contain duplicates, the output file should list each word only once. However, it's easy to leave out the duplicates as the file is being written. If wordList is the variable of type ArrayList<String> that holds the already sorted list of words, then the following code will output the list without duplicates. The idea is that a word is written only if it is different from the previous word in the list. Word number 0 is a special case, because there is no previous location in the list in that case:
for (int i = 0; i < wordlist.size(); i++) {
if (i == 0 || ! wordlist.get(i).equals(wordList.get(i-1) )
TextIO.putln(wordlist.get(i));
}
A second approach to the problem would be to keep the list of words in sorted order as it is being constructed. This can be done by applying the insert() routine from Subsection 7.4.3 to insert each word into the list.
A third approach, and the one that I use in my solution, is to eliminate duplicates from the list as it being constructed. Each time a word is read from the input file, I first check whether the word is already in the list. If so, I discard it; if not, I add it to the end of the list. The function wordList.indexOf(word) can be used to test whether a given word is already in the list; this function returns the value -1 if word is not in the list. After the input file has been read, the list contains one copy of each word that was found in the file. At this point the list still has to be sorted. I use a selection sort algorithm (Subsection 7.4.4) to do the sorting. Then, all the elements of the list are output using a for-each loop:
for (String w : wordList)
TextIO.putln(" " + w);
To let the user select the input and the output files, I use the methods TextIO.readUserSelectedFile() and TextIO.writeUserSelectedFile(), which are discussed in Subsection 2.4.5. These methods put up a file dialog where the user can select a file. After the user selects an input file, TextIO reads from that file instead of from the user's input. After the user selects an output file, TextIO writes to that file instead of to standard output. If the user cancels the input file dialog, then there is no input file to process, so I exit the program. If the user cancels the output file dialog, I write the list of words anyway -- it will go to standard output so the user will see it on the screen.
When TextIO is working with files and an error occurs, it will generate an error of type IllegalArgumentException. My program catches the error if one occurs, and it prints an error message. An error is not very likely in this case but one could occur if, for example, the user selects an input file that the user does not have permission to read.
The complete code for my solution is shown below.
import java.util.ArrayList;
/**
* Makes an alphabetical list of all the words in a file selected
* by the user. The list can be written to a file.
*/
public class ListAllWordsFromFile {
public static void main(String[] args) {
System.out.println("\n\nThis program will ask you to select an input file");
System.out.println("It will read that file and make an alphabetical");
System.out.println("list of all the words in the file. After reading");
System.out.println("the file, the program asks you to select an output");
System.out.println("file. If you select a file, the list of words will");
System.out.println("be written to that file; if you cancel, the list");
System.out.println("be written to standard output. All words are converted");
System.out.println("lower case, and duplicates are eliminated from the list.\n\n");
System.out.print("Press return to begin.");
TextIO.getln(); // Wait for user to press return.
try {
if (TextIO.readUserSelectedFile() == false) {
System.out.println("No input file selected. Exiting.");
System.exit(1);
}
ArrayList<String> wordList = new ArrayList<String>();
String word = readNextWord();
while (word != null) {
word = word.toLowerCase(); // convert word to lower case
if ( wordList.indexOf(word) == -1 ) {
// This is a new word, so add it to the list
wordList.add(word);
}
word = readNextWord();
}
System.out.println("Number of different words found in file: "
+ wordList.size());
System.out.println();
if (wordList.size() == 0) {
System.out.println("No words found in file.");
System.out.println("Exiting without saving data.");
System.exit(0);
}
selectionSort(wordList);
TextIO.writeUserSelectedFile(); // If user cancels, output automatically
// goes to standard output.
TextIO.putln(wordList.size() + " words found in file:\n");
for (String w : wordList)
TextIO.putln(" " + w);
System.out.println("\n\nDone.\n\n");
}
catch (Exception e) {
System.out.println("Sorry, an error has occurred.");
System.out.println("Error Message: " + e.getMessage());
}
System.exit(0); // Might be necessary, because of use of file dialogs.
}
/**
* Sorts a list of strings into lexicographical order, using
* selection sort and treating the list much like an array. In this
* program, the list only contains words made up of lower case
* letters, so lexicographic order is the same as alphabetical order.
*/
private static void selectionSort(ArrayList<String> list) {
for (int top = list.size() - 1; top > 0; top--) {
int indexOfBiggest = 0;
for (int j = 0; j < top; j++) {
String str = list.get(j);
if (str.compareTo( list.get(indexOfBiggest) ) > 0) {
indexOfBiggest = j;
}
}
String temp = list.get(top);
list.set( top, list.get(indexOfBiggest) );
list.set( indexOfBiggest, temp );
}
}
/**
* Read the next word from TextIO, if there is one. First, skip past
* any non-letters in the input. If an end-of-file is encountered before
* a word is found, return null. Otherwise, read and return the word.
* A word is defined as a sequence of letters. Also, a word can include
* an apostrophe if the apostrophe is surrounded by letters on each side.
* @return the next word from TextIO, or null if an end-of-file is encountered
*/
private static String readNextWord() {
char ch = TextIO.peek(); // Look at next character in input.
while (ch != TextIO.EOF && ! Character.isLetter(ch)) {
TextIO.getAnyChar(); // Read the character.
ch = TextIO.peek(); // Look at the next character.
}
if (ch == TextIO.EOF) // Encountered end-of-file
return null;
// At this point, we know that the next character, so read a word.
String word = ""; // This will be the word that is read.
while (true) {
word += TextIO.getAnyChar(); // Append the letter onto word.
ch = TextIO.peek(); // Look at next character.
if ( ch == '\'' ) {
// The next character is an apostrophe. Read it, and
// if the following character is a letter, add both the
// apostrophe and the letter onto the word and continue
// reading the word. If the character after the apostrophe
// is not a letter, the word is done, so break out of the loop.
TextIO.getAnyChar(); // Read the apostrophe.
ch = TextIO.peek(); // Look at char that follows apostrophe.
if (Character.isLetter(ch)) {
word += "\'" + TextIO.getAnyChar();
ch = TextIO.peek(); // Look at next char.
}
else
break;
}
if ( ! Character.isLetter(ch) ) {
// If the next character is not a letter, the word is
// finished, so bread out of the loop.
break;
}
// If we haven't broken out of the loop, next char is a letter.
}
return word; // Return the word that has been read.
}
}