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Symbol Tables
Sets Inside a Map
Using a Comparator
Word Counting

Section 10.4

Programming with the Collection Framework

In this section, we'll look at some programming examples that use classes from the Java Collection Framework. The Collection Framework is easy to use, especially compared to the difficulty of programming new data structures from scratch.

10.4.1  Symbol Tables

We begin with a straightforward but important application of maps. When a compiler reads the source code of a program, it encounters definitions of variables, subroutines, and classes. The names of these things can be used later in the program. The compiler has to remember the definition of each name, so that it can recognize the name and apply the definition when the name is encountered later in the program. This is a natural application for a Map. The name can be used as a key in the map. The value associated to the key is the definition of the name, encoded somehow as an object. A map that is used in this way is called a symbol table.

In a compiler, the values in a symbol table can be quite complicated, since the compiler has to deal with names for various sorts of things, and it needs a different type of information for each different type of name. We will keep things simple by looking at a symbol table in another context. Suppose that we want a program that can evaluate expressions entered by the user, and suppose that the expressions can contain variables, in addition to operators, numbers, and parentheses. For this to make sense, we need some way of assigning values to variables. When a variable is used in an expression, we need to retrieve the variable's value. A symbol table can be used to store the data that we need. The keys for the symbol table are variable names. The value associated with a key is the value of that variable, which is of type double. The symbol table will be an object of type Map<String,Double>. (Remember that primitive types such as double can't be used as type parameters; a wrapper class such as Double must be used instead. See Subsection 10.1.7.)

To demonstrate the idea, we'll use a rather simple-minded program in which the user types commands such as:

let x = 3 + 12
print 2 + 2
print 10*x +17
let rate = 0.06
print 1000*(1+rate)

The program is an interpreter for a very simple language. The only two commands that the program understands are "print" and "let". When a "print" command is executed, the computer evaluates the expression and displays the value. If the expression contains a variable, the computer has to look up the value of that variable in the symbol table. A "let" command is used to give a value to a variable. The computer has to store the value of the variable in the symbol table. (Note: The "variables" I am talking about here are not variables in the Java program. The Java program is executing a sort of program typed in by the user. I am talking about variables in the user's program. The user gets to make up variable names, so there is no way for the Java program to know in advance what the variables will be.)

In Subsection 9.5.2, we saw how to write a program, SimpleParser2.java, that can evaluate expressions that do not contain variables. Here, I will discuss another example program, SimpleInterpreter.java, that is based on the older program. I will only talk about the parts that are relevant to the symbol table. Here is an applet that simulates the program:

The program uses a HashMap as the symbol table. A TreeMap could also be used, but since the program does not need to access the variables in alphabetical order, we don't need to have the keys stored in sorted order. The symbol table in the program is represented by a variable named symbolTable of type HashMap<String,Double>. At the beginning of the program, the symbol table object is created with the command:

symbolTable = new HashMap<String,Double>();

This creates a map that initially contains no key/value associations. To execute a "let" command, the program uses the symbol table's put() method to associate a value with the variable name. Suppose that the name of the variable is given by a String, varName, and the value of the variable is stored in a variable val of type double. The following command would then set the value associated with the variable in the symbol table:

symbolTable.put( varName, val );

In the program SimpleInterpreter.java, you'll find this in the method named doLetCommand(). The actual value that is stored in the symbol table is an object of type Double. We can use the double value val in the call to put because Java does an automatic conversion of type double to Double when necessary. The double value is "wrapped" in an object of type Double, so that, in effect, the above statement is equivalent to

symbolTable.put( varName, new Double(val) );

Just for fun, I decided to pre-define two variables named "pi" and "e" whose values are the usual mathematical constants π and e. In Java, the values of these constants are given by Math.PI and Math.E. To make these variables available to the user of the program, they are added to the symbol table with the commands:

symbolTable.put( "pi", Math.PI );
symbolTable.put( "e", Math.E );

When the program encounters a variable while evaluating an expression, the symbol table's get() method is used to retrieve its value. The function symbolTable.get(varName) returns a value of type Double. It is possible that the return value is null; this will happen if no value has ever been assigned to varName in the symbol table. It's important to check this possibility. It indicates that the user is trying to use a variable that the user has not defined. The program considers this to be an error, so the processing looks something like this:

Double val = symbolTable.get(varName);
if (val == null) {
   ... // Throw an exception:  Undefined variable.
// The value associated to varName is val.doubleValue()

You will find this code, more or less, in a method named primaryValue() in SimpleInterpreter.java.

As you can see from this example, Maps are very useful and are really quite easy to use.

10.4.2  Sets Inside a Map

The objects in a collection or map can be of any type. They can even be collections. Here's an example where it's natural to store sets as the value objects in a map.

Consider the problem of making an index for a book. An index consists of a list of terms that appear in the book. Next to each term is a list of the pages on which that term appears. To represent an index in a program, we need a data structure that can hold a list of terms, along with a list of pages for each term. Adding new data should be easy and efficient. When it's time to print the index, it should be easy to access the terms in alphabetical order. There are many ways this could be done, but I'd like to use Java's generic data structures and let them do as much of the work as possible.

We can think of an index as a Map that associates a list of page references to each term. The terms are keys, and the value associated with a given key is the list of page references for that term. A Map can be either a TreeMap or a HashMap, but only a TreeMap will make it easy to access the terms in sorted order. The value associated with a term is a list of page references. How can we represent such a value? If you think about it, you see that it's not really a list in the sense of Java's generic classes. If you look in any index, you'll see that a list of page references has no duplicates, so it's really a set rather than a list. Furthermore, the page references for a given term are always printed in increasing order, so we want a sorted set. This means that we should use a TreeSet to represent each list of page references. The values that we really want to put in this set are of type int, but once again we have to deal with the fact that generic data structures can only hold objects, so we must use the wrapper class, Integer, for the objects in the set.

To summarize, an index will be represented by a TreeMap. The keys for the map will be terms, which are of type String. The values in the map will be TreeSets that contain the Integers which give the page numbers of every page on which a term appears. The parameterized type that we should use for the sets is TreeSet<Integer>. For the TreeMap that represents the index as a whole, the key type is String and the value type is TreeSet<Integer>. This means that the index has type

TreeMap< String, TreeSet<Integer> >

This is just the usual TreeMap<T,S> with T=String and S=TreeSet<Integer>. A type name as complicated as this one can look intimidating (especially, I think, when used in a constructor with the new operator), but if you think about the data structure that we want to represent, it makes sense. Given a little time and practice, you can get used to types like this one.

To make an index, we need to start with an empty TreeMap, look through the book, inserting every reference that we want to be in the index into the map, and then print out the data from the map. Let's leave aside the question of how we find the references to put in the index, and just look at how the TreeMap is used. It can be created with the commands:

TreeMap<String,TreeSet<Integer>>  index;         // Declare the variable.
index = new TreeMap<String,TreeSet<Integer>>();  // Create the map object.

Now, suppose that we find a reference to some term (of type String) on some pageNum (of type int). We need to insert this information into the index. To do this, we should look up the term in the index, using index.get(term). The return value is either null or is the set of page references that we have previously found for the term. If the return value is null, then this is the first page reference for the term, so we should add the term to the index, with a new set that contains the page reference we've just found. If the return value is non-null, we already have a set of page references, and we should just add the new page reference to the set. Here is a subroutine that does this:

 * Add a page reference to the index.
void addReference(String term, int pageNum) {
   TreeSet<Integer> references; // The set of page references that we
                                //    have so far for the term.
   references = index.get(term);
   if (references == null){
          // This is the first reference that we have
          // found for the term.  Make a new set containing
          // the page number and add it to the index, with
          // the term as the key.
       TreeSet<Integer> firstRef = new TreeSet<Integer>();
       firstRef.add( pageNum );  // pageNum is "autoboxed" to give an Integer!
   else {
         // references is the set of page references
         // that we have found previously for the term.
         // Add the new page number to that set.  This
         // set is already associated to term in the index.
      references.add( pageNum ); // pageNum is "autoboxed" to give an Integer!

The only other thing we need to do with the index is print it out. We want to iterate through the index and print out each term, together with the set of page references for that term. We could use an Iterator to iterate through the index, but it's much easier to do it with a for-each loop. The loop will iterate through the entry set of the map (see Subsection 10.3.2). Each "entry" is a key/value pair from the map; the key is a term and the value is the associated set of page references. Inside the for-each loop, we will have to print out a set of Integers, which can also be done with a for-each loop. So, here we have an example of nested for-each loops. (You might try to do the same thing entirely with iterators; doing so should give you some appreciation for the for-each loop!) Here is a subroutine that will print the index:

 * Print each entry in the index.
void printIndex() {
    for ( Map.Entry<String,TreeSet<Integer>>  entry :  index.entrySet() ) {
        String term = entry.getKey();
        TreeSet<Integer> pageSet = entry.getValue();
        System.out.print( term + " " );
        for ( int page : pageSet ) {
            System.out.print( page + " " );

The hardest thing here is the name of the type Map.Entry<String,TreeSet<Integer>>! Remember that the entries in a map of type Map<T,S> have type Map.Entry<T,S>, so the type parameters in Map.Entry<String,TreeSet<Integer>> are simply copied from the declaration of index. Another thing to note is that I used a loop control variable, page, of type int to iterate through the elements of pageSet, which is of type TreeSet<Integer>. You might have expected page to be of type Integer, not int, and in fact Integer would have worked just as well here. However, int does work, because of automatic type conversion: it's legal to assign a value of type Integer to a variable of type int. (To be honest, I was sort of surprised that this worked when I first tried it!)

This is not a lot of code, considering the complexity of the operations. I have not written a complete indexing program, but Exercise 10.5 presents a problem that is almost identical to the indexing problem.

By the way, in this example, I would prefer to print each list of page references with the integers separated by commas. In the printIndex() method given above,they are separated by spaces. There is an extra space after the last page reference in the list, but it does no harm since it's invisible in the printout. An extra comma at the end of the list would be annoying. The lists should be in a form such as "17,42,105" and not "17,42,105,". The problem is, how to leave that last comma out. Unfortunately, this is not so easy to do with a for-each loop. It might be fun to look at a few ways to solve this problem. One alternative is to use an iterator:

Iterator<Integer>  iter = pageSet.iterator();
int firstPage = iter.next();  // In this program, we know the set has at least
                              // one element.  Note also that this statement
                              // uses an auto-conversion from Integer to int.
while ( iter.hasNext() ) {
   int nextPage = iter.next();
   System.out.print("," + nextPage);

Another possibility is to use the fact that the TreeSet class defines a method first() that returns the first item in the set, that is, the one that is smallest in terms of the ordering that is used to compare items in the set. (It also defines the method last().) We can solve our problem using this method and a for-each loop:

int firstPage = pageSet.first();  // Find out the first page number in the set.
for ( int page : pageSet ) {
   if ( page != firstPage )
      System.out.print(","); // Output comma only if this is not the first page.

Finally, here is an elegant solution using a subset view of the tree. (See Subsection 10.3.2.) Actually, this solution might be a bit extreme:

int firstPage = pageSet.first();  // Get first item, which we know exists.
System.out.print(firstPage);      // Print first item, with no comma.
for ( int page : pageSet.tailSet( firstPage+1 ) ) // Process remaining items.
   System.out.print( "," + page );

10.4.3  Using a Comparator

There is a potential problem with our solution to the indexing problem. If the terms in the index can contain both upper case and lower case letters, then the terms will not be in alphabetical order! The ordering on String is not alphabetical. It is based on the Unicode codes of the characters in the string. The codes for all the upper case letters are less than the codes for the lower case letters. So, for example, terms beginning with "Z" come before terms beginning with "a". If the terms are restricted to use lower case letters only (or upper case only), then the ordering would be alphabetical. But suppose that we allow both upper and lower case, and that we insist on alphabetical order. In that case, our index can't use the usual ordering for Strings. Fortunately, it's possible to specify a different method to be used for comparing the keys of a map. This is a typical use for a Comparator.

Recall that an object that implements the interface Comparator<T> defines a method for comparing two objects of type T:

public int compare( T obj1, T obj2 )

This method should return an integer that is positive, zero, or negative, depending on whether obj1 is less than, equal to, or greater than obj2. We need an object of type Comparator<String> that will compare two Strings based on alphabetical order. The easiest way to do this is to convert the Strings to lower case and use the default comparison on the lower case Strings. The following class defines such a comparator:

 * Represents a Comparator that can be used for comparing two
 * strings based on alphabetical order.
class AlphabeticalOrder implements Comparator<String> {
   public int compare(String str1, String str2) {
      String s1 = str1.toLowerCase();  // Convert to lower case.
      String s2 = str2.toLowerCase();
      return s1.compareTo(s2);  // Compare lower-case Strings.

To solve our indexing problem, we just need to tell our index to use an object of type AlphabeticalOrder for comparing keys. This is done by providing a Comparator object as a parameter to the constructor. We just have to create the index in our example with the command:

index = new TreeMap<String,TreeSet<Integer>>( new AlphabeticalOrder() );

This does work. However, I've been concealing one technicality. Suppose, for example, that the indexing program calls addReference("aardvark",56) and that it later calls addReference("Aardvark",102). The words "aardvark" and "Aardvark" differ only in that one of them begins with an upper case letter; when converted to lower case, they are the same. When we insert them into the index, do they count as two different terms or as one term? The answer depends on the way that a TreeMap tests objects for equality. In fact, TreeMaps and TreeSets always use a Comparator object or a compareTo method to test for equality. They do not use the equals() method for this purpose. The Comparator that is used for the TreeMap in this example returns the value zero when it is used to compare "aardvark" and "Aardvark", so the TreeMap considers them to be the same. Page references to "aardvark" and "Aardvark" are combined into a single list. This is probably the correct behavior in this example. If not, some other technique must be used to sort the terms into alphabetical order.

10.4.4  Word Counting

The final example in this section also deals with storing information about words. The problem here is to make a list of all the words that occur in a file, along with the number of times that each word occurs. The file will be selected by the user. The output of the program will consist of two lists. Each list contains all the words from the file, along with the number of times that the word occurred. One list is sorted alphabetically, and the other is sorted according to the number of occurrences, with the most common words at the top and the least common at the bottom. The problem here is a generalization of Exercise 7.6, which asked you to make an alphabetical list of all the words in a file, without counting the number of occurrences.

My word counting program can be found in the file WordCount.java. As the program reads an input file, it must keep track of how many times it encounters each word. We could simply throw all the words, with duplicates, into a list and count them later. But that would require a lot of storage and would not be very efficient. A better method is to keep a counter for each word. The first time the word is encountered, the counter is initialized to 1. On subsequent encounters, the counter is incremented. To keep track of the data for one word, the program uses a simple class that holds a word and the counter for that word. The class is a static nested class:

 * Represents the data we need about a word:  the word and
 * the number of times it has been encountered.
private static class WordData { 
   String word;
   int count;
   WordData(String w) {
         // Constructor for creating a WordData object when
         // we encounter a new word.
      word = w;
      count = 1;  // The initial value of count is 1.
} // end class WordData

The program has to store all the WordData objects in some sort of data structure. We want to be able to add new words efficiently. Given a word, we need to check whether a WordData object already exists for that word, and if it does, we need to find that object so that we can increment its counter. A Map can be used to implement these operations. Given a word, we want to look up a WordData object in the Map. This means that the word is the key, and the WordData object is the value. (It might seem strange that the key is also one of the instance variables in the value object, but in fact this is probably the most common situation: The value object contains all the information about some entity, and the key is one of those pieces of information; the partial information in the key is used to retrieve the full information in the value object.) After reading the file, we want to output the words in alphabetical order, so we should use a TreeMap rather than a HashMap. This program converts all words to lower case so that the default ordering on Strings will put the words in alphabetical order. The data is stored in a variable named words of type TreeMap<String,WordData>. The variable is declared and the map object is created with the statement:

TreeMap<String,WordData> words = new TreeMap<String,WordData>();

When the program reads a word from a file, it calls words.get(word) to find out if that word is already in the map. If the return value is null, then this is the first time the word has been encountered, so a new WordData object is created and inserted into the map with the command words.put(word, new WordData(word)). If words.get(word) is not null, then its value is the WordData object for this word, and the program only has to increment the counter in that object. The program uses a method readNextWord(), which was given in Exercise 7.6, to read one word from the file. This method returns null when the end of the file is encountered. Here is the complete code segment that reads the file and collects the data:

String word = readNextWord();
while (word != null) {
   word = word.toLowerCase();  // convert word to lower case
   WordData data = words.get(word);
   if (data == null)
      words.put( word, new WordData(word) );
   word = readNextWord();

After reading the words and printing them out in alphabetical order, the program has to sort the words by frequency and print them again. To do the sorting using a generic algorithm, I defined a simple Comparator class for comparing two word objects according to their frequency counts. The class implements the interface Comparator<WordData>, since it will be used to compare two objects of type WordData:

 * A comparator class for comparing objects of type WordData according to 
 * their counts.  This is used for sorting the list of words by frequency.
private static class CountCompare implements Comparator<WordData> {
   public int compare(WordData data1, WordData data2) {
      return data2.count - data1.count;
          // The return value is positive if data2.count > data1.count.
          // I.E., data1 comes after data2 in the ordering if there
          // were more occurrences of data2.word than of data1.word.
          // The words are sorted according to decreasing counts.
} // end class CountCompare

Given this class, we can sort the WordData objects according to frequency by first copying them into a list and then using the generic method Collections.sort(list,comparator). The WordData objects that we need are the values in the map, words. Recall that words.values() returns a Collection that contains all the values from the map. The constructor for the ArrayList class lets you specify a collection to be copied into the list when it is created. So, we can use the following commands to create a list of type ArrayList<WordData> containing the word data and then sort that list according to frequency:

ArrayList<WordData> wordsByFrequency = new ArrayList<WordData>( words.values() );
Collections.sort( wordsByFrequency, new CountCompare() );

You should notice that these two lines replace a lot of code! It requires some practice to think in terms of generic data structures and algorithms, but the payoff is significant in terms of saved time and effort.

The only remaining problem is to print the data. We have to print the data from all the WordData objects twice, first in alphabetical order and then sorted according to frequency count. The data is in alphabetical order in the TreeMap, or more precisely, in the values of the TreeMap. We can use a for-each loop to print the data in the collection words.values(), and the words will appear in alphabetical order. Another for-each loop can be used to print the data in the list wordsByFrequency, and the words will be printed in order of decreasing frequency. Here is the code that does it:

TextIO.putln("List of words in alphabetical order" 
      + " (with counts in parentheses):\n");
for ( WordData data : words.values() )
   TextIO.putln("   " + data.word + " (" + data.count + ")");

TextIO.putln("\n\nList of words by frequency of occurrence:\n");
for ( WordData data : wordsByFrequency )
   TextIO.putln("   " + data.word + " (" + data.count + ")");

You can find the complete word-counting program in the file WordCount.java. Note that for reading and writing files, it uses the file I/O capabilities of TextIO.java, which were discussed in Subsection 2.4.5.

By the way, if you run the WordCount program on a reasonably large file and take a look at the output, it will illustrate something about the Collections.sort() method. The second list of words in the output is ordered by frequency, but if you look at a group of words that all have the same frequency, you will see that the words in that group are in alphabetical order. The method Collections.sort() was applied to sort the words by frequency, but before it was applied, the words were already in alphabetical order. When Collections.sort() rearranged the words, it did not change the ordering of words that have the same frequency, so they were still in alphabetical order within the group of words with that frequency. This is because the algorithm used by Collections.sort() is a stable sorting algorithm. A sorting algorithm is said to be stable if it satisfies the following condition: When the algorithm is used to sort a list according to some property of the items in the list, then the sort does not change the relative order of items that have the same value of that property. That is, if item B comes after item A in the list before the sort, and if both items have the same value for the property that is being used as the basis for sorting, then item B will still come after item A after the sorting has been done. Neither SelectionSort nor QuickSort are stable sorting algorithms. Insertion sort is stable, but is not very fast. The sorting algorithm used by Collections.sort() is both stable and fast.

I hope that the programming examples in this section have convinced you of the usefulness of the Java Collection Framework!

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