Arrange lines of a text file

Question

This program orders the lines of a text file alphabetically and prints them out to a new text file.

Usage:

java Arranger input.txt output.txt

Is this the best and especially the most performant way?

import java.io.File;
import java.io.FileReader;
import java.io.BufferedReader;
import java.io.FileWriter;
import java.io.BufferedWriter;
import java.io.IOException;
import java.util.ArrayList;

/*
 * this program arranges the lines of a text file in it's alphabetical order
 * usage: java Arranger input.txt output.txt
 */

public class Arranger { 

    public static void main(String[] args) {

        File inputFile = new File(args[0]);
        File outputFile = new File(args[1]);
        ArrayList<String> words = new ArrayList<>();

        try (BufferedReader reader = new BufferedReader(new FileReader(inputFile))) {
            // read lines of file and order them
            String line = new String();
            while ((line = reader.readLine()) != null) {
                words.add(line);
            }
            words.sort(String::compareToIgnoreCase);
        } catch (IOException e) {
            e.printStackTrace();
        }

        // write ordered list of words to file
        try (BufferedWriter writer = new BufferedWriter(new FileWriter(outputFile))) {
            for (String word : words) {
                writer.write(word + "\n");
            }
        } catch (IOException e) {
            e.printStackTrace();
        }

    }
}

Answer 1

The most significant issues in your code are:

• the handling of inputs.
• the use of helper functions from the standard libraries.

The algorithm you have chose in is fine, and both other answer recommend using streams, but I disagree (hence my answer).

So, your user-inputs are not validated, and your code can throw more than just IOExceptions, but also NoSuchElementException and so on if the file name arguments are not given on the commandline. You8 need to validate these and throw appropriate exceptions.

Further, you just print stack traces for IOException, but the program exists with code 0 - a success condition. If you're just going to print the stack trace it makes more sense to declare that the exception is thrown from the main method (and that will automagically print the trace and return with a non-zero code).

Other answers have pointed to the Files.lines(...) method, but I think you should consider the Files.readAllLines instead. Note that Files.Lines(...) and Files.readAllLines(...) methods will both trim the whitespace at the end-of-line. This may be a problem.

Regardless, Files is a good class to know about.

Consider this code:

public static void main(String[] args) throws IOException {
    if (args.length < 2) {
        throw new IllegalArgumentException("Expect 2 file-name command-line arguments");
    }

    Path source = Paths.get(args[0]);
    Path target = Paths.get(args[1]);

    List<String> lines = Files.readAllLines(source);
    lines.sort(String::compareToIgnoreCase);
    Files.write(target, lines);
}

I prefer the read-the-whole-file concept to the stream concept. It makes it clear that there are memory-requirements. Additionally, it makes the logic clear.

I would also consider a mechanism for handling raw lines without messing with the line termination padding and characters. Your code strips new-line/carriage-return characters and replaces them with just newline characters. I would prefer to see the line's end-of-line sequence unaltered in the transform. To do this requires a more careful consideration of what methods to use.... None of the Files methods, nor the default BufferedReader.readLine() nor Scanner methods do. You have to override these things. Consider the code:

private static final Pattern EOL = Pattern.compile("$", Pattern.MULTILINE);

private static final List<String> getLines(Path source) throws IOException {
    try (Scanner scanner = new Scanner(Files.newBufferedReader(source))) {
        scanner.useDelimiter(EOL);
        List<String> lines = new ArrayList<>();
        while (scanner.hasNext()) {
            lines.add(scanner.next());
        }
        return lines;
    }
}

private static final void writeLines(Path target, List<String> lines) throws IOException {
    try (BufferedWriter writer = Files.newBufferedWriter(target)) {
        for (String line : lines) {
            writer.write(line);
        }
    }
}

public static void main(String[] args) throws IOException {
    if (args.length < 2) {
        throw new IllegalArgumentException("Expect 2 file-name command-line arguments");
    }

    Path source = Paths.get(args[0]);
    Path target = Paths.get(args[1]);

    List<String> lines = getLines(source);
    lines.sort(String::compareToIgnoreCase);
    writeLines(target, lines);
}

Note how the IOException is still thrown out the main method, but also note that I am using the try-with-resource options for a try-block to ensure the files/streams are closed. The above code does not strip any line-terminators, and writes the output with the same termination as the input.

Answer 2

You can probably do something like this as well:

Files.lines(path)

It gets all the lines from the file as a stream, then you can sort the string based on your logic and then collect the same in a list and write to the output file.

This answer gives you a functional programming approach. Most part of the method is self-explanatory.

Files.write(
            Paths.get(outFile), 
            Files.lines(Paths.get(path))
                .sorted(String::compareToIgnoreCase)
                .collect(Collectors.toList())
    );

Answer 3

I can't think of a way with better time complexity, you have to know all values before you can sort. Though with some Java 8 stream capability added to the BufferedReader class, you could implement it quite cleanly as follows:

public class Arranger {

    public static void main(String[] args) throws IOException {
        File inputFile = new File(args[0]);
        File outputFile = new File(args[1]);

        try (BufferedReader reader = new BufferedReader(new FileReader(inputFile));
             PrintWriter writer = new PrintWriter(outputFile)) {
            reader.lines()
                    .sorted(String::compareToIgnoreCase)
                    .forEachOrdered(writer::println);
        }
}

This is a simple example. Purely performance wise I do not know the implications. Measure!

Answer 4

Use Java nio library for faster file read/writes:

java.nio.*;

nio is the latest library designed and developed by Oracle, which uses non-blocking I/O for reading and writing.

Let's take a look at two main methods of java.nio.Files class:

readAllBytes(Path path): 

This method reads all the bytes from the file at given path and returns the byte array containing the bytes read from the file.

readAllLines(Path path,Charset cs): 

This method read all lines from the file at given path and returns the List containing the lines from the file.

For your example, you can use as follows(I have used Java-8 streams as well for more readability)

List<String> lines = Files.readAllLines(Paths.get(args[0])).stream().sort(String::compareToIgnoreCase).collect(Collectors.toList());