这是我迄今为止发现的最快的版本，大约比readLines快6倍。对于150MB的日志文件，这需要0.35秒，而在使用readLines()时需要2.40秒。只是为了好玩，linux的wc -l命令需要0.15秒。

public static int countLinesOld(String filename) throws IOException {
    InputStream is = new BufferedInputStream(new FileInputStream(filename));
    try {
        byte[] c = new byte[1024];
        int count = 0;
        int readChars = 0;
        boolean empty = true;
        while ((readChars = is.read(c)) != -1) {
            empty = false;
            for (int i = 0; i < readChars; ++i) {
                if (c[i] == '\n') {
                    ++count;
                }
            }
        }
        return (count == 0 && !empty) ? 1 : count;
    } finally {
        is.close();
    }
}

编辑，9年半后:我几乎没有java经验，但无论如何，我试图将这段代码与下面的LineNumberReader解决方案进行基准测试，因为没有人这样做让我感到困扰。似乎对于大文件，我的解决方案更快。虽然它似乎需要几次运行，直到优化器做一个像样的工作。我已经玩了一些代码，并产生了一个新版本，始终是最快的:

public static int countLinesNew(String filename) throws IOException {
    InputStream is = new BufferedInputStream(new FileInputStream(filename));
    try {
        byte[] c = new byte[1024];
        
        int readChars = is.read(c);
        if (readChars == -1) {
            // bail out if nothing to read
            return 0;
        }
        
        // make it easy for the optimizer to tune this loop
        int count = 0;
        while (readChars == 1024) {
            for (int i=0; i<1024;) {
                if (c[i++] == '\n') {
                    ++count;
                }
            }
            readChars = is.read(c);
        }
        
        // count remaining characters
        while (readChars != -1) {
            for (int i=0; i<readChars; ++i) {
                if (c[i] == '\n') {
                    ++count;
                }
            }
            readChars = is.read(c);
        }
        
        return count == 0 ? 1 : count;
    } finally {
        is.close();
    }
}

1.3GB文本文件的基准测试结果，y轴以秒为单位。我已经对同一个文件执行了100次运行，并使用System.nanoTime()对每次运行进行了测量。您可以看到countLinesOld有一些异常值，而countLinesNew没有异常值，虽然它只是稍微快一点，但差异在统计上是显著的。LineNumberReader显然更慢。

/**
 * Count file rows.
 *
 * @param file file
 * @return file row count
 * @throws IOException
 */
public static long getLineCount(File file) throws IOException {

    try (Stream<String> lines = Files.lines(file.toPath())) {
        return lines.count();
    }
}

在JDK8_u31上测试。但与此方法相比，性能确实较慢:

/**
 * Count file rows.
 *
 * @param file file
 * @return file row count
 * @throws IOException
 */
public static long getLineCount(File file) throws IOException {

    try (BufferedInputStream is = new BufferedInputStream(new FileInputStream(file), 1024)) {

        byte[] c = new byte[1024];
        boolean empty = true,
                lastEmpty = false;
        long count = 0;
        int read;
        while ((read = is.read(c)) != -1) {
            for (int i = 0; i < read; i++) {
                if (c[i] == '\n') {
                    count++;
                    lastEmpty = true;
                } else if (lastEmpty) {
                    lastEmpty = false;
                }
            }
            empty = false;
        }

        if (!empty) {
            if (count == 0) {
                count = 1;
            } else if (!lastEmpty) {
                count++;
            }
        }

        return count;
    }
}

经过测试，非常快。

我知道这是一个老问题，但公认的解决方案并不完全符合我所需要的。因此，我将其改进为接受各种行结束符(而不仅仅是换行)并使用指定的字符编码(而不是ISO-8859-n)。所有在一个方法(适当重构):

public static long getLinesCount(String fileName, String encodingName) throws IOException {
    long linesCount = 0;
    File file = new File(fileName);
    FileInputStream fileIn = new FileInputStream(file);
    try {
        Charset encoding = Charset.forName(encodingName);
        Reader fileReader = new InputStreamReader(fileIn, encoding);
        int bufferSize = 4096;
        Reader reader = new BufferedReader(fileReader, bufferSize);
        char[] buffer = new char[bufferSize];
        int prevChar = -1;
        int readCount = reader.read(buffer);
        while (readCount != -1) {
            for (int i = 0; i < readCount; i++) {
                int nextChar = buffer[i];
                switch (nextChar) {
                    case '\r': {
                        // The current line is terminated by a carriage return or by a carriage return immediately followed by a line feed.
                        linesCount++;
                        break;
                    }
                    case '\n': {
                        if (prevChar == '\r') {
                            // The current line is terminated by a carriage return immediately followed by a line feed.
                            // The line has already been counted.
                        } else {
                            // The current line is terminated by a line feed.
                            linesCount++;
                        }
                        break;
                    }
                }
                prevChar = nextChar;
            }
            readCount = reader.read(buffer);
        }
        if (prevCh != -1) {
            switch (prevCh) {
                case '\r':
                case '\n': {
                    // The last line is terminated by a line terminator.
                    // The last line has already been counted.
                    break;
                }
                default: {
                    // The last line is terminated by end-of-file.
                    linesCount++;
                }
            }
        }
    } finally {
        fileIn.close();
    }
    return linesCount;
}

这个解决方案在速度上与公认的解决方案相当，在我的测试中大约慢了4%(尽管Java中的计时测试是出了名的不可靠)。

一个直接的方式使用扫描器

static void lineCounter (String path) throws IOException {

        int lineCount = 0, commentsCount = 0;

        Scanner input = new Scanner(new File(path));
        while (input.hasNextLine()) {
            String data = input.nextLine();

            if (data.startsWith("//")) commentsCount++;

            lineCount++;
        }

        System.out.println("Line Count: " + lineCount + "\t Comments Count: " + commentsCount);
    }

如果你用这个

public int countLines(String filename) throws IOException {
    LineNumberReader reader  = new LineNumberReader(new FileReader(filename));
    int cnt = 0;
    String lineRead = "";
    while ((lineRead = reader.readLine()) != null) {}

    cnt = reader.getLineNumber(); 
    reader.close();
    return cnt;
}

你不能运行到大num行，比如100K行，因为从读取器返回。getLineNumber是int。你需要长类型的数据来处理最多的行。

我的结论是wc -l:s计算换行的方法是好的，但是在最后一行不以换行符结束的文件上返回非直观的结果。

和@。基于LineNumberReader的vikas解决方案，但在行数中添加一个，在最后一行以换行符结束的文件上返回非直观的结果。

因此我做了一个算法，处理如下:

@Test
public void empty() throws IOException {
    assertEquals(0, count(""));
}

@Test
public void singleNewline() throws IOException {
    assertEquals(1, count("\n"));
}

@Test
public void dataWithoutNewline() throws IOException {
    assertEquals(1, count("one"));
}

@Test
public void oneCompleteLine() throws IOException {
    assertEquals(1, count("one\n"));
}

@Test
public void twoCompleteLines() throws IOException {
    assertEquals(2, count("one\ntwo\n"));
}

@Test
public void twoLinesWithoutNewlineAtEnd() throws IOException {
    assertEquals(2, count("one\ntwo"));
}

@Test
public void aFewLines() throws IOException {
    assertEquals(5, count("one\ntwo\nthree\nfour\nfive\n"));
}

它是这样的:

static long countLines(InputStream is) throws IOException {
    try(LineNumberReader lnr = new LineNumberReader(new InputStreamReader(is))) {
        char[] buf = new char[8192];
        int n, previousN = -1;
        //Read will return at least one byte, no need to buffer more
        while((n = lnr.read(buf)) != -1) {
            previousN = n;
        }
        int ln = lnr.getLineNumber();
        if (previousN == -1) {
            //No data read at all, i.e file was empty
            return 0;
        } else {
            char lastChar = buf[previousN - 1];
            if (lastChar == '\n' || lastChar == '\r') {
                //Ending with newline, deduct one
                return ln;
            }
        }
        //normal case, return line number + 1
        return ln + 1;
    }
}

如果你想要直观的结果，你可以用这个。如果您只想要wc -l兼容性，只需使用@er即可。Vikas解决方案，但不添加一个到结果，并重试跳过:

try(LineNumberReader lnr = new LineNumberReader(new FileReader(new File("File1")))) {
    while(lnr.skip(Long.MAX_VALUE) > 0){};
    return lnr.getLineNumber();
}