Java文件中的行数

我使用巨大的数据文件，有时我只需要知道这些文件中的行数，通常我会打开它们并逐行读取它们，直到到达文件末尾

我想知道是否有更聪明的方法可以做到这一点

这是到目前为止我发现的最快的版本，比readLines快6倍。在150MB的日志文件上，这需要0.35秒，而使用readLines（）则需要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，在9 1/2年后：我几乎没有Java经验，但是无论如何我都尝试根据LineNumberReader下面的解决方案对该代码进行基准测试，因为它困扰着我没有人做。似乎特别是对于大文件，我的解决方案更快。尽管似乎要花一些时间才能使优化程序完成不错的工作。我已经玩了一些代码，并产生了一个始终最快的新版本：

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) {
            System.out.println(readChars);
            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文本文件的基准结果，y轴以秒为单位。我用相同的文件执行了100次运行，并用System.nanoTime()。您会看到它countLinesOld有一些异常值，并且countLinesNew没有异常值，虽然速度更快一点，但是差异在统计上是显着的。LineNumberReader显然慢一些。

我已经实现了该问题的另一种解决方案，我发现它在计数行时更有效：

try
(
   FileReader       input = new FileReader("input.txt");
   LineNumberReader count = new LineNumberReader(input);
)
{
   while (count.skip(Long.MAX_VALUE) > 0)
   {
      // Loop just in case the file is > Long.MAX_VALUE or skip() decides to not read the entire file
   }

   result = count.getLineNumber() + 1;                                    // +1 because line index starts at 0
}

对于不以换行符结尾的多行文件，可接受的答案有一个错误。一个没有换行符结束的一行文件将返回1，但是一个没有换行符结束的两行文件也将返回1。这是解决此问题的公认解决方案的实现。除了最后的读取，这些endsWithoutNewLine检查会浪费一切，但与整个函数相比，在时间上应该是微不足道的。

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

用 Java-8，您可以使用流：

try (Stream<String> lines = Files.lines(path, Charset.defaultCharset())) {
  long numOfLines = lines.count();
  ...
}

如果文件的末尾没有换行符，则上述方法count（）的答案给了我行错误计数-它无法计算文件中的最后一行。

这种方法对我来说更好：

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;
}

我知道这是一个老问题，但是公认的解决方案与我需要它做的并不完全匹配。因此，我对其进行了改进，使其可以接受各种行终止符（而不仅仅是换行符），并使用指定的字符编码（而不是ISO-8859- n）。一机多用（适当地重构）：

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;
}

该解决方案的速度与公认的解决方案相当，但在我的测试中却慢了约4％（尽管众所周知，Java中的时序测试并不可靠）。

我测试了上述计算行数的方法，这是我在系统上测试的不同方法的观察结果

文件大小：1.6 Gb方法：

1. 使用扫描仪：约35秒
2. 使用BufferedReader：大约5秒钟
3. 使用Java 8：大约5秒
4. 使用LineNumberReader：大约5秒

而且Java8方法似乎很方便：

Files.lines(Paths.get(filePath), Charset.defaultCharset()).count()
[Return type : long]

/**
 * 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。但是与这种方法相比，确实性能较慢：

/**
 * 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。

使用Scanner的直接方式

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);
    }

我得出结论wc -l：：s计数换行符的方法很好，但是在最后一行不以换行符结尾的文件上返回非直观结果。

基于LineNumberReader的@ er.vikas解决方案但在行数中添加了一个，则在最后一行以换行符结尾的文件上返回了非直观的结果。

因此，我做了一个算法，其处理如下：

@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"));
}

它看起来像这样：

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;
    }
}

如果您想获得直观的结果，则可以使用它。如果只需要wc -l兼容性，请简单使用@ er.vikas解决方案，但不要在结果中添加一个并重试跳过：

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

如何在Java代码中使用Process类？然后读取命令的输出。

Process p = Runtime.getRuntime().exec("wc -l " + yourfilename);
p.waitFor();

BufferedReader b = new BufferedReader(new InputStreamReader(p.getInputStream()));
String line = "";
int lineCount = 0;
while ((line = b.readLine()) != null) {
    System.out.println(line);
    lineCount = Integer.parseInt(line);
}

需要尝试一下。将发布结果。

如果您没有任何索引结构，那么您将无法读取整个文件。但是您可以通过避免逐行读取并使用正则表达式匹配所有行终止符来对其进行优化。

这个有趣的解决方案实际上真的很好！

public static int countLines(File input) throws IOException {
    try (InputStream is = new FileInputStream(input)) {
        int count = 1;
        for (int aChar = 0; aChar != -1;aChar = is.read())
            count += aChar == '\n' ? 1 : 0;
        return count;
    }
}

在基于Unix的系统上，在wc命令行上使用命令。

知道文件中有多少行的唯一方法是对它们进行计数。当然，您可以根据数据创建一个指标，平均长度为一行，然后获取文件大小并将其除以avg。长度，但这并不准确。

EOF处没有换行符（'\ n'）的多行文件的最佳优化代码。

/**
 * 
 * @param filename
 * @return
 * @throws IOException
 */
public static int countLines(String filename) throws IOException {
    int count = 0;
    boolean empty = true;
    FileInputStream fis = null;
    InputStream is = null;
    try {
        fis = new FileInputStream(filename);
        is = new BufferedInputStream(fis);
        byte[] c = new byte[1024];
        int readChars = 0;
        boolean isLine = false;
        while ((readChars = is.read(c)) != -1) {
            empty = false;
            for (int i = 0; i < readChars; ++i) {
                if ( c[i] == '\n' ) {
                    isLine = false;
                    ++count;
                }else if(!isLine && c[i] != '\n' && c[i] != '\r'){   //Case to handle line count where no New Line character present at EOF
                    isLine = true;
                }
            }
        }
        if(isLine){
            ++count;
        }
    }catch(IOException e){
        e.printStackTrace();
    }finally {
        if(is != null){
            is.close();    
        }
        if(fis != null){
            fis.close();    
        }
    }
    LOG.info("count: "+count);
    return (count == 0 && !empty) ? 1 : count;
}

带有正则表达式的扫描器：

public int getLineCount() {
    Scanner fileScanner = null;
    int lineCount = 0;
    Pattern lineEndPattern = Pattern.compile("(?m)$");  
    try {
        fileScanner = new Scanner(new File(filename)).useDelimiter(lineEndPattern);
        while (fileScanner.hasNext()) {
            fileScanner.next();
            ++lineCount;
        }   
    }catch(FileNotFoundException e) {
        e.printStackTrace();
        return lineCount;
    }
    fileScanner.close();
    return lineCount;
}

还没计时。

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;
}

您无法运行到大行，例如10万行，因为从reader.getLineNumber返回的是int。您需要长类型的数据来处理最多的行。