Java:在每个第309个字符后插入换行符

Question

让我先说一下我是Java的新手.

我有一个包含一行的文件.文件大小约为200MB.我需要在每个第309个字符后插入一个换行符.我相信我有正确执行此操作的代码,但我一直遇到内存错误.我试过增加堆空间无济于事.

是否有一个内存密集度较低的处理方式？

BufferedReader r = new BufferedReader(new FileReader(fileName));

String line;

while ((line=r.readLine()) != null) {
  System.out.println(line.replaceAll("(.{309})", "$1\n"));
}

Answer 1

您的代码有两个问题:

1. 你将整个文件一次加载到内存中,假设它是一行,所以你需要至少200MB的堆空间; 和

2. 添加新行来使用像这样的正则表达式是一种非常低效的方法.直接的代码解决方案将快一个数量级.

这两个问题都很容易解决.

使用a FileReader和一次FileWriter加载309个字符,附加换行符并将其写出来.

更新:添加了逐个字符和缓冲读取的测试.缓冲读数实际上增加了很多复杂性,因为您需要满足可能(但通常非常罕见)的情况,即read()返回的字节数比您要求的少,并且仍有字节需要读取.

首先是简单版本:

private static void charRead(boolean verifyHash) {
  Reader in = null;
  Writer out = null;
  long start = System.nanoTime();
  long wrote = 0;
  MessageDigest md = null;
  try {
    if (verifyHash) {
      md = MessageDigest.getInstance("SHA1");
    }
    in = new BufferedReader(new FileReader(IN_FILE));
    out = new BufferedWriter(new FileWriter(CHAR_FILE));
    int count = 0;
    for (int c = in.read(); c != -1; c = in.read()) {
      if (verifyHash) {
        md.update((byte) c);
      }
      out.write(c);
      wrote++;
      if (++count >= COUNT) {
        if (verifyHash) {
          md.update((byte) '\n');
        }
        out.write("\n");
        wrote++;
        count = 0;
      }
    }
  } catch (IOException e) {
    throw new RuntimeException(e);
  } catch (NoSuchAlgorithmException e) {
    throw new RuntimeException(e);
  } finally {
    safeClose(in);
    safeClose(out);
    long end = System.nanoTime();
    System.out.printf("Created %s size %,d in %,.3f seconds. Hash: %s%n",
        CHAR_FILE, wrote, (end - start) / 1000000000.0d, hash(md, verifyHash));
  }
}

和"块"版本:

private static void blockRead(boolean verifyHash) {
  Reader in = null;
  Writer out = null;
  long start = System.nanoTime();
  long wrote = 0;
  MessageDigest md = null;
  try {
    if (verifyHash) {
      md = MessageDigest.getInstance("SHA1");
    }
    in = new BufferedReader(new FileReader(IN_FILE));
    out = new BufferedWriter(new FileWriter(BLOCK_FILE));
    char[] buf = new char[COUNT + 1]; // leave a space for the newline
    int lastRead = in.read(buf, 0, COUNT); // read in 309 chars at a time
    while (lastRead != -1) { // end of file
      // technically less than 309 characters may have been read
      // this is very unusual but possible so we need to keep
      // reading until we get all the characters we want
      int totalRead = lastRead;
      while (totalRead < COUNT) {
        lastRead = in.read(buf, totalRead, COUNT - totalRead);
        if (lastRead == -1) {
          break;
        } else {
          totalRead++;
        }
      }

      // if we get -1, it'll eventually signal an exit but first
      // we must write any characters we have read
      // note: it is assumed that the trailing number, which may be
      // less than 309 will still have a newline appended. this may
      // note be the case
      if (totalRead == COUNT) {
        buf[totalRead++] = '\n';
      }
      if (totalRead > 0) {
        out.write(buf, 0, totalRead);
        if (verifyHash) {
          md.update(new String(buf, 0, totalRead).getBytes("UTF-8"));
        }
        wrote += totalRead;
      }

      // don't try and read again if we've already hit EOF
      if (lastRead != -1) {
        lastRead = in.read(buf, 0, 309);
      }
    }
  } catch (IOException e) {
    throw new RuntimeException(e);
  } catch (NoSuchAlgorithmException e) {
    throw new RuntimeException(e);
  } finally {
    safeClose(in);
    safeClose(out);
    long end = System.nanoTime();
    System.out.printf("Created %s size %,d in %,.3f seconds. Hash: %s%n",
        CHAR_FILE, wrote, (end - start) / 1000000000.0d, hash(md, verifyHash));
  }
}

以及创建测试文件的方法:

private static void createFile() {
  Writer out = null;
  long start = System.nanoTime();
  try {
    out = new BufferedWriter(new FileWriter(IN_FILE));
    Random r = new Random();
    for (int i = 0; i < SIZE; i++) {
      out.write(CHARS[r.nextInt(CHARS.length)]);
    }
  } catch (IOException e) {
    throw new RuntimeException(e);
  } finally {
    safeClose(out);
    long end = System.nanoTime();
    System.out.printf("Created %s size %,d in %,.3f seconds%n",
      IN_FILE, SIZE, (end - start) / 1000000000.0d);
  }
}

这些都假设:

private static final int SIZE = 200000000;
private static final int COUNT = 309;
private static final char[] CHARS;
private static final char[] BYTES = new char[]{'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b', 'c', 'd', 'e', 'f'};
private static final String IN_FILE = "E:\\temp\\in.dat";
private static final String CHAR_FILE = "E:\\temp\\char.dat";
private static final String BLOCK_FILE = "E:\\temp\\block.dat";

static {
  char[] chars = new char[1000];
  int nchars = 0;
  for (char c = 'a'; c <= 'z'; c++) {
    chars[nchars++] = c;
    chars[nchars++] = Character.toUpperCase(c);
  }
  for (char c = '0'; c <= '9'; c++) {
    chars[nchars++] = c;
  }
  chars[nchars++] = ' ';
  CHARS = new char[nchars];
  System.arraycopy(chars, 0, CHARS, 0, nchars);
}

运行此测试:

public static void main(String[] args) {
  if (!new File(IN_FILE).exists()) {
    createFile();
  }
  charRead(true);
  charRead(true);
  charRead(false);
  charRead(false);
  blockRead(true);
  blockRead(true);
  blockRead(false);
  blockRead(false);
}

给出了这个结果(Intel Q9450,Windows 7 64位,8GB RAM,在7200rpm 1.5TB驱动器上运行测试):

Created E:\temp\char.dat size 200,647,249 in 29.690 seconds. Hash: 0x22ce9e17e17a67e5ea6f8fe929d2ce4780e8ffa4
Created E:\temp\char.dat size 200,647,249 in 18.177 seconds. Hash: 0x22ce9e17e17a67e5ea6f8fe929d2ce4780e8ffa4
Created E:\temp\char.dat size 200,647,249 in 7.911 seconds. Hash: (not calculated)
Created E:\temp\char.dat size 200,647,249 in 7.867 seconds. Hash: (not calculated)
Created E:\temp\char.dat size 200,647,249 in 8.018 seconds. Hash: 0x22ce9e17e17a67e5ea6f8fe929d2ce4780e8ffa4
Created E:\temp\char.dat size 200,647,249 in 7.949 seconds. Hash: 0x22ce9e17e17a67e5ea6f8fe929d2ce4780e8ffa4
Created E:\temp\char.dat size 200,647,249 in 3.958 seconds. Hash: (not calculated)
Created E:\temp\char.dat size 200,647,249 in 3.909 seconds. Hash: (not calculated)

结论: SHA1哈希验证非常昂贵,这就是我运行和不运行版本的原因.基本上在热身之后,"高效"版本的速度只有2倍.我猜这个时候文件实际上是在内存中.

如果我颠倒了块和char读取的顺序,结果是:

Created E:\temp\char.dat size 200,647,249 in 8.071 seconds. Hash: 0x22ce9e17e17a67e5ea6f8fe929d2ce4780e8ffa4
Created E:\temp\char.dat size 200,647,249 in 8.087 seconds. Hash: 0x22ce9e17e17a67e5ea6f8fe929d2ce4780e8ffa4
Created E:\temp\char.dat size 200,647,249 in 4.128 seconds. Hash: (not calculated)
Created E:\temp\char.dat size 200,647,249 in 3.918 seconds. Hash: (not calculated)
Created E:\temp\char.dat size 200,647,249 in 18.020 seconds. Hash: 0x22ce9e17e17a67e5ea6f8fe929d2ce4780e8ffa4
Created E:\temp\char.dat size 200,647,249 in 17.953 seconds. Hash: 0x22ce9e17e17a67e5ea6f8fe929d2ce4780e8ffa4
Created E:\temp\char.dat size 200,647,249 in 7.879 seconds. Hash: (not calculated)
Created E:\temp\char.dat size 200,647,249 in 8.016 seconds. Hash: (not calculated)

有趣的是,逐字符版本在第一次读取文件时需要更大的初始命中.

因此,按照惯例,它是效率和简单性之间的选择.