33

我正在努力将 Perl 程序移植到 Java,并在学习过程中学习 Java。原始程序的一个核心组件是一个Perl 模块,它使用二进制搜索在 +500 GB 排序的文本文件中进行字符串前缀查找(本质上,“寻找”到文件中间的字节偏移量,回溯到最近的换行符,比较带有搜索字符串的行前缀,“寻找”到该字节偏移量的一半/两倍,重复直到找到......)

我已经尝试了几种数据库解决方案,但发现在这种大小的数据集的绝对查找速度上没有什么比这更好的了。您知道任何现有的实现此类功能的 Java 库吗?如果做不到这一点,您能否指出一些在文本文件中进行随机访问读取的惯用示例代码?

或者,我不熟悉新的(?)Java I/O 库,但它是否可以选择对 500 GB 文本文件进行内存映射(我在 64 位机器上,有可用内存)并执行二进制搜索内存映射的字节数组?我很想听听您分享有关此问题和类似问题的任何经验。

4

8 回答 8

29

对于这种情况,我是Java 的忠实粉丝MappedByteBuffers。它燃烧得很快。下面是我为您整理的一个片段,它将缓冲区映射到文件,寻找到中间,然后向后搜索到换行符。这应该足以让你去吧?

我在自己的应用程序中有类似的代码(查找、读取、重复直到完成), 在生产环境中对流进行基准测试,并将结果发布在我的博客上(java.ioGeekomatic帖子标记为“java.nio”),其中包含原始数据、图表和所有内容。 MappedByteBuffer

两秒总结? 基于我MappedByteBuffer的实现快了大约 275%。 YMMV。

为了处理大于 ~2GB 的文件,这是一个问题,因为 cast 和.position(int pos),我制作了由MappedByteBuffers 数组支持的分页算法。您需要在 64 位系统上工作才能处理大于 2-4GB 的文件,因为 MBB 使用操作系统的虚拟内存系统来发挥它们的魔力。

public class StusMagicLargeFileReader  {
    private static final long PAGE_SIZE = Integer.MAX_VALUE;
    private List<MappedByteBuffer> buffers = new ArrayList<MappedByteBuffer>();
    private final byte raw[] = new byte[1];

    public static void main(String[] args) throws IOException {
        File file = new File("/Users/stu/test.txt");
        FileChannel fc = (new FileInputStream(file)).getChannel(); 
        StusMagicLargeFileReader buffer = new StusMagicLargeFileReader(fc);
        long position = file.length() / 2;
        String candidate = buffer.getString(position--);
        while (position >=0 && !candidate.equals('\n')) 
            candidate = buffer.getString(position--);
        //have newline position or start of file...do other stuff    
    }
    StusMagicLargeFileReader(FileChannel channel) throws IOException {
        long start = 0, length = 0;
        for (long index = 0; start + length < channel.size(); index++) {
            if ((channel.size() / PAGE_SIZE) == index)
                length = (channel.size() - index *  PAGE_SIZE) ;
            else
                length = PAGE_SIZE;
            start = index * PAGE_SIZE;
            buffers.add(index, channel.map(READ_ONLY, start, length));
        }    
    }
    public String getString(long bytePosition) {
        int page  = (int) (bytePosition / PAGE_SIZE);
        int index = (int) (bytePosition % PAGE_SIZE);
        raw[0] = buffers.get(page).get(index);
        return new String(raw);
    }
}
于 2009-04-10T07:03:44.467 回答
4

我也有同样的问题。我正在尝试在已排序的文件中查找以某个前缀开头的所有行。

这是我编写的一种方法,主要是在这里找到的 Python 代码端口:http ://www.logarithmic.net/pfh/blog/01186620415

我已经对其进行了测试,但还没有彻底测试。但是,它不使用内存映射。

public static List<String> binarySearch(String filename, String string) {
    List<String> result = new ArrayList<String>();
    try {
        File file = new File(filename);
        RandomAccessFile raf = new RandomAccessFile(file, "r");

        long low = 0;
        long high = file.length();

        long p = -1;
        while (low < high) {
            long mid = (low + high) / 2;
            p = mid;
            while (p >= 0) {
                raf.seek(p);

                char c = (char) raf.readByte();
                //System.out.println(p + "\t" + c);
                if (c == '\n')
                    break;
                p--;
            }
            if (p < 0)
                raf.seek(0);
            String line = raf.readLine();
            //System.out.println("-- " + mid + " " + line);
            if (line.compareTo(string) < 0)
                low = mid + 1;
            else
                high = mid;
        }

        p = low;
        while (p >= 0) {
            raf.seek(p);
            if (((char) raf.readByte()) == '\n')
                break;
            p--;
        }

        if (p < 0)
            raf.seek(0);

        while (true) {
            String line = raf.readLine();
            if (line == null || !line.startsWith(string))
                break;
            result.add(line);
        }

        raf.close();
    } catch (IOException e) {
        System.out.println("IOException:");
        e.printStackTrace();
    }
    return result;
}
于 2009-06-15T21:17:45.380 回答
2

我不知道任何具有该功能的库。但是,Java 中外部二进制搜索的正确代码应该类似于:

class ExternalBinarySearch {
final RandomAccessFile file;
final Comparator<String> test; // tests the element given as search parameter with the line. Insert a PrefixComparator here
public ExternalBinarySearch(File f, Comparator<String> test) throws FileNotFoundException {
    this.file = new RandomAccessFile(f, "r");
    this.test = test;
}
public String search(String element) throws IOException {
    long l = file.length();
    return search(element, -1, l-1);
}
/**
 * Searches the given element in the range [low,high]. The low value of -1 is a special case to denote the beginning of a file.
 * In contrast to every other line, a line at the beginning of a file doesn't need a \n directly before the line
 */
private String search(String element, long low, long high) throws IOException {
    if(high - low < 1024) {
        // search directly
        long p = low;
        while(p < high) {
            String line = nextLine(p);
            int r = test.compare(line,element);
            if(r > 0) {
                return null;
            } else if (r < 0) {
                p += line.length();
            } else {
                return line;
            }
        }
        return null;
    } else {
        long m  = low + ((high - low) / 2);
        String line = nextLine(m);
        int r = test.compare(line, element);
        if(r > 0) {
            return search(element, low, m);
        } else if (r < 0) {
            return search(element, m, high);
        } else {
            return line;
        }
    }
}
private String nextLine(long low) throws IOException {
    if(low == -1) { // Beginning of file
        file.seek(0);           
    } else {
        file.seek(low);
    }
    int bufferLength = 65 * 1024;
    byte[] buffer = new byte[bufferLength];
    int r = file.read(buffer);
    int lineBeginIndex = -1;

    // search beginning of line
    if(low == -1) { //beginning of file
        lineBeginIndex = 0;
    } else {
        //normal mode
        for(int i = 0; i < 1024; i++) {
        if(buffer[i] == '\n') {
            lineBeginIndex = i + 1;
            break;
        }
        }
    }
    if(lineBeginIndex == -1) {
        // no line begins within next 1024 bytes
        return null;
    }
    int start = lineBeginIndex;
        for(int i = start; i < r; i++) {
            if(buffer[i] == '\n') {
                // Found end of line
                return new String(buffer, lineBeginIndex, i - lineBeginIndex + 1);
                return line.toString();
            }
        }
        throw new IllegalArgumentException("Line to long");
}
}

请注意:我专门编写了这段代码:角落案例的测试还不够好,代码假设没有单行大于 64K,等等。

我还认为建立行开始的偏移量索引可能是一个好主意。对于 500 GB 的文件,该索引应存储在索引文件中。您应该使用该索引获得一个不那么小的常数因子,因为不需要在每个步骤中搜索下一行。

我知道这不是问题,但是构建一个前缀树数据结构,如 (Patrica) Tries(在磁盘/SSD 上)可能是进行前缀搜索的好主意。

于 2009-04-10T11:56:41.843 回答
1

这是您想要实现的目标的简单示例。我可能会首先索引文件,跟踪每个字符串的文件位置。我假设字符串由换行符(或回车符)分隔:

    RandomAccessFile file = new RandomAccessFile("filename.txt", "r");
    List<Long> indexList = new ArrayList();
    long pos = 0;
    while (file.readLine() != null)
    {
        Long linePos = new Long(pos);
        indexList.add(linePos);
        pos = file.getFilePointer();
    }
    int indexSize = indexList.size();
    Long[] indexArray = new Long[indexSize];
    indexList.toArray(indexArray);

最后一步是在进行大量查找时转换为数组以稍微提高速度。我可能会将其转换Long[]long[]也,但我没有在上面显示。最后是从给定索引位置读取字符串的代码:

    int i; // Initialize this appropriately for your algorithm.
    file.seek(indexArray[i]);
    String line = file.readLine();
            // At this point, line contains the string #i.
于 2009-04-10T03:06:46.593 回答
1

如果您正在处理一个 500GB 的文件,那么您可能希望使用比二进制搜索更快的查找方法 - 即基数排序,它本质上是散列的一种变体。执行此操作的最佳方法实际上取决于您的数据分布和查找类型,但如果您正在寻找字符串前缀,那么应该有一个很好的方法来执行此操作。

我发布了一个整数的基数排序解决方案的示例,但是您可以使用相同的想法 - 基本上是通过将数据分成桶来减少排序时间,然后使用 O(1) 查找来检索相关的数据桶.

Option Strict On
Option Explicit On

Module Module1

Private Const MAX_SIZE As Integer = 100000
Private m_input(MAX_SIZE) As Integer
Private m_table(MAX_SIZE) As List(Of Integer)
Private m_randomGen As New Random()
Private m_operations As Integer = 0

Private Sub generateData()
    ' fill with random numbers between 0 and MAX_SIZE - 1
    For i = 0 To MAX_SIZE - 1
        m_input(i) = m_randomGen.Next(0, MAX_SIZE - 1)
    Next

End Sub

Private Sub sortData()
    For i As Integer = 0 To MAX_SIZE - 1
        Dim x = m_input(i)
        If m_table(x) Is Nothing Then
            m_table(x) = New List(Of Integer)
        End If
        m_table(x).Add(x)
        ' clearly this is simply going to be MAX_SIZE -1
        m_operations = m_operations + 1
    Next
End Sub

 Private Sub printData(ByVal start As Integer, ByVal finish As Integer)
    If start < 0 Or start > MAX_SIZE - 1 Then
        Throw New Exception("printData - start out of range")
    End If
    If finish < 0 Or finish > MAX_SIZE - 1 Then
        Throw New Exception("printData - finish out of range")
    End If
    For i As Integer = start To finish
        If m_table(i) IsNot Nothing Then
            For Each x In m_table(i)
                Console.WriteLine(x)
            Next
        End If
    Next
End Sub

' run the entire sort, but just print out the first 100 for verification purposes
Private Sub test()
    m_operations = 0
    generateData()
    Console.WriteLine("Time started = " & Now.ToString())
    sortData()
    Console.WriteLine("Time finished = " & Now.ToString & " Number of operations = " & m_operations.ToString())
    ' print out a random 100 segment from the sorted array
    Dim start As Integer = m_randomGen.Next(0, MAX_SIZE - 101)
    printData(start, start + 100)
End Sub

Sub Main()
    test()
    Console.ReadLine()
End Sub

End Module
于 2009-06-15T21:42:20.423 回答
1

我发布了一个要点https://gist.github.com/mikee805/c6c2e6a35032a3ab74f643a1d0f8249c

这是一个相当完整的例子,基于我在堆栈溢出和一些博客上发现的内容,希望其他人可以使用它

import static java.nio.file.Files.isWritable;
import static java.nio.file.StandardOpenOption.READ;
import static org.apache.commons.io.FileUtils.forceMkdir;
import static org.apache.commons.io.IOUtils.closeQuietly;
import static org.apache.commons.lang3.StringUtils.isBlank;
import static org.apache.commons.lang3.StringUtils.trimToNull;

import java.io.File;
import java.io.IOException;
import java.nio.Buffer;
import java.nio.MappedByteBuffer;
import java.nio.channels.FileChannel;
import java.nio.file.Path;

public class FileUtils {

    private FileUtils() {
    }

    private static boolean found(final String candidate, final String prefix) {
        return isBlank(candidate) || candidate.startsWith(prefix);
    }

    private static boolean before(final String candidate, final String prefix) {
        return prefix.compareTo(candidate.substring(0, prefix.length())) < 0;
    }

    public static MappedByteBuffer getMappedByteBuffer(final Path path) {
        FileChannel fileChannel = null;
        try {
            fileChannel = FileChannel.open(path, READ);
            return fileChannel.map(FileChannel.MapMode.READ_ONLY, 0, fileChannel.size()).load();
        } 
        catch (Exception e) {
            throw new RuntimeException(e);
        }
        finally {
            closeQuietly(fileChannel);
        }
    }

    public static String binarySearch(final String prefix, final MappedByteBuffer buffer) {
        if (buffer == null) {
            return null;
        }
        try {
            long low = 0;
            long high = buffer.limit();
            while (low < high) {
                int mid = (int) ((low + high) / 2);
                final String candidate = getLine(mid, buffer);
                if (found(candidate, prefix)) {
                    return trimToNull(candidate);
                } 
                else if (before(candidate, prefix)) {
                    high = mid;
                } 
                else {
                    low = mid + 1;
                }
            }
        } 
        catch (Exception e) {
            throw new RuntimeException(e);
        } 
        return null;
    }

    private static String getLine(int position, final MappedByteBuffer buffer) {
        // search backwards to the find the proceeding new line
        // then search forwards again until the next new line
        // return the string in between
        final StringBuilder stringBuilder = new StringBuilder();
        // walk it back
        char candidate = (char)buffer.get(position);
        while (position > 0 && candidate != '\n') {
            candidate = (char)buffer.get(--position);
        }
        // we either are at the beginning of the file or a new line
        if (position == 0) {
            // we are at the beginning at the first char
            candidate = (char)buffer.get(position);
            stringBuilder.append(candidate);
        }
        // there is/are char(s) after new line / first char
        if (isInBuffer(buffer, position)) {
            //first char after new line
            candidate = (char)buffer.get(++position);
            stringBuilder.append(candidate);
            //walk it forward
            while (isInBuffer(buffer, position) && candidate != ('\n')) {
                candidate = (char)buffer.get(++position);
                stringBuilder.append(candidate);
            }
        }
        return stringBuilder.toString();
    }

    private static boolean isInBuffer(final Buffer buffer, int position) {
        return position + 1 < buffer.limit();
    }

    public static File getOrCreateDirectory(final String dirName) { 
        final File directory = new File(dirName);
        try {
            forceMkdir(directory);
            isWritable(directory.toPath());
        } 
        catch (IOException e) {
            throw new RuntimeException(e);
        }
        return directory;
    }
}
于 2017-09-01T05:54:45.657 回答
0

我有类似的问题,所以我从这个线程中提供的解决方案创建了(Scala)库:

https://github.com/avast/BigMap

它包含用于在此排序文件中排序大文件和二进制搜索的实用程序...

于 2014-12-09T18:20:30.217 回答
-1

如果你真的想尝试内存映射文件,我找到了一个关于如何在 Java nio 中使用内存映射的教程。

于 2009-04-10T03:09:26.650 回答