算法：5 亿整数的大文件如何排序

偏执的太偏执、 2023-07-10 08:58 24阅读 0赞

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2Jhb2JpbmdqaQ_size_16_color_FFFFFF_t_70]  
内部排序  
3路快排：

private final int cutoff = 8;
    
    public <T> void perform(Comparable<T>[] a) { 
            perform(a,0,a.length - 1);
        }
    
        private <T> int median3(Comparable<T>[] a,int x,int y,int z) { 
            if(lessThan(a[x],a[y])) { 
                if(lessThan(a[y],a[z])) { 
                    return y;
                }
                else if(lessThan(a[x],a[z])) { 
                    return z;
                }else { 
                    return x;
                }
            }else { 
                if(lessThan(a[z],a[y])){ 
                    return y;
                }else if(lessThan(a[z],a[x])) { 
                    return z;
                }else { 
                    return x;
                }
            }
        }
    
        private <T> void perform(Comparable<T>[] a,int low,int high) { 
            int n = high - low + 1;
            //当序列非常小，用插入排序
            if(n <= cutoff) { 
                InsertionSort insertionSort = SortFactory.createInsertionSort();
                insertionSort.perform(a,low,high);
                //当序列中小时，使用median3
            }else if(n <= 100) { 
                int m = median3(a,low,low + (n >>> 1),high);
                exchange(a,m,low);
                //当序列比较大时，使用ninther
            }else { 
                int gap = n >>> 3;
                int m = low + (n >>> 1);
                int m1 = median3(a,low,low + gap,low + (gap << 1));
                int m2 = median3(a,m - gap,m,m + gap);
                int m3 = median3(a,high - (gap << 1),high - gap,high);
                int ninther = median3(a,m1,m2,m3);
                exchange(a,ninther,low);
            }
    
            if(high <= low)
                return;
            //lessThan
            int lt = low;
            //greaterThan
            int gt = high;
            //中心点
            Comparable<T> pivot =  a[low];
            int i = low + 1;
    
            /* * 不变式： * a[low..lt-1] 小于pivot -> 前部(first) * a[lt..i-1] 等于 pivot -> 中部(middle) * a[gt+1..n-1] 大于 pivot -> 后部(final) * * a[i..gt] 待考察区域 */
    
            while (i <= gt) { 
                if(lessThan(a[i],pivot)) { 
                    //i-> ,lt ->
                    exchange(a,lt++,i++);
                }else if(lessThan(pivot,a[i])) { 
                    exchange(a,i,gt--);
                }else{ 
                    i++;
                }
            }
    
            // a[low..lt-1] < v = a[lt..gt] < a[gt+1..high].
            perform(a,low,lt - 1);
            perform(a,gt + 1,high);
        }

归并排序：

/** * 小于等于这个值的时候，交给插入排序 */
        private final int cutoff = 8;
    
        /** * 对给定的元素序列进行排序 * * @param a 给定元素序列 */
        @Override
        public <T> void perform(Comparable<T>[] a) { 
            Comparable<T>[] b = a.clone();
            perform(b, a, 0, a.length - 1);
        }
    
        private <T> void perform(Comparable<T>[] src,Comparable<T>[] dest,int low,int high) { 
            if(low >= high)
                return;
    
            //小于等于cutoff的时候,交给插入排序
            if(high - low <= cutoff) { 
                SortFactory.createInsertionSort().perform(dest,low,high);
                return;
            }
    
            int mid = low + ((high - low) >>> 1);
            perform(dest,src,low,mid);
            perform(dest,src,mid + 1,high);
    
            //考虑局部有序 src[mid] <= src[mid+1]
            if(lessThanOrEqual(src[mid],src[mid+1])) { 
                System.arraycopy(src,low,dest,low,high - low + 1);
            }
    
            //src[low .. mid] + src[mid+1 .. high] -> dest[low .. high]
            merge(src,dest,low,mid,high);
        }
    
        private <T> void merge(Comparable<T>[] src,Comparable<T>[] dest,int low,int mid,int high) { 
    
            for(int i = low,v = low,w = mid + 1; i <= high; i++) { 
                if(w > high || v <= mid && lessThanOrEqual(src[v],src[w])) { 
                    dest[i] = src[v++];
                }else { 
                    dest[i] = src[w++];
                }
            }
        }

数据太多，递归太深 ->栈溢出？加大Xss？  
数据太多，数组太长 -> OOM？加大Xmx?

耐心不足，没跑出来.而且要将这么大的文件读入内存，在堆中维护这么大个数据量，还有内排中不断的拷贝，对栈和堆都是很大的压力，不具备通用性。

sort命令来跑  
跑了多久呢？24分钟.

为什么这么慢？

粗略的看下我们的资源：  
内存 jvm-heap/stack，native-heap/stack,page-cache，block-buffer 外存 swap + 磁盘 数据量很大，函数调用很多，系统调用很多，内核/用户缓冲区拷贝很多，脏页回写很多，io-wait很高，io很繁忙，堆栈数据不断交换至swap，线程切换很多，每个环节的锁也很多.  
总之，内存吃紧，问磁盘要空间，脏数据持久化过多导致cache频繁失效，引发大量回写，回写线程高，导致cpu大量时间用于上下文切换，一切，都很糟糕，所以24分钟不细看了，无法忍受.

位图法

private BitSet bits;
    
        public void perform(
                String largeFileName,
                int total,
                String destLargeFileName,
                Castor<Integer> castor,
                int readerBufferSize,
                int writerBufferSize,
                boolean asc) throws IOException { 
    
            System.out.println("BitmapSort Started.");
            long start = System.currentTimeMillis();
            bits = new BitSet(total);
            InputPart<Integer> largeIn = PartFactory.createCharBufferedInputPart(largeFileName, readerBufferSize);
            OutputPart<Integer> largeOut = PartFactory.createCharBufferedOutputPart(destLargeFileName, writerBufferSize);
            largeOut.delete();
    
            Integer data;
            int off = 0;
            try { 
                while (true) { 
                    data = largeIn.read();
                    if (data == null)
                        break;
                    int v = data;
                    set(v);
                    off++;
                }
                largeIn.close();
                int size = bits.size();
                System.out.println(String.format("lines : %d ,bits : %d", off, size));
    
                if(asc) { 
                    for (int i = 0; i < size; i++) { 
                        if (get(i)) { 
                            largeOut.write(i);
                        }
                    }
                }else { 
                    for (int i = size - 1; i >= 0; i--) { 
                        if (get(i)) { 
                            largeOut.write(i);
                        }
                    }
                }
    
                largeOut.close();
                long stop = System.currentTimeMillis();
                long elapsed = stop - start;
                System.out.println(String.format("BitmapSort Completed.elapsed : %dms",elapsed));
            }finally { 
                largeIn.close();
                largeOut.close();
            }
        }
    
        private void set(int i) { 
            bits.set(i);
        }
    
        private boolean get(int v) { 
            return bits.get(v);
        }

nice!跑了190秒，3分来钟.  
以核心内存4663M/32大小的空间跑出这么个结果，而且大量时间在用于I/O，不错.

问题是，如果这个时候突然内存条坏了1、2根，或者只有极少的内存空间怎么搞？

外部排序

该外部排序上场了,外部排序干嘛的？  
该外部排序上场了,外部排序干嘛的？

内存极少的情况下，利用分治策略，利用外存保存中间结果，再用多路归并来排序;  
map-reduce的嫡系；  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2Jhb2JpbmdqaQ_size_16_color_FFFFFF_t_70 1]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2Jhb2JpbmdqaQ_size_16_color_FFFFFF_t_70 2]  
1.分

内存中维护一个极小的核心缓冲区memBuffer，将大文件bigdata按行读入，搜集到memBuffer满或者大文件读完时，对memBuffer中的数据调用内排进行排序，排序后将有序结果写入磁盘文件bigdata.xxx.part.sorted.  
循环利用memBuffer直到大文件处理完毕，得到n个有序的磁盘文件：  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2Jhb2JpbmdqaQ_size_16_color_FFFFFF_t_70 3]  
2.合

现在有了n个有序的小文件，怎么合并成1个有序的大文件？  
把所有小文件读入内存，然后内排？  
(⊙o⊙)…  
no!

利用如下原理进行归并排序：  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2Jhb2JpbmdqaQ_size_16_color_FFFFFF_t_70 4]  
我们举个简单的例子：  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2Jhb2JpbmdqaQ_size_16_color_FFFFFF_t_70 5]  
最终的时间，跑了771秒，13分钟左右.

less bigdata.sorted.text
    ...
    9999966
    9999967
    9999968
    9999969
    9999970
    9999971
    9999972
    9999973
    9999974
    9999975
    9999976
    9999977
    9999978
    ...

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