大文件,5亿整数,怎么排?
问题给你1个文件 6196302
3557681
6121580
2039345
2095006
1746773
7934312
2016371
7123302
8790171
2966901
...
7005375
现在要对这个文件进行排序,怎么搞? 内部排序先尝试内排,选2种排序方式: 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 + (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 + gap,low + (gap << 1));
int m2 = median3(a,m - gap,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,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,a.length - 1);
}
private <T> void perform(Comparable<T>[] src,Comparable<T>[] dest,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,high);
return;
}
int mid = low + ((high - low) >>> 1);
perform(dest,src,mid);
perform(dest,mid + 1,high);
//考虑局部有序 src[mid] <= src[mid+1]
if(lessThanOrEqual(src[mid],src[mid+1])) {
System.arraycopy(src,dest,high - low + 1);
}
//src[low .. mid] + src[mid+1 .. high] -> dest[low .. high]
merge(src,mid,high);
}
private <T> void merge(Comparable<T>[] src,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? 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sort -n bigdata -o bigdata.sorted
跑了多久呢?24分钟. 为什么这么慢?
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. 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.
1.分内存中维护一个极小的核心缓冲区 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? 利用如下原理进行归并排序: 我们举个简单的例子:
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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