按多个数字排序Java String数组
我有一个像这样的txt文件中的数据列表
Date,Lat,Lon,Depth,Mag 20000101,34.6920,-116.3550,12.30,1.21 20000101,34.4420,-116.2280,7.32,1.01 20000101,37.4172,-121.7667,5.88,1.14 20000101,-41.1300,174.7600,27.00,1.90 20000101,37.6392,-119.0482,2.40,1.03 20000101,32.1790,-115.0730,6.00,2.44 20000101,59.7753,-152.2192,86.34,1.48 20000101,34.5230,-116.2410,11.63,1.61 20000101,59.5369,-153.1360,100.15,1.62 20000101,44.7357,-110.7932,4.96,2.20 20000101,34.6320,-116.2950,9.00,1.73 20000101,44.7370,-110.7938,5.32,1.75 20000101,35.7040,-117.6320,4.15,1.45 20000101,41.9270,20.5430,10.00,4.80 我的任务是按照每个标准对这些数据进行排序，例如按日期，纬度和经度排序
我试过像这样的冒泡
if ( Double.parseDouble(a[0].split(",")[1]) < Double.parseDouble(a[1].split(",")[1]))
这可行，但需要太多时间
txt文件中有40000数据
有没有其他方法来排序这些数据？
我可能会破坏一些学生的家庭作业，但是这里……
正如问题所示，Java中的自然方式是创建一个表示数据的类。 然后实现一个Comparator传递给实用程序方法Collections.sort 。
在运行带有Java 8的Parallels虚拟机的MacBook Pro 2.3 GHz Intel Core i7上，42,000个元素的数据集需要45-90毫秒才能进行排序。
我将您的示例数据更改为更有趣，引入了一些不同的日期和重复的纬度。
20000101,34.6920,-116.3550,12.30,1.21 20000101,34.4420,-116.2280,7.32,1.01 20000101,34.6920,-121.7667,5.88,1.14 20000101,-41.1300,174.7600,27.00,1.90 20000101,37.6392,-119.0482,2.40,1.03 20000101,32.1790,-115.0730,6.00,2.44 20000101,34.6920,-152.2192,86.34,1.48 20000102,34.6320,-116.2410,11.63,1.61 20000102,59.5369,-153.1360,100.15,1.62 20000102,44.7357,-110.7932,4.96,2.20 20000102,34.6320,-116.2950,9.00,1.73 20000102,34.6320,-110.7938,5.32,1.75 20000102,34.6320,-117.6320,4.15,1.45 20000102,41.9270,20.5430,10.00,4.80
我的GeoReading类来表示数据。
class GeoReading { LocalDate localDate = null; BigDecimal latitude = null; BigDecimal longitude = null; BigDecimal depth = null; BigDecimal magnitude = null; public GeoReading( String arg ) { // String is comma-separated values of: Date,Lat,Lon,Depth,Mag List items = Arrays.asList( arg.split( "\\s*,\\s*" ) ); // Regex explained here: http://stackoverflow.com/a/7488676/642706 this.localDate = ISODateTimeFormat.basicDate().parseLocalDate( items.get( 0 ) ); this.latitude = new BigDecimal( items.get( 1 ) ); this.longitude = new BigDecimal( items.get( 2 ) ); this.depth = new BigDecimal( items.get( 3 ) ); this.magnitude = new BigDecimal( items.get( 4 ) ); } @Override public String toString() { return "GeoReading{" + "localDate=" + localDate + ", latitude=" + latitude + ", longitude=" + longitude + ", depth=" + depth + ", magnitude=" + magnitude + '}'; } }
这是比较器实现。
class GeoReadingAscendingComparator implements Comparator { @Override public int compare( GeoReading o1 , GeoReading o2 ) { int localDateCompare = o1.localDate.compareTo( o2.localDate ); if ( localDateCompare != 0 ) { // If not equal on this component, so compare on this. return localDateCompare; } int latitudeCompare = o1.latitude.compareTo( o2.latitude ); if ( latitudeCompare != 0 ) { // If not equal on this component, so compare on this. return latitudeCompare; } return o1.longitude.compareTo( o2.longitude ); } }
主要代码。
Path path = Paths.get( "/Users/basil/lat-lon.txt" ); // Path for Mac OS X. try { List list = new ArrayList<>(); Stream lines = Files.lines( path ); lines.forEach( line -> list.add( new GeoReading( line ) ) ); // Take those 14 lines and multiply to simulate large text file. 14 * 3,000 = 42,000. int count = 3000; List bigList = new ArrayList<>( list.size() * count ); // Initialze capacite to expected number of elements. for ( int i = 0 ; i < count ; i++ ) { bigList.addAll( list ); } long start = System.nanoTime(); Collections.sort( bigList , new GeoReadingAscendingComparator() ); long elapsed = ( System.nanoTime() - start ); System.out.println( "Done sorting the GeoReading list. Sorting " + bigList.size() + " took: " + TimeUnit.MILLISECONDS.convert( elapsed , TimeUnit.NANOSECONDS ) + " ms ( " + elapsed + " nanos )." ); System.out.println( "Dump…" ); for ( GeoReading g : bigList ) { System.out.println( g ); } } catch ( IOException ex ) { System.out.println( "ERROR - ex: " + ex ); }
在现实世界中，我会添加一些防御性编程代码来validation传入的数据。 来自外部来源的数据总是有缺陷和/或变化。
尝试合并排序 。 合并排序的最差情况是O（n log n）。 冒泡排序的最坏情况时间是O（n ^ 2）。