我已经从double类型的文本文件中打印了数据,并将其转换为如下所示的double数组
[-2.0, -2.0, -2.0, 2.0, 2.0, 2.0, 2.0, 1.39E-04
-2.0, 0.0, -2.0, -2.0, 0.0, -2.0, 2.0, 0.020446
0.0, -2.0, 2.0, 2.0, -2.0, -2.0, -2.0, 0.032339
2.0, -2.0, -2.0, 2.0, 2.0, -2.0, 2.0, 0.026673
0.0, -2.0, -2.0, 0.0, 2.0, 0.0, 2.0, 0.094135
0.0, 0.0, -2.0, 0.0, 2.0, 2.0, 0.0, 0.045922
-2.0, 0.0, -2.0, 0.0, 2.0, 0.0, -2.0, 0.117043
-2.0, -2.0, 2.0, 2.0, 2.0, 2.0, -2.0, 0.425709
-2.0, -2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 0.156286]
最后一列是根据特定公式计算的每一行的分数。我的问题是如何根据最后一列的最大值对该数组列表排序? 排序后的数组的输出应如下所示
-2.0, -2.0, -2.0, 2.0, 2.0, 2.0, 2.0, 1.39E-04
-2.0, -2.0, 2.0, 2.0, 2.0, 2.0, -2.0, 0.425709
-2.0, -2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 0.156286
-2.0, 0.0, -2.0, 0.0, 2.0, 0.0, -2.0, 0.117043
0.0, -2.0, -2.0, 0.0, 2.0, 0.0, 2.0, 0.094135
0.0, 0.0, -2.0, 0.0, 2.0, 2.0, 0.0, 0.045922
0.0, -2.0, 2.0, 2.0, -2.0, -2.0, -2.0, 0.032339
2.0, -2.0, -2.0, 2.0, 2.0, -2.0, 2.0, 0.026673
-2.0, 0.0, -2.0, -2.0, 0.0, -2.0, 2.0, 0.020446
到目前为止我尝试过的代码
ArrayList<String> a2 = new ArrayList<String>();
File file7 = new File("kk.txt");
BufferedWriter output7 = new BufferedWriter(new FileWriter(file7));
output7.write(array+"");
output7.close();
Scanner s = new Scanner(new FileReader("kk.txt"));
while (s.hasNextLine()) {
String line1 = s.nextLine();
//store this line to string [] here
line1 = line1.replaceAll("\\[", "");
line1 = line1.replaceAll("\\]", "");
line1= line1.replaceAll("\\;,","\r\n"+"");
line1= line1.replaceAll("\\;","\r\n"+"");
a2.add(line1);
}
Collections.sort(a2,Collections.reverseOrder());
System.out.println("Sorted List : " + a2);
,但是代码返回数组而不进行排序。感谢您的帮助
答案 0 :(得分:1)
以下是您遇到的问题的可能解决方案之一:
List<String> a2 = ...; //your list
Collections.sort(a2,
Collections.reverseOrder((s1, s2) -> {
String[] d1 = s1.split(",");
String[] d2 = s2.split(",");
return Double.compare(Double.parseDouble(d1[d1.length-1]),
Double.parseDouble(d2[d2.length-1]));
}));
a2.forEach(System.out::println);
对于Java 1.7:
Collections.sort(a2,
Collections.reverseOrder(new Comparator<String>() {
@Override
public int compare(String s1, String s2) {
String[] d1 = s1.split(",");
String[] d2 = s2.split(",");
return Double.compare(Double.parseDouble(d1[d1.length-1]),
Double.parseDouble(d2[d2.length-1]));
}
}));
输出:
-2.0, -2.0, 2.0, 2.0, 2.0, 2.0, -2.0, 0.425709
-2.0, -2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 0.156286
-2.0, 0.0, -2.0, 0.0, 2.0, 0.0, -2.0, 0.117043
0.0, -2.0, -2.0, 0.0, 2.0, 0.0, 2.0, 0.094135
0.0, 0.0, -2.0, 0.0, 2.0, 2.0, 0.0, 0.045922
0.0, -2.0, 2.0, 2.0, -2.0, -2.0, -2.0, 0.032339
2.0, -2.0, -2.0, 2.0, 2.0, -2.0, 2.0, 0.026673
-2.0, 0.0, -2.0, -2.0, 0.0, -2.0, 2.0, 0.020446
-2.0, -2.0, -2.0, 2.0, 2.0, 2.0, 2.0, 1.39E-04
注意:1.39E-04
实际上是Scientific Notation,其值为0.000139
。因此,它似乎位于列表的末尾。
答案 1 :(得分:0)
Comparator<String> cmp = (o1, o2) -> {
final Double a1Double = Double.valueOf(o1.substring(o1.lastIndexOf(" ")));
final Double a2Double = Double.valueOf(o2.substring(o2.lastIndexOf(" ")));
return a1Double < a2Double ? 0 : -1;
};