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java - 使用比较器对字符串列表进行排序

问题描述

我有字符串。BOM结果：

1|00022954|41.418\n 2|00022951|1.0\n 3|00022945|41.575\n 3|00022944|41.684\n 3|00022944|41.778\n 3|00022944|41.871|624.|000223|000223 |00022944|41.918\n 3|00022944|41.825\n 3|00022944|41.731\n 3|00022945|41.621\n 3|00022953|41.512\n 4|00022957|0.0|\n 5.|0002\n2947 00022942|42.918\n 5|00022948|43.918\n 5|00022947|42.871\n 5|00022950|42.746\n 4|00022952|1.0\n 5|00022941|41.246|10.084|41.246|4.0884\n\20472 |0.0\n 3|00022945|39.621\n 3|00022944|39.731\n 3|00022944|39.84 \n 3|00022944|39.949\n 3|00022944|39.887\n 3|00022944|02.793\n 39.684\n 3|00022956|39.512\n 4|00022959|1.0\n 5|00022941|40.762\n 5|00022943|40.699\n 4|00022957|0.0\n 5|0002|2947|4802.809\n42.5| \n 5|00022948|43.918\n 5|00022947|42.871\n 5|00022950|42.746\n 3|00022949|40.996\n 3|00022944|39.996\n

在此处输入图像描述

当竖线（|）被分隔时，

第一个字符串是水平的。

最后一个字符串是排序的基础。

如果级别相等，则对字符串进行排序。

List<String> rowList = Arrays.asList(BOMResult.split("\n"));
Collections.sort(rowList, new Comparator<String>() {
    @Override
    public int compare(String s1, String s2) {
        String array1[] = s1.split("\\|");
        String array2[] = s2.split("\\|");
        int i1 = Integer.parseInt(array1[0]);
        int i2 = Integer.parseInt(array2[0]);

        if (i1 == i2) {
            return Double.valueOf(array1[array1.length-1]).compareTo(Double.valueOf(array2[array2.length-1]));
        } else {
            return 0;
        }
    }
});
BOMResult = String.join("\n", rowList);

当前结果：

1|00022954|41.418
2|00022951|1.0
3|00022953|41.512
3|00022945|41.575
3|00022945|41.621
3|00022944|41.684
3|00022944|41.731
3|00022944|41.778
3|00022944|41.825
3|00022944|41.871
3|00022944|41.918
3|00022946|42.918
4|00022957|0.0
5|00020472|41.184
5|00022941|41.246
5|00022950|42.746
5|00022947|42.809
5|00022947|42.871
5|00022942|42.918
5|00022948|43.918
4|00022952|1.0
2|00022958|0.0
3|00022956|39.512
3|00022945|39.621
3|00022945|39.684
3|00022944|39.731
3|00022944|39.793
3|00022944|39.84
3|00022944|39.887
3|00022944|39.949
4|00022957|0.0
4|00022959|1.0
5|00022943|40.699
5|00022941|40.762
5|00022950|42.746
5|00022947|42.809
5|00022947|42.871
5|00022942|42.918
5|00022948|43.918
3|00022944|39.996
3|00022949|40.996

我想在同一级别排序，即使行不同。

预期结果：

1|00022954|41.418
2|00022958|0.0
3|00022956|39.512
3|00022945|39.621
3|00022945|39.684
3|00022944|39.731
3|00022944|39.793
3|00022944|39.84
3|00022944|39.887
3|00022944|39.949
4|00022957|0.0
4|00022959|1.0
5|00022943|40.699
5|00022941|40.762
5|00022950|42.746
5|00022947|42.809
5|00022947|42.871
5|00022942|42.918
5|00022948|43.918
3|00022944|39.996
3|00022949|40.996
2|00022951|1.0
3|00022953|41.512
3|00022945|41.575
3|00022945|41.621
3|00022944|41.684
3|00022944|41.731
3|00022944|41.778
3|00022944|41.825
3|00022944|41.871
3|00022944|41.918
3|00022946|42.918
4|00022957|0.0
5|00020472|41.184
5|00022941|41.246
5|00022950|42.746
5|00022947|42.809
5|00022947|42.871
5|00022942|42.918
5|00022948|43.918
4|00022952|1.0

标签： javasortingcomparecomparator

解决方案

首先为记录创建一个数据对象
（用实际属性替换名称，即first-> level）：

 public static class Data {
    private final int first;
    private final String second;
    private final double third;

    public Data(int first, String second, double third) {
        this.first = first;
        this.second = second;
        this.third = third;
    }

    public int getFirst() {  return first; }
    public String getSecond() { return second; }
    public double getThird() {  return third; }

    @Override
    public String toString() {
        return first + "|" + second + "|" + third;
    }
}

之后，您可以通过链接 Comparator对此类记录的列表进行排序。示例（按第一个，然后是第二个，然后是第三个）：

List<Data> list = new LinkedList<>();
list.add(new Data(1, "00022954", 41.418));
list.add(new Data(2, "00022951", 1.0));
list.add(new Data(3, "00022953", 41.512));
list.add(new Data(3, "00022945", 41.575));
list.add(new Data(3, "00022945", 41.621));

// shuffle to test sorting
Collections.shuffle(list);

// sort the data
Collections.sort(list, 
    Comparator.comparing(Data::getFirst)
        .thenComparing(Data::getSecond)
        .thenComparing(Data::getThird));

// output the data
list.forEach(System.out::println);

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