搜索对象数组列表的特定属性

时间:2018-10-05 07:57:56

标签: java

我有一个对象ArrayList<Tile> list的数组列表Tile的属性为new Tile("colour", value),我想创建一个搜索功能,在其中迭代Tile的每个Colour属性和每个{{1} } arraylist中每种颜色内的属性(就像每个循环嵌套一样),是否有一种简单的方法?

1 个答案:

答案 0 :(得分:3)

假设Tile类具有两个属性String colourint value。它具有toStringjava.lang.Object class的替代方法),如下所示:

@Override public String toString() {
    return colour + ":" + value;
}

制作一些瓷砖:

Tile t1 = new Tile("red", 7); // constructor takes a colour and a value
Tile t2 = new Tile("red", 2);
Tile t3 = new Tile("blue", 9);
Tile t4 = new Tile("white", 17);
Tile t5 = new Tile("blue", 3);
Tile t6 = new Tile("red", 15);
Tile t7 = new Tile("white", 10);


方案1:

该函数将Tile对象的列表和String颜色作为输入,并返回具有输入颜色(及其值)的所有图块。有两种方法可以实现,并且在两种方法中显示出来:

private static List<Tile> getTilesWithColor1(List<Tile> tilesList, String searchColor) {
    return tilesList.stream()
                     .filter(tile -> tile.getColour().equals(searchColor))
                     .collect(Collectors.toList());
}

private static List<Tile> getTilesWithColor2(List<Tile> tilesList, String searchColor) {
    List<Tile> result = new ArrayList<>();
    for (Tile t : tilesList) {
        if (t.getColour().equals(searchColor)) {
            result.add(t);
        }
    }
    return result;
}
  • 输入:tilesListcolour="red"
  • (两种方法的输出相同):[red:7, red:2, red:15]
  

我想做一个搜索功能,在其中我迭代每种颜色   Tile的属性和每个颜色中的每个Value属性   arraylist(就像每个循环的嵌套)一样,有没有一种简单的方法   这样吗?

可以更改此功能以添加其他条件或过滤器以获得所需的结果。


方案2:

获取所有颜色及其值:

private static Map<String, List<Integer>> getTileColorsAndValues(List<Tile> tilesList) {
    return tilesList.stream()
                     .collect(Collectors.groupingBy(Tile::getColour,
                         Collectors.mapping(Tile::getValue, Collectors.toList())));
}
  • 输入:tilesList
  • 输出:{red=[7, 2, 15], white=[17, 10], blue=[9, 3]}

请注意,可以从生成的地图中像这样在“红色”图块中获取值:

List<Integer> valuesList = map.get("red");


方案3:

按颜色获取所有图块:

private static Map<String, List<Tile>> getTilesByColorsAndValues(List<Tile> tilesList) {
    return tilesList.stream()
                    .collect(Collectors.groupingBy(Tile::getColour));
}
  • 输入:tilesList
  • 输出:{red=[red:7, red:2, red:15], white=[white:17, white:10], blue=[blue:9, blue:3]}

请注意,可以从生成的地图中像这样在“红色”图块中获取图块:

List<Tile> tilesList = map.get("red");



示例代码:

import java.util.*;
import java.util.stream.*;
import java.util.function.*;

public class TilesExample {

    public static void main(String [] args) {

        Tile t1 = new Tile("red", 7);
        Tile t2 = new Tile("red", 2);
        Tile t3 = new Tile("blue", 9);
        Tile t4 = new Tile("white", 17);
        Tile t5 = new Tile("blue", 3);
        Tile t6 = new Tile("red", 15);
        Tile t7 = new Tile("white", 10);
        List<Tile> tilesList = Arrays.asList(t1, t2, t3, t4, t5, t6, t7);

        System.out.println(getTilesWithColor1(tilesList, "red"));
        System.out.println(getTilesWithColor2(tilesList, "red"));

        System.out.println(getTileColorsAndValues(tilesList));

        System.out.println(getTilesByColorsAndValues(tilesList));
    }

    private static Map<String, List<Tile>> getTilesByColorsAndValues(List<Tile> tilesList) {
        return tilesList.stream()
                        .collect(Collectors.groupingBy(Tile::getColour));
    }

    private static Map<String, List<Integer>> getTileColorsAndValues(List<Tile> tilesList) {
        return tilesList.stream()
                        .collect(Collectors.groupingBy(Tile::getColour,
                            Collectors.mapping(Tile::getValue, Collectors.toList())));
    }

    private static List<Tile> getTilesWithColor1(List<Tile> tilesList, String searchColor) {
        return tilesList.stream()
                         .filter(tile -> tile.getColour().equals(searchColor))
                         .collect(Collectors.toList());
    }

    private static List<Tile> getTilesWithColor2(List<Tile> tilesList, String searchColor) {
        List<Tile> result = new ArrayList<>();
        for (Tile t : tilesList) {
            if (t.getColour().equals(searchColor)) {
                result.add(t);
            }
        }
        return result;
    }
}