根据属性过滤重复的元素

参考地址

1
2
3
4
5
6
7
/**
 * 根据属性过滤重复的元素
 */
public static <T> Predicate<T> distinctByKey(Function<? super T,Object> keyExtractor) {
    Map<Object,Boolean> seen = new ConcurrentHashMap<>();
    return t -> seen.putIfAbsent(keyExtractor.apply(t), Boolean.TRUE) == null;
}

测试

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
static String[] data = new String[]{
        "1,2,a,4,5,6",
        "2,2,b,4,5,6",
        "3,2,a,4,5,6",
        "4,2,d,4,5,6",
        "5,2,b,5,5,6",
        "6,2,f,4,5,6",
        "7,2,c,4,5,6",
        "8,2,g,4,5,6",
        "9,2,d,4,5,6",
        "10,2,g,5,5,6",
};

/**
 * 根据单个属性过滤
 */
@Test
public void test1() {
    int[] arrays = Stream.of(data).map(d -> d.split(","))
            .filter(array -> !array[0].equals("1"))
            .filter(distinctByKey(array -> array[2]))
            .mapToInt(array -> Integer.valueOf(array[0]))
            .toArray();
    Assert.assertArrayEquals(new int[]{2, 3, 4, 6, 7, 8}, arrays);
}

/**
 * 根据多个属性过滤
 */
@Test
public void test2() {
    int[] arrays = Stream.of(data).map(d -> d.split(","))
            .filter(distinctByKey(array -> array[2] + array[3]))
            .mapToInt(array -> Integer.valueOf(array[0]))
            .toArray();
    Assert.assertArrayEquals(new int[]{1, 2, 4, 5, 6, 7, 8, 10}, arrays);
}