Java 中使用 Guava 库的布隆过滤器的交集和并集

Question

布隆过滤器似乎很有希望解决我正在处理的现实问题。一个流行的 java 实现似乎是 Google 的 guava 库。

我需要使用 bloom filter 的并集运算，如果有交集运算就好了，但我可以解决它。

在 java 文档中，我找不到执行交集的方法，而 putAll 方法似乎像并集操作一样工作。

所以我的问题是：

putAll() 是获取两个布隆过滤器并集的正确方法吗？
是否有非反射的方式来获取两个或多个布隆过滤器的交集？
万一反射没问题，我们可以在 'data' 字段上安全地执行双向运算（或，和）吗？

如果有人可以推荐任何其他流行且经过良好测试的库，该库在 maven 的存储库中可用，并且具有交集和联合，那也可以解决我的需求。

Answer 1

Is putAll() the correct way to get the union of two bloom filters?

是的，BloomFilter#putAll(BloomFilter) JavaDoc 说：

Combines this Bloom filter with another Bloom filter by performing a bitwise OR of the underlying data. The mutations happen to this instance. Callers must ensure the Bloom filters are appropriately sized to avoid saturating them.

问题是它会抛出 IllegalArgumentException - if isCompatible(that) == false，所以如果你想使用这个方法，你必须确保 isCompatible returns true:

Determines whether a given Bloom filter is compatible with this Bloom filter. For two Bloom filters to be compatible, they must:

not be the same instance

have the same number of hash functions

have the same bit size

have the same strategy

have equal funnels

也就是说，让我们尝试回答第二个问题并找到更简单的解决方案。

Is there a non-reflective way to get the intersection of two or more bloom filters?

是的，您可以使用 BloomFilters 实现 Predicate 并使用其 .and() 和 .or() 方法这一事实，记住 mightContain 行为取决于关于 BloomFilter 的参数。

示例：

// bloomFilter1: [0, 100], expected size 200, FPP 1%
BloomFilter<Integer> bloomFilter1 = IntStream.rangeClosed(0, 100)
        .boxed()
        .collect(toBloomFilter(Funnels.integerFunnel(), 200L, 0.01d));

// bloomFilter2: [80, 200], expected size 200, FPP 1%
BloomFilter<Integer> bloomFilter2 = IntStream.rangeClosed(80, 200)
        .boxed()
        .collect(toBloomFilter(Funnels.integerFunnel(), 200L, 0.01d));

// bloomFilter2: [80, 200], expected size 1000, FPP 0.1%
BloomFilter<Integer> bloomFilter3 = IntStream.rangeClosed(80, 200)
        .boxed()
        .collect(toBloomFilter(Funnels.integerFunnel(), 1000L, 0.001d));

assertThat(bloomFilter1.isCompatible(bloomFilter2)).isTrue(); // same parameters
assertThat(bloomFilter1.isCompatible(bloomFilter3)).isFalse(); // different parameters

BloomFilter<Integer> bloomFilterUnion = bloomFilter1.copy(); // preserves BF parameters
bloomFilterUnion.putAll(bloomFilter2);
// note that bloomFilterUnion.putAll(bloomFilter3) would throw IAE

// using AssertJ Predicate asserts https://joel-costigliola.github.io/assertj/core-8/api/org/assertj/core/api/PredicateAssert.html

// or == union == putAll
assertThat(bloomFilterUnion).accepts(90);
assertThat(bloomFilter1.or(bloomFilter2)).accepts(90);
assertThat(bloomFilter1.or(bloomFilter3)).accepts(90);
assertThat(bloomFilterUnion).accepts(42);
assertThat(bloomFilter1.or(bloomFilter2)).accepts(42);
assertThat(bloomFilter1.or(bloomFilter3)).accepts(42);
assertThat(bloomFilterUnion).accepts(180);
assertThat(bloomFilter1.or(bloomFilter2)).accepts(180);
assertThat(bloomFilter1.or(bloomFilter3)).accepts(180);
assertThat(bloomFilterUnion).rejects(300);
assertThat(bloomFilter1.or(bloomFilter2)).rejects(300);
assertThat(bloomFilter1.or(bloomFilter3)).rejects(300);

// and == intersection
assertThat(bloomFilter1.and(bloomFilter2)).accepts(90);
assertThat(bloomFilter1.and(bloomFilter3)).accepts(90);
assertThat(bloomFilter1.and(bloomFilter2)).rejects(42);
assertThat(bloomFilter1.and(bloomFilter3)).rejects(42);
assertThat(bloomFilter1.and(bloomFilter2)).rejects(180);
assertThat(bloomFilter1.and(bloomFilter3)).rejects(180);
assertThat(bloomFilter1.and(bloomFilter2)).rejects(300);
assertThat(bloomFilter1.and(bloomFilter3)).rejects(300);

In case one is okay with reflection, can we safely perform biwise operations (or, and) on the 'data' field?

我个人不会弄乱任何 class 的内部结构，但在我看来，这里没有必要这样做。

Java 中使用 Guava 库的布隆过滤器的交集和并集

Intersections and unions of Bloom filters in Java using the Guava library

java

data-structures

set

guava

bloom-filter