Guava 函数简介

2019-05-08 486 words One minute

Contents

1. 概述

这篇文章被组织成小而集中的示例和代码片段，用于使用 Guava 函数式元素 - 谓词和函数。

2. 示例

按条件过滤集合（自定义谓词）

List<Integer> numbers = Lists.newArrayList(1, 2, 3, 6, 10, 34, 57, 89);
Predicate<Integer> acceptEven = new Predicate<Integer>() {
    @Override
    public boolean apply(Integer number) {
        return (number % 2) == 0;
    }
};
List<Integer> evenNumbers = Lists.newArrayList(Collections2.filter(numbers, acceptEven));
Integer found = Collections.binarySearch(evenNumbers, 57);
assertThat(found, lessThan(0));

从集合中过滤掉空值

List<String> withNulls = Lists.newArrayList("a", "bc", null, "def");
Iterable<String> withoutNuls = Iterables.filter(withNulls, Predicates.notNull());
assertTrue(Iterables.all(withoutNuls, Predicates.notNull()));

检查集合中所有元素的条件

List<Integer> evenNumbers = Lists.newArrayList(2, 6, 8, 10, 34, 90);
Predicate<Integer> acceptEven = new Predicate<Integer>() {
    @Override
    public boolean apply(Integer number) {
        return (number % 2) == 0;
    }
};
assertTrue(Iterables.all(evenNumbers, acceptEven));

否定谓词

List<Integer> evenNumbers = Lists.newArrayList(2, 6, 8, 10, 34, 90);
Predicate<Integer> acceptOdd = new Predicate<Integer>() {
    @Override
    public boolean apply(Integer number) {
        return (number % 2) != 0;
    }
};
assertTrue(Iterables.all(evenNumbers, Predicates.not(acceptOdd)));

应用一个简单的功能

List<Integer> numbers = Lists.newArrayList(1, 2, 3);
List<String> asStrings = Lists.transform(numbers, Functions.toStringFunction());
assertThat(asStrings, contains("1", "2", "3"));

通过首先应用中间函数对集合进行排序

List<Integer> numbers = Arrays.asList(2, 1, 11, 100, 8, 14);
Ordering<Object> ordering = Ordering.natural().onResultOf(Functions.toStringFunction());
List<Integer> inAlphabeticalOrder = ordering.sortedCopy(numbers);
List<Integer> correctAlphabeticalOrder = Lists.newArrayList(1, 100, 11, 14, 2, 8);
assertThat(correctAlphabeticalOrder, equalTo(inAlphabeticalOrder));

复杂示例——链接谓词和函数

List<Integer> numbers = Arrays.asList(2, 1, 11, 100, 8, 14);
Predicate<Integer> acceptEvenNumber = new Predicate<Integer>() {
    @Override
    public boolean apply(Integer number) {
        return (number % 2) == 0;
    }
};
Function<Integer, Integer> powerOfTwo = new Function<Integer, Integer>() {
    @Override
    public Integer apply(Integer input) {
        return (int) Math.pow(input, 2);
    }
};
FluentIterable<Integer> powerOfTwoOnlyForEvenNumbers = 
FluentIterable.from(numbers).filter(acceptEvenNumber).transform(powerOfTwo);
assertThat(powerOfTwoOnlyForEvenNumbers, contains(4, 10000, 64, 196));

组合两个函数

List<Integer> numbers = Arrays.asList(2, 3);
Function<Integer, Integer> powerOfTwo = new Function<Integer, Integer>() {
    @Override
    public Integer apply(Integer input) {
        return (int) Math.pow(input, 2);
    }
};
List<Integer> result = Lists.transform(numbers, 
    Functions.compose(powerOfTwo, powerOfTwo));
assertThat(result, contains(16, 81));

创建由 Set 和 Function 支持的 Map

Function<Integer, Integer> powerOfTwo = new Function<Integer, Integer>() {
    @Override
    public Integer apply(Integer input) {
        return (int) Math.pow(input, 2);
    }
};
Set<Integer> lowNumbers = Sets.newHashSet(2, 3, 4);
Map<Integer, Integer> numberToPowerOfTwoMuttable = Maps.asMap(lowNumbers, powerOfTwo);
Map<Integer, Integer> numberToPowerOfTwoImuttable = Maps.toMap(lowNumbers, powerOfTwo);
assertThat(numberToPowerOfTwoMuttable.get(2), equalTo(4));
assertThat(numberToPowerOfTwoImuttable.get(2), equalTo(4));

从谓词中创建一个函数

List<Integer> numbers = Lists.newArrayList(1, 2, 3, 6);
Predicate<Integer> acceptEvenNumber = new Predicate<Integer>() {
    @Override
    public boolean apply(Integer number) {
        return (number % 2) == 0;
    }
};
Function<Integer, Boolean> isEventNumberFunction = Functions.forPredicate(acceptEvenNumber);
List<Boolean> areNumbersEven = Lists.transform(numbers, isEventNumberFunction);
assertThat(areNumbersEven, contains(false, true, false, true));