Using zip operator in Java Stream

Rust中的zip算子很方便，能对两个集合进行平行处理，例如：

fn main() {
    let vec1 = vec![1, 2, 3, 4, 5, 6, 7, 8, 9];
    let vec2 = vec![11, 22, 33, 44, 55, 66, 77, 88, 99];
    let rest: Vec<i32> = vec1.iter().zip(vec2).map(|(e1, e2)| e1 + e2).collect();
    println!("{:?}", rest);
}

// 输出结果：[12, 24, 36, 48, 60, 72, 84, 96, 108]

那么Java中可以进行类似的操作吗？

答案显然是标准库中没有直接提供类似的方案，可以操作，但会比较蛋疼。

但是！但是！google的guava库中以工具类的方式提供了这个算子，让我们欢呼吧🥰

package com.demos;

import com.google.common.collect.Maps;
import com.google.common.collect.Streams;

import java.util.LinkedHashMap;
import java.util.List;
import java.util.Map;
import java.util.stream.Collectors;

public final class Main {
    public static void main(String[] args) {
        Map<String, List<String>> map1 = LinkedHashMap.newLinkedHashMap(3);
        Map<String, List<String>> map2 = LinkedHashMap.newLinkedHashMap(3);

        map1.put("One", List.of("AAA", "BBB", "CCC"));
        map1.put("Two", List.of("DDD", "EEE", "FFF"));
        map1.put("Three", List.of("GGG", "HHH", "III"));

        map2.put("first", List.of("J00", "K01", "L02"));
        map2.put("second", List.of("M03", "N04", "O05"));
        map2.put("third", List.of("P06", "Q07", "R08"));

        Map<String, List<String>> result = Streams.zip(
                map1.entrySet().stream(),
                map2.entrySet().stream(),
                (e1, e2) -> Maps.immutableEntry(
                        e1.getKey() + "-" + e2.getKey(),
                        Streams.zip(
                                e1.getValue().stream(),
                                e2.getValue().stream(),
                                (e01, e02) -> e01 + "-" + e02
                        ).toList()
                )
        ).collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));

        System.out.println(result);
    }
}

// 输出结果：{One-first=[AAA-J00, BBB-K01, CCC-L02], Two-second=[DDD-M03, EEE-N04, FFF-O05], Three-third=[GGG-P06, HHH-Q07, III-R08]}

完美 撒花🎉🎉🎉🎉🎉🎉🎉🎉