如何根据数组数组中的前一个元素过滤元素答案

【问题标题】：How to filter element depending on the previous element in an array of arrays如何根据数组数组中的前一个元素过滤元素
【发布时间】：2020-07-15 02:49:30
【问题描述】：

假设您有以下代码来过滤动物，以便仅保存比数组中前一个动物更重的那些（第一个对象将永远不会被保存，因为没有前一个动物）：

    const animals = [
        { id: 3, weight: 300, type: 'pig' },
        { id: 1, weight: 200, type: "cow" },
        { id: 7, weight: 400, type: "horse" },
        { id: 6, weight: 100, type: "pig" },
    ];


const filteredAnimals = animals.filter((animal, index) => 
    animals[index - 1] && animal.weight > animals[index - 1].weight);

console.log('filtered animals: ', filteredAnimals);

这按预期工作，但我不知道如何应用具有数组数组的相同过滤器：

const animals = [
    [
        { id: 5, weight: 300, type: 'pig' },
        { id: 1, weight: 200, type: "cow" },
        { id: 3, weight: 400, type: "horse" },
        { id: 4, weight: 350, type: "pig" },
    ],
    [
        { id: 2, weight: 250, type: "horse" },
        { id: 6, weight: 350, type: 'pig' },
        { id: 8, weight: 250, type: "cow" },
        { id: 7, weight: 400, type: "pig" },
    ]
]

在这种情况下，预期的结果应该是：

const filteredAnimals = [
    [
        { id: 3, weight: 400, type: "horse" },
    ],
    [
        { id: 6, weight: 350, type: "pig" },
        { id: 7, weight: 400, type: "pig" },
    ]
]

关于如何实现这一点的任何建议？

【问题讨论】：

标签： javascript arrays algorithm data-structures ecmascript-6

【解决方案1】：

只需将您的filter 包围在.map 中，即可遍历所有子数组：

const animals = [
    [
        { id: 5, weight: 300, type: 'pig' },
        { id: 1, weight: 200, type: "cow" },
        { id: 3, weight: 400, type: "horse" },
        { id: 4, weight: 350, type: "pig" },
    ],
    [
        { id: 2, weight: 250, type: "horse" },
        { id: 6, weight: 350, type: 'pig' },
        { id: 8, weight: 250, type: "cow" },
        { id: 7, weight: 400, type: "pig" },
    ]
]


const filteredAnimals = animals.map(
  subarr => subarr.filter((animal, index) => 
    subarr[index - 1] && animal.weight > subarr[index - 1].weight
  )
);
console.log(filteredAnimals);

【讨论】：