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首页 > 解决方案 > 如何将数组中的值（即对象数组内部）与对象的属性进行比较？

javascript - 如何将数组中的值（即对象数组内部）与对象的属性进行比较？

问题描述

我有一个可以连接到多个其他节点的节点图。

每个节点都是数组中的一个对象。在每个节点的对象中都有一个数组，其中包含链接到该节点的所有节点的 id 及其深度：

const nodes = [
    { "id": 37, "depth": 0, "children": [210, 395, 265], "next": [] },
    { "id": 210, "depth": 1, "children": [37, 260, 259, 391],"next": [] },
    { "id": 256, "depth": 2, "children": [265], "next": [] },
    { "id": 259, "depth": 2, "children": [210, 397, 396], "next": [] },
    { "id": 260, "depth": 2, "children": [210], "next": [] },
    { "id": 265, "depth": 1, "children": [37, 256, 388, 394, 271, 269], "next": [] },
    { "id": 269, "depth": 2, "children": [265], "next": [] },
    { "id": 271, "depth": 2, "children": [265], "next": [] },
    { "id": 388, "depth": 2, "children": [265], "next": [] },
    { "id": 391, "depth": 2, "children": [210], "next": [] },
    { "id": 394, "depth": 2, "children": [265], "next": [] },
    { "id": 395, "depth": 1, "children": [37], "next": [] },
    { "id": 396, "depth": 3, "children": [259, 413], "next": [] },
    { "id": 397, "depth": 3, "children": [259], "next": [] },
    { "id": 413, "depth": 4, "children": [396], "next": [] }
];

我想遍历深度为 0 的节点是根的图。

问题是节点的子数组包含链接到它的所有节点。深度为 2 的节点指向深度为 1 的节点。

所以我想在节点的对象中创建一个新数组，比如说 nodes.next 并摆脱指向深度低于自身的节点的子节点。

一段时间后，我通过两种方式让它工作了。首先，我检查了子数组的长度是否大于 1。然后我依靠这样一个事实，即子数组中不应该被推送到下一个的节点恰好位于索引 0。那不是很可靠的。

我在第二种解决方案中发现困难的是检查子数组中节点的深度是否高于当前迭代中节点的深度。如果是，则将其推送到节点的下一个数组。我希望你能以更好的方式展示如何做到这一点，因为这个解决方案无论如何都不漂亮：

let currentDepth;
let childDepth;
let currentID;
let childID;

const getChildDepth = (childID) => {
    for (let i = 0; i < nodes.length; i++) {
        if (childID === nodes[i].id) {
            childDepth = nodes[i].depth
        }
    }
};

for (let i = 0; j < nodes.length; j++) {
    currentDepth = nodes[j].depth;
    currentID = nodes[j].id;
    if (nodes[j].children.length > 1) {
        for (let i = 0; i < nodes[j].children.length; i++) {
            childID = nodes[j].children[i];
            getChildDepth(childID);
            if (childDepth > currentDepth) {
                nodes[j].next.push(childID)
            }
        }
    }
}

样本输出：

const nodes = [
    { "id": 37, "depth": 0, "children": [210, 395, 265], "next": [210, 395, 265] },
    { "id": 210, "depth": 1, "children": [37, 260, 259, 391],"next": [260, 259, 391] },
    { "id": 256, "depth": 2, "children": [265], "next": [] },
    { "id": 259, "depth": 2, "children": [210, 397, 396], "next": [397, 396] },
    { "id": 260, "depth": 2, "children": [210], "next": [] },
    { "id": 265, "depth": 1, "children": [37, 256, 388, 394, 271, 269], "next": [256, 388, 394, 271, 269] },
    { "id": 269, "depth": 2, "children": [265], "next": [] },
    { "id": 271, "depth": 2, "children": [265], "next": [] },
    { "id": 388, "depth": 2, "children": [265], "next": [] },
    { "id": 391, "depth": 2, "children": [210], "next": [] },
    { "id": 394, "depth": 2, "children": [265], "next": [] },
    { "id": 395, "depth": 1, "children": [37], "next": [] },
    { "id": 396, "depth": 3, "children": [259, 413], "next": [413] },
    { "id": 397, "depth": 3, "children": [259], "next": [] },
    { "id": 413, "depth": 4, "children": [396], "next": [] }
];

标签： javascriptarrays

解决方案

您可以将Map节点作为参考并next通过使用深度检查过滤子节点来更新。

var nodes = [{ id: 37, depth: 0, children: [210, 395, 265], next: [] }, { id: 210, depth: 1, children: [37, 260, 259, 391], next: [] }, { id: 256, depth: 2, children: [265], next: [] }, { id: 259, depth: 2, children: [210, 397, 396], next: [] }, { id: 260, depth: 2, children: [210], next: [] }, { id: 265, depth: 1, children: [37, 256, 388, 394, 271, 269], next: [] }, { id: 269, depth: 2, children: [265], next: [] }, { id: 271, depth: 2, children: [265], next: [] }, { id: 388, depth: 2, children: [265], next: [] }, { id: 391, depth: 2, children: [210], next: [] }, { id: 394, depth: 2, children: [265], next: [] }, { id: 395, depth: 1, children: [37], next: [] }, { id: 396, depth: 3, children: [259, 413], next: [] }, { id: 397, depth: 3, children: [259], next: [] }, { id: 413, depth: 4, children: [396], next: [] }],
    references = new Map(nodes.map(n => [n.id, n]));

nodes.forEach(node => node.next = node.children.filter(
    id => references.get(id).depth > node.depth
));

console.log(nodes);
.as-console-wrapper { max-height: 100% !important; top: 0; }

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