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[英]filter an array of objects based on an object with an array as it's property value in react
[英]match 2 properties' values in array of objects and filter matches based on a 3rd property's value
我有一个图,其中的节点可以连接到多个其他节点。
每个节点代表数组中的一个对象。 每个节点的对象内都有一个数组,其中包含链接到该节点的所有节点的ID及其深度:
nodes: [
{"id":1, "depth":0, "next":[], "children":[2, 3]}, // nodes.next = [2, 3]
{"id":2, "depth":1, "next":[], "children":[1, 4, 5]}, // nodes.next = [4, 5]
{"id":3, "depth":1, "next":[], "children":[1, 6, 7]}, // nodes.next = [6, 7]
{"id":4, "depth":2, "next":[], "children":[2, 8]}, // nodes.next = [8]
{"id":5, "depth":2, "next":[], "children":[2, 9]} // nodes.next = [9]
]
我想从某个节点遍历图。
问题在于节点的子节点数组包含链接到该节点的所有节点。 深度为2的节点指向深度为1的节点。
因此,我想在节点的对象内创建一个新的数组,例如nodes.next
并删除指向深度小于其自身深度的节点的子级。
真正nodes.children
我的部分是检查nodes.children
节点的深度。 我甚至还没有接近的部分,我会看一下一个节点的深度nodes.children
高于nodes[i].depth
推nodes[i].children[i]
到nodes[i].next
。
如果有解决此问题的更好方法,我将很高兴知道。 我的尝试在许多方面都是徒劳的:
let childDepth;
for (let i = 0; i < nodes.length; i++) {
for (let child in nodes[i].children) {
if (nodes.id === child) {
childDepth = nodes[i].depth;
}
if (childDepth > graph.nodes[i].depth) {
nodes[i].next.push(child)
}
}
}
更新的数组:
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": [] }
];
请看下面的代码,看它是否在寻找
const array = [
{id:1, depth:0, next:[], children:[2, 3]},
{id:2, depth:1, next:[], children:[1, 4, 5]},
{id:3, depth:1, next:[], children:[1, 6, 7]},
{id:4, depth:2, next:[], children:[2, 8]},
{id:5, depth:2, next:[], children:[2, 9]}
]
array.forEach(x => {
let { children, depth } = x;
for(let i=depth; i< children.length; i++){
x.next.push(children[i]);
}
});
输出如下:
[
{"id":1,"depth":0,"next":[2,3],"children":[2,3]},
{"id":2,"depth":1,"next":[4,5],"children":[1,4,5]},
{"id":3,"depth":1,"next":[6,7],"children":[1,6,7]},
{"id":4,"depth":2,"next":[],"children":[2,8]},
{"id":5,"depth":2,"next":[],"children":[2,9]}
]
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