用这个撞墙,以为我会把它贴在这里以防万一有些灵魂碰到了类似的灵魂。我有一些看起来像这样的数据:
const input = [
{
value: 'Miss1',
children: [
{ value: 'Miss2' },
{ value: 'Hit1', children: [ { value: 'Miss3' } ] }
]
},
{
value: 'Miss4',
children: [
{ value: 'Miss5' },
{ value: 'Miss6', children: [ { value: 'Hit2' } ] }
]
},
{
value: 'Miss7',
children: [
{ value: 'Miss8' },
{ value: 'Miss9', children: [ { value: 'Miss10' } ] }
]
},
{
value: 'Hit3',
children: [
{ value: 'Miss11' },
{ value: 'Miss12', children: [ { value: 'Miss13' } ] }
]
},
{
value: 'Miss14',
children: [
{ value: 'Hit4' },
{ value: 'Miss15', children: [ { value: 'Miss16' } ] }
]
},
];
我在运行时不知道层次结构的深度,即有多少级别的对象将拥有子阵列。我稍微简化了示例,实际上我需要将值属性与搜索项数组进行匹配。让我们暂时假设我匹配value.includes('Hit')
的位置。
我需要一个返回新数组的函数,例如:
输出对象中不应存在每个没有子项的非匹配对象,或子项层次结构中没有匹配项
具有包含匹配对象的后代的每个对象都应保留
匹配对象的所有后代都应保留
我正在考虑一个匹配对象'是一个value
属性,在这种情况下包含字符串Hit
,反之亦然。
输出应如下所示:
const expected = [
{
value: 'Miss1',
children: [
{ value: 'Hit1', children: [ { value: 'Miss3' } ] }
]
},
{
value: 'Miss4',
children: [
{ value: 'Miss6', children: [ { value: 'Hit2' } ] }
]
},
{
value: 'Hit3',
children: [
{ value: 'Miss11' },
{ value: 'Miss12', children: [ { value: 'Miss13' } ] }
]
},
{
value: 'Miss14',
children: [
{ value: 'Hit4' },
]
}
];
非常感谢任何花时间阅读这篇文章的人,如果我先到那儿就会发布我的解决方案。
答案 0 :(得分:30)
使用.filter()
进行递归调用,就像我在上面的评论中所描述的那样,基本上就是你所需要的。您只需要在返回之前使用递归调用的结果更新每个.children
属性。
返回值只是生成的.length
集合的.children
,因此,如果至少有一个,则保留该对象。
var res = input.filter(function f(o) {
if (o.value.includes("Hit")) return true
if (o.children) {
return (o.children = o.children.filter(f)).length
}
})
const input = [
{
value: 'Miss1',
children: [
{ value: 'Miss2' },
{ value: 'Hit1', children: [ { value: 'Miss3' } ] }
]
},
{
value: 'Miss4',
children: [
{ value: 'Miss5' },
{ value: 'Miss6', children: [ { value: 'Hit2' } ] }
]
},
{
value: 'Miss7',
children: [
{ value: 'Miss8' },
{ value: 'Miss9', children: [ { value: 'Miss10' } ] }
]
},
{
value: 'Hit3',
children: [
{ value: 'Miss11' },
{ value: 'Miss12', children: [ { value: 'Miss13' } ] }
]
},
{
value: 'Miss14',
children: [
{ value: 'Hit4' },
{ value: 'Miss15', children: [ { value: 'Miss16' } ] }
]
},
];
var res = input.filter(function f(o) {
if (o.value.includes("Hit")) return true
if (o.children) {
return (o.children = o.children.filter(f)).length
}
})
console.log(JSON.stringify(res, null, 2))

请注意,String上的.includes()
是ES7,因此可能需要针对旧版浏览器进行修补。您可以使用传统的.indexOf("Hit") != -1
。
要不改变原始内容,请创建一个复制对象的地图函数,并在过滤器之前使用该函数。
function copy(o) {
return Object.assign({}, o)
}
var res = input.map(copy).filter(function f(o) {
if (o.value.includes("Hit")) return true
if (o.children) {
return (o.children = o.children.map(copy).filter(f)).length
}
})
要真正压缩代码,你可以这样做:
var res = input.filter(function f(o) {
return o.value.includes("Hit") ||
o.children && (o.children = o.children.filter(f)).length
})
虽然它有点难以阅读。
答案 1 :(得分:4)
这是一个能够完成您正在寻找的功能的功能。基本上它会测试arr
中的每个项目的匹配,然后递归调用其children
上的过滤器。此外,还使用Object.assign
,以便不更改基础对象。
function filter(arr, term) {
var matches = [];
if (!Array.isArray(arr)) return matches;
arr.forEach(function(i) {
if (i.value.includes(term)) {
matches.push(i);
} else {
let childResults = filter(i.children, term);
if (childResults.length)
matches.push(Object.assign({}, i, { children: childResults }));
}
})
return matches;
}
答案 2 :(得分:2)
我认为这将是一个递归的解决方案。这是我试过的一个。
function find(obj, key) {
if (obj.value && obj.value.indexOf(key) > -1){
return true;
}
if (obj.children && obj.children.length > 0){
return obj.children.reduce(function(obj1, obj2){
return find(obj1, key) || find(obj2, key);
}, {});
}
return false;
}
var output = input.filter(function(obj){
return find(obj, 'Hit');
});
console.log('Result', output);
答案 3 :(得分:1)
const input = [
{
value: 'Miss1',
children: [
{ value: 'Miss1' },
{ value: 'Hit1', children: [ { value: 'Miss3' } ] }
]
},
{
value: 'Miss4',
children: [
{ value: 'Miss5' },
{ value: 'Miss6', children: [ { value: 'Hit2' } ] }
]
},
{
value: 'Miss7',
children: [
{ value: 'Miss8' },
{ value: 'Miss9', children: [ { value: 'Miss10' } ] }
]
},
{
value: 'Hit3',
children: [
{ value: 'Miss11' },
{ value: 'Miss12', children: [ { value: 'Miss13' } ] }
]
},
{
value: 'Miss14asds',
children: [
{ value: 'Hit4sdas' },
{ value: 'Miss15', children: [ { value: 'Miss16' } ] }
]
},
];
function filter(arr, term) {
var matches = [];
if (!Array.isArray(arr)) return matches;
arr.forEach(function(i) {
if (i.value === term) {
const filterData = (i.children && Array.isArray(i.children))? i.children.filter(values => values.value ===term):[];
console.log(filterData)
i.children =filterData;
matches.push(i);
} else {
// console.log(i.children)
let childResults = filter(i.children, term);
if (childResults.length)
matches.push(Object.assign({}, i, { children: childResults }));
}
})
return matches;
}
const filterData= filter(input,'Miss1');
console.log(filterData)
下面的代码使用递归函数过滤父数组和子数组数据
const input = [
{
value: 'Miss1',
children: [
{ value: 'Miss2' },
{ value: 'Hit1', children: [ { value: 'Miss3' } ] }
]
},
{
value: 'Miss4',
children: [
{ value: 'Miss5' },
{ value: 'Miss6', children: [ { value: 'Hit2' } ] }
]
},
{
value: 'Miss7',
children: [
{ value: 'Miss8' },
{ value: 'Miss9', children: [ { value: 'Miss10' } ] }
]
},
{
value: 'Hit3',
children: [
{ value: 'Miss11' },
{ value: 'Miss12', children: [ { value: 'Miss13' } ] }
]
},
{
value: 'Miss14',
children: [
{ value: 'Hit4' },
{ value: 'Miss15', children: [ { value: 'Miss16' } ] }
]
},
];
var res = input.filter(function f(o) {
if (o.value.includes("Hit")) return true
if (o.children) {
return (o.children = o.children.filter(f)).length
}
})
console.log(JSON.stringify(res, null, 2))
答案 4 :(得分:1)
Array.prototype.flatMap
非常适合进行递归过滤。与map
,filter
和reduce
类似,使用flatMap
不会不修改原始输入-
const findHits = (t = []) =>
t.flatMap(({ value, children }) => {
if (value.startsWith("Hit"))
return [{ value, children }]
else {
const r = findHits(children)
return r.length ? [{ value, children: r }] : []
}
})
const input =
[{value:'Miss1',children:[{value:'Miss2'},{value:'Hit1', children:[{value:'Miss3'}]}]},{value:'Miss4',children:[{value:'Miss5'},{value:'Miss6', children:[{value:'Hit2'}]}]},{value:'Miss7',children:[{value:'Miss8'},{value:'Miss9', children:[{value:'Miss10'}]}]},{value:'Hit3',children:[{value:'Miss11'},{value:'Miss12', children:[{value:'Miss13'}]}]},{value:'Miss14',children:[{value:'Hit4'},{value:'Miss15', children:[{value:'Miss16'}]}]}]
const result =
findHits(input)
console.log(JSON.stringify(result, null, 2))
[
{
"value": "Miss1",
"children": [
{
"value": "Hit1",
"children": [
{
"value": "Miss3"
}
]
}
]
},
{
"value": "Miss4",
"children": [
{
"value": "Miss6",
"children": [
{
"value": "Hit2"
}
]
}
]
},
{
"value": "Hit3",
"children": [
{
"value": "Miss11"
},
{
"value": "Miss12",
"children": [
{
"value": "Miss13"
}
]
}
]
},
{
"value": "Miss14",
"children": [
{
"value": "Hit4"
}
]
}
]
答案 5 :(得分:0)
或者您可以将deepdash扩展名的_.filterDeep
方法用于lodash
:
var keyword = 'Hit';
var foundHit = _.filterDeep(
input,
function(value) {
return value.value.includes(keyword);
},
{
tree: true,
onTrue: { skipChildren: true },
}
);
这里是您的情况的full test
答案 6 :(得分:0)
这是一个很好的解决方案,它利用了数组reduce
的功能,因此比其他解决方案可读性更高的代码。我认为它也更具可读性。我们递归调用filter
函数来过滤数组及其子元素
const input = [
{
value: "Miss1",
children: [
{ value: "Miss2" },
{ value: "Hit1", children: [{ value: "Miss3" }] },
],
},
{
value: "Miss4",
children: [
{ value: "Miss5" },
{ value: "Miss6", children: [{ value: "Hit2" }] },
],
},
{
value: "Miss7",
children: [
{ value: "Miss8" },
{ value: "Miss9", children: [{ value: "Miss10" }] },
],
},
{
value: "Hit3",
children: [
{ value: "Miss11" },
{ value: "Miss12", children: [{ value: "Miss13" }] },
],
},
{
value: "Miss14",
children: [
{ value: "Hit4" },
{ value: "Miss15", children: [{ value: "Miss16" }] },
],
},
];
function recursiveFilter(arr) {
return arr.reduce(function filter(prev, item) {
if (item.value.includes("Hit")) {
if (item.children && item.children.length > 0) {
return prev.concat({
...item,
children: item.children.reduce(filter, []),
});
} else {
return item;
}
}
return prev;
}, []);
}
console.log(recursiveFilter(input));
答案 7 :(得分:0)
这是一种使用 object-scan 的不同类型的解决方案。
这个解决方案是迭代的而不是递归的。它起作用是因为对象扫描以删除安全顺序迭代。基本上我们遍历树并在任何匹配上中断。然后我们跟踪不同深度的匹配(确保我们在遍历相邻分支时适当地重置该信息)。
这主要是一种学术练习,因为递归方法更简洁、更快捷。但是,如果需要进行额外的处理或者堆栈深度错误是一个问题,这个答案可能会很有趣。
// const objectScan = require('object-scan');
const myInput = [{ value: 'Miss1', children: [{ value: 'Miss2' }, { value: 'Hit1', children: [{ value: 'Miss3' }] }] }, { value: 'Miss4', children: [{ value: 'Miss5' }, { value: 'Miss6', children: [{ value: 'Hit2' }] }] }, { value: 'Miss7', children: [{ value: 'Miss8' }, { value: 'Miss9', children: [{ value: 'Miss10' }] }] }, { value: 'Hit3', children: [{ value: 'Miss11' }, { value: 'Miss12', children: [{ value: 'Miss13' }] }] }, { value: 'Miss14', children: [{ value: 'Hit4' }, { value: 'Miss15', children: [{ value: 'Miss16' }] }] }];
const myFilterFn = ({ value }) => value.includes('Hit');
const rewrite = (input, filterFn) => objectScan(['**(^children$)'], {
breakFn: ({ isMatch, value }) => isMatch && filterFn(value),
filterFn: ({
parent, property, key, value, context
}) => {
const len = (key.length - 1) / 2;
if (context.prevLen <= len) {
context.matches.length = context.prevLen + 1;
}
context.prevLen = len;
if (context.matches[len + 1] === true || filterFn(value)) {
context.matches[len] = true;
return false;
}
parent.splice(property, 1);
return true;
},
useArraySelector: false,
rtn: 'count'
})(input, { prevLen: 0, matches: [] });
console.log(rewrite(myInput, myFilterFn)); // returns number of deletions
// => 8
console.log(myInput);
// => [ { value: 'Miss1', children: [ { value: 'Hit1', children: [ { value: 'Miss3' } ] } ] }, { value: 'Miss4', children: [ { value: 'Miss6', children: [ { value: 'Hit2' } ] } ] }, { value: 'Hit3', children: [ { value: 'Miss11' }, { value: 'Miss12', children: [ { value: 'Miss13' } ] } ] }, { value: 'Miss14', children: [ { value: 'Hit4' } ] } ]
.as-console-wrapper {max-height: 100% !important; top: 0}
<script src="https://bundle.run/object-scan@16.0.0"></script>
免责声明:我是object-scan
的作者