[英]JS Array of Objects - Remove Duplicates and Merge nested objects
仍在学习 JS,我一直在为这个问题苦苦挣扎,并在 Stack Overflow 链接上找到了一个类似的解决方案,我试图在此之后为我的解决方案建模,但我似乎无法为我的用例弄清楚......感谢有关如何修复我拥有的代码的建议,甚至只是在正确方向上的一些帮助
所以我试图从基于“objectID”的对象数组(见下文)中删除重复项,并将每个级别(lvl0、lvl1、lvl2)的嵌套“hierarchicalCategories”合并到一个数组中,如果它们是唯一的。
let objArray = [
{
"objectID": "1234",
"hierarchicalCategories": {
"lvl0": "Women's"
}
},
{
"objectID": "1234",
"hierarchicalCategories": {
"lvl0": "Women's",
"lvl1": "Women's > Jewelry",
"lvl2": "Women's > Jewelry > New"
}
},
{
"objectID": "1234",
"hierarchicalCategories": {
"lvl0": "New",
"lvl1": "New > Jewelry"
}
},
{
"objectID": "5678",
"hierarchicalCategories": {
"lvl0": "Men's",
"lvl1": "Men's > Shoes",
}
},
{
"objectID": "5678",
"hierarchicalCategories": {
"lvl0": "New",
"lvl1": "New > Shoes"
}
}
]
预期结果应如下所示:每个“objectID”都有一个实例,然后合并“hierarchicalCategories”(如果每个级别都有唯一值)
let newArray = [
{
"objectID": "1234",
"hierarchicalCategories": {
"lvl0": ["Women's","New"],
"lvl1": ["Women's > Jewelry","New > Jewelry"],
"lvl2": ["Women's > Jewelry > New"]
}
},
{
"objectID": "5678",
"hierarchicalCategories": {
"lvl0": ["Men's", "New"],
"lvl1": ["Men's > Shoes","New > Shoes"]
}
}
]
这是我使用的代码,它在一定程度上起作用但不能完全起作用。 基本上在每个级别(lvl0,lvl1,lvl2),我正在创建一个数组,然后仅在以前未包含它时才推送。 但是,如果没有在"objectID": "5678"
定义的级别没有在任何重复项中定义"lvl2"
,则该插槽中的过滤数组中将有一个我不想要的空数组,但似乎无法在不完全破坏它的情况下修复它。 也欢迎其他建议 + 学习以改进代码或其他方法。
const filteredArr = objArray.reduce((acc, current) => {
const x = acc.find(item => item.objectID === current.objectID);
if (!x) {
current.hierarchicalCategories.lvl0 ? current.hierarchicalCategories.lvl0 = [current.hierarchicalCategories.lvl0] : current.hierarchicalCategories.lvl0 = []
current.hierarchicalCategories.lvl1 ? current.hierarchicalCategories.lvl1 = [current.hierarchicalCategories.lvl1] : current.hierarchicalCategories.lvl1 = []
current.hierarchicalCategories.lvl2 ? current.hierarchicalCategories.lvl2 = [current.hierarchicalCategories.lvl2] : current.hierarchicalCategories.lvl2 = []
acc.push(current)
} else {
if (current.hierarchicalCategories.lvl0 && !x.hierarchicalCategories.lvl0.includes(current.hierarchicalCategories.lvl0)) {
x.hierarchicalCategories.lvl0.push(current.hierarchicalCategories.lvl0)
}
if (current.hierarchicalCategories.lvl1 && !x.hierarchicalCategories.lvl1.includes(current.hierarchicalCategories.lvl1)) {
x.hierarchicalCategories.lvl1.push(current.hierarchicalCategories.lvl1)
}
if (current.hierarchicalCategories.lvl2 && !x.hierarchicalCategories.lvl2.includes(current.hierarchicalCategories.lvl2)) {
x.hierarchicalCategories.lvl2.push(current.hierarchicalCategories.lvl2)
}
}
return acc;
}, []);
我得到这个作为回应,因为你可以在 lvl2 看到空数组
[
{
"objectID": "1234",
"hierarchicalCategories": {
"lvl0": ["Women's","New"],
"lvl1": ["Women's > Jewelry","New > Jewelry"],
"lvl2": ["Women's > Jewelry > New"]
}
},
{
"objectID": "5678",
"hierarchicalCategories": {
"lvl0": ["Men's", "New"],
"lvl1": ["Men's > Shoes","New > Shoes"],
"lvl2": []
}
}
]
感谢任何愿意提供帮助的人!
有几种方法可以做到这一点,但我认为使用Map
和Set
可以帮助将事物分组并使它们独一无二:
let objArray = [{"objectID": "1234","hierarchicalCategories": {"lvl0": "Women's"}},{"objectID": "1234","hierarchicalCategories": {"lvl0": "Women's","lvl1": "Women's > Jewelry","lvl2": "Women's > Jewelry > New"}},{"objectID": "1234","hierarchicalCategories": {"lvl0": "New","lvl1": "New > Jewelry"}},{"objectID": "5678","hierarchicalCategories": {"lvl0": "Men's","lvl1": "Men's > Shoes",}},{"objectID": "5678","hierarchicalCategories": {"lvl0": "New","lvl1": "New > Shoes"}}]; let map = new Map(objArray.map(o => [o.objectID, {}] )); for (let obj of objArray) { let cats = map.get(obj.objectID); for (let [key, val] of Object.entries(obj.hierarchicalCategories)) { cats[key] = (cats[key] || new Set).add(val); } } let result = Array.from(map.entries(), ([objectId, cats]) => ({ objectId, hierarchicalCategories: Object.fromEntries(Object.entries(cats).map(([k, v]) => [k, [...v]] )) })); console.log(result);
首先创建一个 Map,以objectID
为键,相应的值初始化为空对象。 Map 构造函数获取一个 [key, value] 对列表,它将从中创建 Map。 它不会抱怨提供给它的重复密钥。
然后对于输入数组中的每个对象,从地图中检索相应的对象并将其分配给cats
。 第一次它将是一个空对象。 然后来自输入对象的hierarchicalCategories
被添加到cats
。 虽然这样做,则确认键(如“LVL2”)是否已经存在cats
。 如果不是,则cats[key]
未定义,并且只有||
运算符将评估正确的操作数,因此创建了一个集合。 否则我们知道它已经是一个集合。 我们将值(如“Woman's”)添加到该集合中。 使用 Set 的优点是重复项将被忽略。
这就是第一个for
循环所做的。 它本质上将输入转换为一个结构,该结构将以有效的方式处理分组和重复。
然后代码的最后一部分会将这些信息转换为所需的输出结构。
map.entries
将给出它拥有的键/值组合。 现在值部分不再是空对象,因为我们在前一个循环中向它们添加了数据。 这些是那些可能有几个“lvl”键和相关集合的cats
对象。
Array.from
将允许我们迭代这些map.entries()
并在 mapper-callback 函数中对每一个做一些事情。 该回调函数为每个条目返回一个对象。 它被括号包围,以避免 JS 解析器将大括号误解为代码块(它实际上会抱怨它)。
使用Object.entries
我们查找每个集合,并使用[...v]
将它们映射到标准数组。 Object.fromEntries
将其组合回一个对象(它与 Object.entries 相反)。
我喜欢简单。
const myArray = [{},{}];
const myArrayMirror = [...myArray];
const sortedArray = myArray.filter(originalObj, originalIndex => {
let objIsDuplicate = false;
myArrayMirror.forEach(duplicate, duplicateIndex => {
if (duplicateIndex > originalIndex && originalObj === duplicate) {
objIsDuplicate = true;
}
});
return objIsDuplicate;
});
我会这样做:
const objArray = [YOUR_OBJECTS_LIVE_HERE]
const mapWithUniqueObjs = new Map()
// loop over properties to set it to your map
for (let i = 0; i < objArray.length; i++) {
// Map's keys are always unique, so in case
// we already have this item in the map, it will be overwritten
uniqueObj.set(objArray[i].objectID, objArray[i])
}
// and make it an array
const arrWithUniqueObjs = [...mapWithUniqueObjs]
将此附加到您的代码末尾(您分配filteredArray):
.map(function(x) {
if (x.hierarchicalCategories.lvl2.length === 0) {
delete x.hierarchicalCategories.lvl2
}
if (x.hierarchicalCategories.lvl1.length === 0) {
delete x.hierarchicalCategories.lvl1
}
if (x.hierarchicalCategories.lvl0.length === 0) {
delete x.hierarchicalCategories.lvl0
}
return x
})
let objArray = [ { "objectID": "1234", "hierarchicalCategories": { "lvl0": "Women's" } }, { "objectID": "1234", "hierarchicalCategories": { "lvl0": "Women's", "lvl1": "Women's > Jewelry", "lvl2": "Women's > Jewelry > New" } }, { "objectID": "1234", "hierarchicalCategories": { "lvl0": "New", "lvl1": "New > Jewelry" } }, { "objectID": "5678", "hierarchicalCategories": { "lvl0": "Men's", "lvl1": "Men's > Shoes", } }, { "objectID": "5678", "hierarchicalCategories": { "lvl0": "New", "lvl1": "New > Shoes" } } ] const filteredArr = objArray.reduce((acc, current) => { const x = acc.find(item => item.objectID === current.objectID); if (!x) { current.hierarchicalCategories.lvl0 ? current.hierarchicalCategories.lvl0 = [current.hierarchicalCategories.lvl0] : current.hierarchicalCategories.lvl0 = [] current.hierarchicalCategories.lvl1 ? current.hierarchicalCategories.lvl1 = [current.hierarchicalCategories.lvl1] : current.hierarchicalCategories.lvl1 = [] current.hierarchicalCategories.lvl2 ? current.hierarchicalCategories.lvl2 = [current.hierarchicalCategories.lvl2] : current.hierarchicalCategories.lvl2 = [] acc.push(current) } else { if (current.hierarchicalCategories.lvl0 && !x.hierarchicalCategories.lvl0.includes(current.hierarchicalCategories.lvl0)) { x.hierarchicalCategories.lvl0.push(current.hierarchicalCategories.lvl0) } if (current.hierarchicalCategories.lvl1 && !x.hierarchicalCategories.lvl1.includes(current.hierarchicalCategories.lvl1)) { x.hierarchicalCategories.lvl1.push(current.hierarchicalCategories.lvl1) } if (current.hierarchicalCategories.lvl2 && !x.hierarchicalCategories.lvl2.includes(current.hierarchicalCategories.lvl2)) { x.hierarchicalCategories.lvl2.push(current.hierarchicalCategories.lvl2) } } return acc; }, []).map(function(x) { if (x.hierarchicalCategories.lvl2.length === 0) { delete x.hierarchicalCategories.lvl2 } if (x.hierarchicalCategories.lvl1.length === 0) { delete x.hierarchicalCategories.lvl1 } if (x.hierarchicalCategories.lvl0.length === 0) { delete x.hierarchicalCategories.lvl0 } return x }) console.log(filteredArr);
这似乎是一个两步过程:
objectID
: 例如变换[{objectID: 1, ...}, {objectID: 1, ...}, {objectID: 2, ...}]
进入[ [{objectID: 1, ...}, {objectID: 1, ...}], [{objectID: 2, ...}] ]
reduce (/deepmerge)
每组: 例如[{objectID: 1, ...merged props}, {objectID: 2, ...merged props}]
这是使用方便的lodash和deepmerge库的groupBy
的示例
deepmerge
是一个非常流行的库,用于合并带有/不带数组/嵌套数组的复杂嵌套对象
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