JS 对象数组 - 删除重复项并合并嵌套对象

Question

仍在学习 JS，我一直在努力解决这个问题，并在 Stack Overflow link 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，但我似乎无法弄清楚它是否适合我的使用案例..因此，我将不胜感激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提供一些帮助

所以我试图从基于“objectID”的对象数组（见下文）中删除重复项，如果它们是唯一的，则将每个级别（lvl0、lvl1、lvl2）的嵌套“hierarchicalCategories”合并到一个数组中。

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"
  }
}
]

预期结果应该是这样的：每个“objectID”都有一个实例，然后如果每个级别都有唯一值，则合并“hierarchicalCategories”

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，它在一定程度上有效，但不能完全有效。基本上在每个级别（lvl0，lvl1，lvl2），我正在创建一个数组，然后仅在之前未包含它的情况下推送。但是，如果没有像"objectID": "5678" 中定义的级别，并且在任何重复项中都没有定义"lvl2"，那么该插槽中的过滤数组中将有一个空数组，但似乎无法在不完全破坏它的情况下修复它。还可以接受其他建议+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;
}, []);

我得到了这个回应，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！

Answer 1

有几种方法可以做到这一点，但我认为使用 Map 和 Set 可以帮助对事物进行分组并使其独一无二：

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);

说明

首先创建一个 Map，以 objectID 为键，并将相应的值初始化为空对象。 Map 构造函数获取 [key, value] 对的列表，它将从中创建 Map。它不会抱怨提供给它的重复键。

然后对于输入数组中的每个对象，从映射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。然后将输入对象中的hierarchicalCategories 添加到cats。在执行此操作时，将验证密钥（如“lvl2”）是否已存在于cats 中。如果不是，则 cats[key] 未定义，只有这样 || 运算符才会计算正确的操作数，因此会创建一个 Set。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 Set。使用 Set 的好处是可以忽略重复项。

这就是第一个for 循环的作用。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。

那么代码的最后一部分会将这些信息转化为想要的输出结构。

map.entries 将给出它拥有的键/值组合。现在值部分不再是 empty 对象，因为我们在之前的循环中向它们添加了数据。那些cats 对象可能有多个“lvl”键和关联的集合。

Array.from 将允许我们迭代那些map.entries() 并在映射器回调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。它用括号括起来，以避免 JS 解析器将大括号误解为代码块（实际上它会抱怨它）。

使用Object.entries，我们查找每个集合，并将它们映射到标准数组，使用[...v]。 Object.fromEntries 将其组合回一个对象（它与 `Object.entries 相反）。

Answer 2

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;
});

Answer 3

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]

Answer 4

将其附加到代码的末尾（分配过滤数组的位置）：

.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);

Answer 5

这似乎是一个两步的过程：

1. groupBy objectID:

例如变换[{objectID: 1, ...}, {objectID: 1, ...}, {objectID: 2, ...}]

转入[ [{objectID: 1, ...}, {objectID: 1, ...}], [{objectID: 2, ...}] ]

1. reduce (/deepmerge)各组：

例如[{objectID: 1, ...merged props}, {objectID: 2, ...merged props}]

this is an example 使用方便的groupBy lodash 和 deepmerge 库

deepmerge 是一个非常流行的库，用于合并带有/不带有数组/嵌套数组的复杂嵌套对象