Question

我仍然在学习JS，我一直在为这个问题苦苦挣扎，并在Stack Overflow link上找到了一个类似的解决方案，我试图在此之后对我的解决方案建模，但是我似乎无法为我的用例..因此，我将很高兴为您提供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的建议

因此，我尝试根据“ objectID”从对象数组（请参见下文）中删除重复项，并将嵌套的“ hierarchicalCategories”在每个级别（lvl0，lvl1，lvl2）合并为一个数组（如果它们是唯一的）。 / p>

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：如果每个级别都有一个唯一值，则每个“ objectID”都有一个实例，然后合并“ 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，可以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），我都会创建一个数组，然后仅在先前未包含它的情况下才进行推送。但是，如果没有在"objectID": "5678"中定义的级别，而在任何重复项中都没有定义"lvl2"，那么在该插槽中过滤的数组中会有一个我不想要的空数组，但似乎无法解决它而不将其完全破坏。也欢迎其他建议+学习改进代码或采用其他方式进行学习。

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;
}, []);

我得到这个响应是因为您可以在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": []
  }
}
]

感谢任何愿意帮助您的人！

Answer 1

有几种方法可以做到这一点，但是我认为使用Map和Set可以帮助对事物进行分组并使它们独特：

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

首先创建一个地图，以objectID为键，并向空对象初始化相应的值。 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密钥。

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。第一次将是一个空对象。然后将来自输入对象的hierarchicalCategories添加到cats。在执行此操作时，将验证密钥（如“ lvl2”）在cats中是否已存在。如果不是，则cats[key]是未定义的，只有||运算符将评估正确的操作数，然后创建一个Set。否则，我们知道它已经是一个Set。然后将值（例如“女人的”）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循环的作用。本质上，它将输入转换为一种结构，该结构将以有效的方式处理分组和重复项。

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将给出其键/值组合。现在，有价部分不再是 empty 对象，因为我们在上一个循环中向它们添加了数据。这些cats对象可能具有多个“ lvl”键和关联的集合。

Array.from将允许我们迭代那些map.entries()并在mapper-callback函数中对每个变量进行处理。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解析器将花括号误解为代码块（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（与Object.entries相反）。

Answer 2

我喜欢它简单

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：

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

将此附加到代码末尾（在其中分配了filterArray）：

.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

这似乎是一个两步过程：

groupBy objectID：

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

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

reduce (/deepmerge)每个组：

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

this is an example使用lodash和deepmerge中方便的groupBy 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

JS对象数组-删除重复项并合并嵌套的对象

5 个答案:

说明