根据多个键的重复值从对象数组中删除元素

Question

我有一个这样的对象数组 -

var arr = [
    { type_id: "3", full_empty:"true", quantity:1},
    { type_id: "3", full_empty:"true", quantity:1},
    { type_id: "9", full_empty:"true", quantity:4},
    { type_id: "9", full_empty:"false", quantity:4},
    { type_id: "9", full_empty:"true", quantity:4},
    { type_id: "9", full_empty:"true", quantity:4},
    { type_id: "9", full_empty:"true", quantity:4}
];

我想删除具有相同type_id和full_empty值的重复项.结果应该是这样的 -

var arr = [
    { type_id: "3", full_empty:"true", quantity:1},
    { type_id: "9", full_empty:"true", quantity:4},
    { type_id: "9", full_empty:"false", quantity:4},
];

我已经搜索并找到了一些解决方案,但其中一些是用于删除重复键或基于仅一个键的重复值删除重复项.一些需要的外部库.还有一些我无法理解的解决方案.有没有简单的方法在纯JavaScript中执行此操作？

编辑以便更好地理解 - 我已阅读此问题.对该问题的接受答案是仅查找一个密钥的重复.在我的情况下,我必须找到多个键的重复.

Answer 1

您可以使用Array.some()和Array.reduce()将纯函数用于将输入数组减少为不同元素的数组,如下所示

    var arr = [
        { type_id: "3", full_empty:"true", quantity:1},
        { type_id: "3", full_empty:"true", quantity:1},
        { type_id: "9", full_empty:"true", quantity:4},
        { type_id: "9", full_empty:"false", quantity:4},
        { type_id: "9", full_empty:"true", quantity:4},
        { type_id: "9", full_empty:"true", quantity:4},
        { type_id: "9", full_empty:"true", quantity:4}
    ];

    var a = arr.reduce(function (accumulator, current) {
      if (checkIfAlreadyExist(current)) {
        return accumulator
      } else {
        return accumulator.concat([current]);
      }
      
      function checkIfAlreadyExist(currentVal) {
        return accumulator.some(function(item){
          return (item.type_id === currentVal.type_id &&
                  item.full_empty === currentVal.full_empty);
        });
      }
    }, []);
        
    console.log(a);

简洁的ES6语法

reduce可以使用ES6箭头函数和扩展运算符编写更简洁的内容,如下所示:

var arr = [
            { type_id: "3", full_empty:"true", quantity:1},
            { type_id: "3", full_empty:"true", quantity:1},
            { type_id: "9", full_empty:"true", quantity:4},
            { type_id: "9", full_empty:"false", quantity:4},
            { type_id: "9", full_empty:"true", quantity:4},
            { type_id: "9", full_empty:"true", quantity:4},
            { type_id: "9", full_empty:"true", quantity:4}
        ];

var a = arr.reduce((accumulator, current) => {
  if (checkIfAlreadyExist(current)) {
    return accumulator;
  } else {
    return [...accumulator, current];
  }

  function checkIfAlreadyExist(currentVal) {
    return accumulator.some((item) => {
      return (item.type_id === currentVal.type_id &&
              item.full_empty === currentVal.full_empty);
    });
  }
}, []);
            
console.log(a);

Answer 2

尽管有其他解决方案，我建议使用带有type_id和full_empty作为键的哈希表，如果找到新的，则将哈希设置为 true。与 一起Array#filter，您将获得一个包含独特项目的新数组。

var arr = [{ type_id: "3", full_empty: "true", quantity: 1 }, { type_id: "3", full_empty: "true", quantity: 1 }, { type_id: "9", full_empty: "true", quantity: 4 }, { type_id: "9", full_empty: "false", quantity: 4 }, { type_id: "9", full_empty: "true", quantity: 4 }, { type_id: "9", full_empty: "true", quantity: 4 }, { type_id: "9", full_empty: "true", quantity: 4 }],
    filtered = arr.filter(function (a) {
        var key = a.type_id + '|' + a.full_empty;
        if (!this[key]) {
            this[key] = true;
            return true;
        }
    }, Object.create(null));

console.log(filtered);

ES6

var arr = [{ type_id: "3", full_empty: "true", quantity: 1 }, { type_id: "3", full_empty: "true", quantity: 1 }, { type_id: "9", full_empty: "true", quantity: 4 }, { type_id: "9", full_empty: "false", quantity: 4 }, { type_id: "9", full_empty: "true", quantity: 4 }, { type_id: "9", full_empty: "true", quantity: 4 }, { type_id: "9", full_empty: "true", quantity: 4 }],
    filtered = arr.filter(
        (temp => a =>
            (k => !temp[k] && (temp[k] = true))(a.type_id + '|' + a.full_empty)
        )(Object.create(null))
    );

console.log(filtered);