arrays - NodeJS : Extracting duplicates from Array

Question

我需要你的帮助...如果问题已经被问到，我很抱歉，但我似乎找不到适合我的问题的答案:我正在尝试提取(而不是删除) )数组中重复项的列表。

最终，主要目标是仅保留一个具有较高利润的重复对象(在数组中)...

这是我的数组的一个简单示例:

var arr = [
  {
    mkBase: "test",
    mkComp: "test1",
    invest: { profit: 10 },
    availability: true,
    option: 1
  },
  {
    mkBase: "test",
    mkComp: "test1",
    invest: { profit: 15 },
    availability: false,
    option: 2
  },
  {
    mkBase: "test1",
    mkComp: "test",
    invest: { profit: 8 },
    availability: true,
    option: 3
  },
  {
    mkBase: "test2",
    mkComp: "test",
    invest: { profit: 6 },
    availability: true,
    option: 4
  },
  {
    mkBase: "test",
    mkComp: "test2",
    invest: { profit: 6 },
    availability: true,
    option: 5
  },
  {
    mkBase: "test",
    mkComp: "test3",
    invest: { profit: 7 },
    availability: true,
    option: 6
  },
  {
    mkBase: "test",
    mkComp: "test3",
    invest: { profit: 10 },
    availability: true,
    option: 7
  },
  {
    mkBase: "test3",
    mkComp: "test4",
    invest: { profit: 10 },
    availability: true,
    option: 8
  }
];

我成功地使用以下方法提取了几乎所有重复项的列表:

for (var i = 0; i < arr.length; i++) {
  if (_.uniqBy(arr, "mkBase").indexOf(arr[i]) == -1) {
    console.log("[SAME BASE]: " + JSON.stringify(arr[i], null, 2));
  } else if (_.uniqBy(arr, "mkComp").indexOf(arr[i]) == -1) {
    console.log("[SAME COMP]: " + JSON.stringify(arr[i], null, 2));
  }
}

结果如下:

[SAME BASE]: {
  "mkBase": "test",
  "mkComp": "test1",
  "invest": {
    "profit": 15
  },
  "availability": false,
  "option": 2
}
[SAME COMP]: {
  "mkBase": "test2",
  "mkComp": "test",
  "invest": {
    "profit": 6
  },
  "availability": true,
  "option": 4
}
[SAME BASE]: {
  "mkBase": "test",
  "mkComp": "test2",
  "invest": {
    "profit": 6
  },
  "availability": true,
  "option": 5
}
[SAME BASE]: {
  "mkBase": "test",
  "mkComp": "test3",
  "invest": {
    "profit": 7
  },
  "availability": true,
  "option": 6
}
[SAME BASE]: {
  "mkBase": "test",
  "mkComp": "test3",
  "invest": {
    "profit": 10
  },
  "availability": true,
  "option": 7
}

Lodash 方法 (_.uniqBy) 在主数组中保留一个重复项，为了最终获得最佳结果 (_.maxBy(arr, 'profit'))的重复项，我需要它与其他重复项。

我不确定我说得是否清楚，但如果您需要任何说明，请告诉我!

提前感谢大家!

************ 编辑 *************正如 stasovlas 所建议的，您将在下面找到预期结果以及删除数组中其他对象的原因:

var result = [
  {
    mkBase: "test",
    mkComp: "test1",
    invest: { profit: 15 },
    availability: false,
    option: 2
  },
  {
    mkBase: "test1",
    mkComp: "test",
    invest: { profit: 8 },
    availability: true,
    option: 3
  },
  {
    mkBase: "test3",
    mkComp: "test4",
    invest: { profit: 10 },
    availability: true,
    option: 8
  }
];

var removed = [
  //Reason: Same Base **and** Comp mk as option 2 && Profit is too low versus option 2
  {
    mkBase: "test",
    mkComp: "test1",
    invest: { profit: 10 },
    availability: true,
    option: 1
  },
  //Reason: Same Comp mk as option 3 && Profit is too low versus option 3
  {
    mkBase: "test2",
    mkComp: "test",
    invest: { profit: 6 },
    availability: true,
    option: 4
    //Reason: Same Base mk as option 2 && Profit is too low versus option 2
  },
  {
    mkBase: "test",
    mkComp: "test2",
    invest: { profit: 6 },
    availability: true,
    option: 5
  },
  //Reason: Same Base mk as option 2 && Profit is too low versus option 2 
  {
    mkBase: "test",
    mkComp: "test3",
    invest: { profit: 7 },
    availability: true,
    option: 6
  },
  //Reason: Same Base mk as option 2 && Profit is too low versus option 2 
  {
    mkBase: "test",
    mkComp: "test3",
    invest: { profit: 10 },
    availability: true,
    option: 7
  }
];

Answer 1

我不确定我的问题理解，但这是我的解决方案:

const res = _.reduce(arr, (result, item) => {
    const same = _.find(result, r => _.some([
        r.mkBase === item.mkBase,
        r.mkComp === item.mkComp
    ])); // find same already added item

    if (same === undefined) {
        return _.concat(result, item); // just push item   
    }

    if (same.invest.profit >= item.invest.profit) {
        return result; // do nothing if profit is less then already added
    }

    return _.chain(result) // remove item with smaller profit and push item with higher profit
        .reject({ mkBase: same.mkBase, mkComp: same.mkComp })
        .concat(item)
        .value();
}, []);