NodeJS：从数组中提取重复项

Question

我需要你的帮助...如果问题已经被问过，但我似乎无法找到适合我的问题的答案，我很抱歉：我正在尝试提取（而不是删除) 我的数组中的重复项列表。

最终，主要目标是只保留一个重复的对象（在数组中）具有更高的利润......

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
  }
];

我设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) 将其中一个副本保留在主数组中，并且，为了最终获得最好的副本 (_.maxBy(arr, 'profit'))，我会需要它与其他副本。

我不确定我是否很清楚，但如果您需要任何澄清，请告诉我！

提前感谢大家！

********** 编辑 ************* 正如 stasovlas 所建议的，您会在预期结果下方找到数组中其他对象被删除的原因：

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，但这里是我的解决方案：

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();
}, []);

Answer 2

var sameBase = {}, sameComp = {};

arr.forEach(item => {
    let existingBase = sameBase[item.mkBase];

    if ( ( existingBase === undefined ) ||  ( existingBase.invest.profit < item.invest.profit ) ) {
            sameBase[item.mkBase] = item;  
    } 

    existingComp = sameComp[item.mkComp];
    if ( ( existingComp === undefined ) ||  ( existingComp.invest.profit < item.invest.profit ) ) {
        sameComp[item.mkComp] = item;  
    }
});

var sameBaseArr = toArr(sameBase);
var sameCompArr = toArr(sameComp);

console.log("SAME BASE: " + JSON.stringify(sameBaseArr, true));
console.log("SAME COMP: " + JSON.stringify(sameCompArr, true));

function toArr(map) {
    let arr = [];
    for (var key in map) {
        arr.push(map[key]);
    }   
    return arr;     
}

Answer 3

试试这个（未经测试，但逻辑可以提供帮助）：

var mkBases = [], mkComps = [];

for (var i = 0; i < arr.length; i++) {
    var actualBase = arr[i].mkBase;
    var actualComp = arr[i].mkComp;

    if (typeof mkBases[actualBase] !== 'undefined') {
        mkBases[actualBase] = {getAllIndexes(arr, actualBase)}
    }

    if (typeof mkComps[actualComp] !== 'undefined') {
        mkComps[actualComp] = {getAllIndexes(arr, actualComp)}
    }
}

function getAllIndexes(arr, val) {
    var indexes = [], i = -1;
    while ((i = arr.indexOf(val, i+1)) != -1){
        indexes.push(i);
    }
    return indexes;
}

现在您可以遍历 mkBases 和 mkComp 来获取您的副本。