Question

我有一个对象数组：

data = [{"origin":"SJU","dest":"JFK","rank":48},{"origin":"JFK","dest":"SJU","rank":21},{"origin":"IAD","dest":"LAX","rank":31},{"origin":"LAS","dest":"SJU","rank":21}]

我正在尝试提取所有副本＆amp;独特的对象BY“origin”＆amp; “目标”。所以这两个是相同的，忽略了等级键

{origin:'JFK',dest:'SJU',rank:21}
{"origin":"SJU","dest":"JFK","rank":48}

基本上我想要2个单独的数组：

duplicates=[{"origin":"SJU","dest":"JFK","rank":48},{"origin":"JFK","dest":"SJU","rank":21}]

unique = [{"origin":"IAD","dest":"LAX","rank":31},{"origin":"LAS","dest":"SJU","rank":21}]

使用下划线，我能够将这样的东西放在一起。但它似乎效率低下，只返回一个重复数组：

duplicates = _.chain(data).map(function (d) {
    var ar = [d.origin, d.dest];
    return ar.sort();
}).sortBy(function (d) {
    return d
}).groupBy(function (d) {return d}).map(function (d) {
    if (d.length > 1) {
        return d[0]
    }
}).compact().value()
single = _.chain(data).map(function (d) {
    var ar = [d.origin, d.dest];
    return ar.sort();
}).sortBy(function (d) {
    return d
}).groupBy(function (d) {
    return d
}).map(function (d) {
    if (d.length == 1) {
        return d[0]
    }
}).compact().value()

我不禁觉得有一种更简单的方法可以解决这个问题。

Answer 1

引入临时变量来保存组可能更容易：

var data = [{"origin":"SJU","dest":"JFK","rank":48},{"origin":"JFK","dest":"SJU","rank":21},{"origin":"IAD","dest":"LAX","rank":31},{"origin":"LAS","dest":"SJU","rank":21}]

var groups = _.groupBy(data, function(item) {
  return [item.origin, item.dest].sort();
});

然后：

var duplicates = [],
singles = [];

_.each(groups, function(group) {
  if (group.length > 1) {
    duplicates.push.apply(duplicates, group);
  } else {
    singles.push(group[0]);
  }
});

Demo

Answer 2

我不理解你的代码。但是我遇到了问题。

我看到的方式是两步：

识别重复项并将其推入数组重复项
通过复制初始数组创建uniquearray并删除所有匹配重复数组的元素。

可能性能较低但可能更清晰。

Answer 3

这是使用普通javascript的算法。如果你有数千条记录，它可能不是最有效的，但它完成了工作。

var data = [
    {"origin": "SJU", "dest": "JFK", "rank":48},
    {"origin": "JFK", "dest": "SJU", "rank":21},
    {"origin": "IAD", "dest": "LAX", "rank":31},
    {"origin": "LAS", "dest": "SJU", "rank":21}
];

var uniques = [];
var doubles = [];

var i, j, l = data.length, origin, dest, foundDouble;

for ( i = 0; i < l; i += 1 ) {
    origin = data[i].origin;
    dest = data[i].dest;
    foundDouble = false;

    for ( j = 0; j < l; j += 1 ) {
        //skip the same row
        if ( i == j ) {
            continue;
        }

        if ( (data[j].origin == origin || data[j].origin == dest) && (data[j].dest == origin || data[j].dest == dest) ) {
            doubles.push( data[i] );
            foundDouble = true;
        }
    }

    if ( !foundDouble ) {
        uniques.push( data[i] );
    }
}

console.log( 'Uniques', uniques );
console.log( 'Doubles', doubles );

Answer 4

有很多方法可以做到这一点，但这里有一个。我从来没有使用下划线，但算法很简单，你应该能够轻松转换它。

var voyages = [{"origin":"SJU","dest":"JFK","rank":48},{"origin":"JFK","dest":"SJU","rank":21},{"origin":"IAD","dest":"LAX","rank":31},{"origin":"LAS","dest":"SJU","rank":21}],
    dupes = [],
    uniques = [],
    countMap = new Map(), //Map is from Harmony, but you can use a plain object
    voyageKeyOf = function (voyage) { return [voyage.origin, voyage.dest].sort().join(''); },
    uniques;

//Create a map that stores how many times every voyage keys were seen
voyages.forEach(function (voyage) {
    var key = voyageKeyOf(voyage),
        hasCount = countMap.get(key);

    if (!hasCount) countMap.set(key, 1);
    else {
        let count = countMap.get(key);
        countMap.set(key, ++count);
    }
});

voyages.forEach(function (voyage) {
    var key = voyageKeyOf(voyage),
        isUnique = countMap.get(key) == 1;

    if (isUnique) uniques.push(voyage)
    else dupes.push(voyage);
});


console.log('uniques', uniques);
console.log('dupes', dupes);

过滤器独特＆amp;使用下划线在数组中复制对象

4 个答案: