過濾器獨特的和重複的對象

Question

我有對象的數組：

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}] 

我試圖提取所有複製&唯一對象BY「出身」 &「目標」。因此，這2是相同的，忽略了排名鍵

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

基本上我想要2個獨立的陣列：

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}]

使用下劃線，我能夠把這樣的東西扔在一起。但似乎效率低下，只返回重複的數組：

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() 

我不禁感到有一種更簡單的方法來得到這個。

Answer 1

可能更容易引入一個臨時變量來保存組：

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(); 
}); 

然後：

var duplicates = [], 
singles = []; 

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

Demo

Answer 2

我不明白你的代碼。但是我遇到了問題。

我看到它的方式是兩個步驟：

• 識別重複並把它們壓在陣列中重複
• 通過複製inital陣列創建uniquearray並刪除所有匹配重複數組的元素。

也許它具有較低的性能，但它可能會更清晰。

Answer 3

下面是使用普通的JavaScript的算法。如果你有成千上萬的記錄，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這一點，但這裏有一個。我從來沒有使用下劃線，但算法非常簡單，你應該能夠輕鬆地進行轉換。

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);