Filter unique & duplicate objects in array using underscore

问题

I have an array of objects:

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}]

I am trying to extract all duplicates & unique objects BY "origin" & "dest". So that these 2 are the same, ignoring the rank key

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

Basically I want 2 seperate arrays:

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}]

using underscore, I was able to throw something like this together. But It seems inefficient and only returns an array of duplicates:

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()

I can't help but feel there is a much easier way to get this.

回答1:

It might be easier to introduce a temporary variable to hold the groups:

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();
});

Then:

var duplicates = [],
singles = [];

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

Demo

回答2:

I don't understand your code. However I got the problem.

The way I see it is 2 steps:

identify duplicates and push them in an array duplicates
create the uniquearray by copying the inital array and remove all the elements that match duplicates array.

Maybe it has lower performance but it could be clearer.

回答3:

Here is the algorithm using plain javascript. It is probably not the most efficient one if you will have thousands of records, but it get's the job done.

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 );

回答4:

There are many ways to do this, but here's one. I never used underscore, but the algorithm is pretty straight forward and you should be able to convert it easily.

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);

来源：https://stackoverflow.com/questions/24327912/filter-unique-duplicate-objects-in-array-using-underscore

标签

javascript

underscore.js