具有很少字段的javascript数组比较

Question

I have two arrays (data and data_not_included).Each elemet of those arrays has attridutes id and name. I fill them this way:

data[i] = {
           name :products.models[i].get('name'),
           id :  products.models[i].get('id')
          };

Now I want do display the elements in data which are not in data_not_included array. For example I have

data=[{name: Sugar}{id: 1},{name: Butter}{id: 2},{name: Cola}{id: 3}]
// and
data_nat_included = [{name: Sugar}{id: 1},{name: Butter}{id: 2}].

It should display {name: Cola}{id: 3} only.

Here is what I have already done:

for(var j=0;j<data_not_icluded.length;j++)
{
    for(var i=0;i<data.length;i++)
    {
        if(data[i].id != data_not_icluded[j].id ){
          //but this doesnt work for me it displayes a lot of element many times
        }
     }
}

Answer 1

Both answers are asymptotically bad. This means they run in suboptimal time. In other words, they are naive approaches to solving the problem. This problem is more widely known in the domain of databases, where join operation is a commonplace. It is also known that the complexity of a join is O(log n * n + log m * m) where n is the number of elements in first table and m is the number of elements in the second table. This is fewer operations then would be required by naive solution offered in other examples O(n^2) .

However, if more is known about your data, as, for example, I would expect that the values are unique and easily serializable to string, you could even reduce the complexity to O(n + m) by simply creating hashes of the objects you want to compare. Here's how to do it:

Where n is the number of elements in the first array and m is the number of elements in the second array.

var data = [{ name: "Sugar" },
            { id: 1 },
            { name: "Butter" },
            { id: 2 },
            { name: "Cola" },
            { id: 3 }];
var dataNatIncluded = [{ name: "Sugar" },
                       { id: 1 },
                       { name: "Butter" },
                       { id: 2 }];

function join(a, b) {
    var hashA = {}, hashB = {}, p, result = [];
    function setter(hash) {
        return function (element) { hash[JSON.stringify(element)] = element; };
    }
    a.forEach(setter(hashA));
    b.forEach(setter(hashB));
    for (p in hashB) delete hashA[p];
    for (p in hashA) result.push(hashA[p]);
    return result;
}
// [{ name: "Cola" }, { id: 3 }]

Answer 2

A simple way to do that:

var vals = [];

for(var i=0;i<data.length;i++)
{
    var found = false;
    for(var j=0;j<data_nat.length;j++)
    {
        if(data[i].id == data_nat[j].id ){
          found = true;
          break;
        }
     }
    if (!found) vals.push(data[i]);
}

JSFiddle

Answer 3

 for(var j=0;j<data_not_icluded.length;j++) for(var i=0;i<data.length;i++) if(data[i].id != data_not_icluded[j].id ) 

Think of what this does: For any not included object, show all objects that have not the same id as the current not included one. This will show many items multiple times, and it will show objects that are in "not included" but at another position.

Instead, loop over data , check each that it is not included in data_not_included , and display it otherwise:

dataloop: for (var i=0; i<data.length; i++) {
    for (var j=0; j<data_not_included.length; j++)
        if (data[i].id == data_not_icluded[j].id)
            continue dataloop;
    display(data[i]);
}

Or, using some iteration methods of Arrays :

data.filter(function(d) {
    return data_not_included.every(function(n) {
        return d.id != n.id;
    });
}).each(display);