按匹配的对象字段数排序对象数组

Question

I have the following object:

var entries = [
  {
    fields: {
      tags: [
        { fields: { id: "football" } },
        { fields: { id: "running" } },
        { fields: { id: "tennis" } }
      ]
    }
  },
  {
    fields: {
      tags: [
        { fields: { id: "tennis" } },
        { fields: { id: "football" } },
        { fields: { id: "rugby" } }
      ]
    }
  },
  {
    fields: {
      tags: [
        { fields: { id: "cricket" } },
        { fields: { id: "running" } }
      ]
    }
  },
  {
    fields: {
      tags: [
        { fields: { id: "football" } },
        { fields: { id: "rugby" } }
      ]
    }
  }
];

Now, I want to sort the array using another array: var testArray = ["football", "rugby"]

So any of the objects that contain both rugby and football in their fields.tags.fields.id should return first in the array, then any that just contain 1 should appear next. So the results I'm after would be:

var sortedEntries = [
  {
    fields: {
      tags: [
        { fields: { id: "tennis" } },
        { fields: { id: "football" } },
        { fields: { id: "rugby" } }
      ]
    }
  },
  {
    fields: {
      tags: [
        { fields: { id: "football" } },
        { fields: { id: "rugby" } }
      ]
    }
  },
  {
    fields: {
      tags: [
        { fields: { id: "football" } },
        { fields: { id: "running" } },
        { fields: { id: "tennis" } }
      ]
    }
  },
  {
    fields: {
      tags: [
        { fields: { id: "cricket" } },
        { fields: { id: "running" } }
      ]
    }
  }
];

I thought this might be possible using the sort() function but then I realised I need to compare my testArray to see how many items are matched. Is this possible? I can't find any examples online of what I'm trying to do and I don't know where to begin.

Answer 1

Custom sort function that filters the tags arrays and checks lengths of those matches

entries.sort(function (a, b) {
    var aMatches = a.fields.tags.filter(function (item) {
        return testArray.indexOf(item.fields.id) > -1
    });
    var bMatches = b.fields.tags.filter(function (item) {
        return testArray.indexOf(item.fields.id) > -1
    });
    return bMatches.length - aMatches.length;
});

DEMO

Answer 2

Solution optimized for large datasets, with temporary storage of count.

 var search = ['football', 'rugby'], entries = [{ fields: { tags: [{ fields: { id: "cricket" } }, { fields: { id: "running" } }] } }, { fields: { tags: [{ fields: { id: "football" } }, { fields: { id: "running" } }, { fields: { id: "tennis" } }] } }, { fields: { tags: [{ fields: { id: "tennis" } }, { fields: { id: "football" } }, { fields: { id: "rugby" } }] } }, { fields: { tags: [{ fields: { id: "football" } }, { fields: { id: "rugby" } }] } }], sortedData = entries.map(function (a) { a.count = a.fields.tags.reduce(function (r, b) { return r + !!~search.indexOf(b.fields.id); }, 0); return a; }, []).sort(function (a, b) { return b.count - a.count; }).map(function (a) { delete a.count; return a; }, []); document.write('<pre>' + JSON.stringify(sortedData, 0, 4) + '</pre>'); 

Solution optimized for small datasets, without temporary storage of count.

 var search = ['football', 'rugby'], entries = [{ fields: { tags: [{ fields: { id: "cricket" } }, { fields: { id: "running" } }] } }, { fields: { tags: [{ fields: { id: "football" } }, { fields: { id: "running" } }, { fields: { id: "tennis" } }] } }, { fields: { tags: [{ fields: { id: "tennis" } }, { fields: { id: "football" } }, { fields: { id: "rugby" } }] } }, { fields: { tags: [{ fields: { id: "football" } }, { fields: { id: "rugby" } }] } }], sortedData = entries.sort(function (a, b) { function f(r, c) { return r + !!~search.indexOf(c.fields.id); } return b.fields.tags.reduce(f, 0) - a.fields.tags.reduce(f, 0); }); document.write('<pre>' + JSON.stringify(sortedData, 0, 4) + '</pre>'); 

Answer 3

It's possible with a custom sort function that would check how many items in tags exist in testArray . Something like this should work:

var testArray = ["football", "rugby"];
entries.sort(function(a, b) {
    var a_score = 0;
    var b_score = 0;
    for (var i=0; i<a.fields.tags.length; i++) {
        if (testArray.indexOf(a.fields.tags[i]) > -1) {
            a_score++;
        }
    }
    for (var i=0; i<b.fields.tags.length; i++) {
        if (testArray.indexOf(b.fields.tags[i]) > -1) {
            b_score++;
        }
    }
    return b_score-a_score;
});