根据匹配的数据返回数组中的对象

Question

I have a json array of about 30 objects. Here's a sample object from the array:

{
    "id": 0,
    "name": "Valle",
    "activities": "night-life",
    "food": "fancy-food",
    "sport" : "baseball",
    "geoProfile": "artsy",
    "priority": 2
}

I am building out another object on a page based on user input. The user will choose between radio buttons and after they have made their choices, I will have an object such as:

{geoProfile: "artsy", activities: "nature", food: "fancy-food", sport: "cricket"}

I am using jQuery's $.each() method to go through each object as follows:

$.each(self.data(), function (i, s) {
    if (s.geoProfile == self.prefLocation() &&
        s.activities == self.prefActivity() &&
        s.food == self.prefFood() &&
        s.sport == self.prefSport()) {
        optionMatched = s;
        return false;
    }
});

This will return an object that has all four matches, but how can I return the json object that has the most matches to the user-built object? If two match, I want to then look at the "priority" property and return the one with the lowest priority.

Answer 1

You could use Array#map and build a new array with the sum of the matched properties.

Later you can sort with map and use the result for sorting and get the first element.

var data = [/* your data here */],
    search = { geoProfile: "artsy", activities: "nature", food: "fancy-food", sport: "cricket" },
    result = data.map(function (a, i) {
        return {
            count: Object.keys(search).reduce(function (r, k) { return r + +(a[k] === search[k]); }, 0),
            priority: a.priority,
            index: i
        }
    });

result.sort(function (a, b) {
    return b.count - a.count || a.priority - b.priority;
});

A single loop solution

var data = [/* your data here */],
    search = { geoProfile: "artsy", activities: "nature", food: "fancy-food", sport: "cricket" },
    result = data.reduce(function (r, a, i) {
        document.write('<pre>' + JSON.stringify(r, 0, 4) + '</pre><hr>');
        var o = {
            count: Object.keys(search).reduce(function (q, k) { return q + +(a[k] === search[k]); }, 0),
            priority: a.priority,
            index: i
        };
        if (!i || o.count > r[0].count || o.count === r[0].count && o.priority < r[0].priority) {
            return [o];
        }
        o.count === r[0].count && o.priority === r[0].priority && r.push(o);
        return r;
    }, []);

Answer 2

Just track the number of matches and update your selected one based on if it has more matches.

var numOfMatches = 0;
$.each(self.data(), function(i, s) {
  var matchingProperties = 0;
  if (s.geoProfile == self.prefLocation()) {
    matchingProperties++;
  }
  if (s.activities == self.prefActivity()) {
    matchingProperties++;
  }
  if (s.food == self.prefFood()) {
    matchingProperties++;
  }
  if (s.sport == self.prefSport()) {
    matchingProperties++;
  }

  if (matchingProperties === 0 || matchingProperties < numOfMatches) {
    return;
  }

  if (!optionMatched // Grab the first match
      || matchingProperties > numOfMatches // or if more properties match
      || s.priority < optionMatched.priority) { // or the one with a lower priority
    optionMatched = s;
    numOfMatches = matchingProperties;
  }
});

Or you could simplify the initial counting using filter :

var numOfMatches = 0;
$.each(self.data(), function(i, s) {
  var matchingProperties = [
    s.geoProfile == self.prefLocation(),
    s.activities == self.prefActivity(),
    s.food == self.prefFood(),
    s.sport == self.prefSport()
  ].filter(function(val) { return val; }).length;

  if (matchingProperties === 0 || matchingProperties < numOfMatches) {
    return;
  }

  if (!optionMatched // Grab the first match
      || matchingProperties > numOfMatches // or if more properties match
      || s.priority < optionMatched.priority) { // or the one with a lower priority
    optionMatched = s;
    numOfMatches = matchingProperties;
  }
});