按对象数组中的值为对象分配排名-Assignrankstoobjectsbyvaluesinanarrayofobjects

1  

this is how I achieved it.

这就是我实现它的方式。

1. collect all keythree into an array and sort them (to assign rankall based on index).

   将所有keythree收集到一个数组中并对它们进行排序（根据索引分配rankall）。

   var all = [];
   _.forEach(data, function (a, key) {
       _.forEach(a, function(n, k){
       all.push(n.keythree);
     });
   });
   all.sort(function(a,b){
       return a-b;
   });
   

2. assign ranks

   分配排名

   _.forEach(data, function (a, key) {
       var sorted = _.sortBy(a, 'keythree');
       _.forEach(sorted, function(n, k) {
         var index = _.findIndex(data[key], {keyone: n.keyone});
         data[key][index]['rankgroup'] = k+1;
         data[key][index]['rankall'] = all.indexOf(n.keythree)+1;
       });
   });
   

check this fiddle

检查这个小提琴

EDIT

i'm creating another array for dupes

我正在为dupes创建另一个数组

_.forEach(a, function(n, k) {
    if (all.indexOf(n.keythree) !== -1) {
        dupes.push(n.keythree);
    }
    all.push(n.keythree);
});

and for getting the global rank for these dupe items

并获得这些欺骗项目的全球排名

    function getGlobalRank(n) {
    var val = n.keythree;
    if (sorted_dupes[val] === undefined) {
        sorted_dupes[val] = [];
        _.forEach(data, function(a, key) {
            _.forEach(_.where(a, {
                keythree: val
            }), function(b) {
                sorted_dupes[val].push(b);
            });
        });
        sorted_dupes[val] = _.sortByAll(sorted_dupes[val], ['keyfour', 'keytwo', 'keyone']);
    }
    return _.findIndex(sorted_dupes[val], {
        keyone: n.keyone,
        keytwo: n.keytwo,
        keythree: n.keythree,
        keyfour: n.keyfour
    }) + 1 + all.indexOf(val);
}

see that the items are sorted based on all the properties in the order keythree, keyfour, keytwo, keyone (you can change the order inside _.sortByAll if you want to)

看到项目是根据订单keythree，keyfour，keytwo，keyone中的所有属性进行排序的（如果你愿意，你可以改变_.sortByAll中的顺序）

the code looking uglier than i thought. will update the refactored code soon

代码看起来比我想象的更丑。将很快更新重构的代码

check the fiddle

检查小提琴

1  

Just in plain Javascript:

只是简单的Javascript：

For ranking, the data must be sorted, somehow.

对于排名，必须以某种方式对数据进行排序。

This solution features an array with objects and the references to the given objects in data. Then all items are sorted and the original data gets their rankgroup and rankall property.

此解决方案具有一个包含对象的数组以及对数据中给定对象的引用。然后对所有项目进行排序，并且原始数据获得其rankgroup和rankall属性。

Edit: Now with same rank for duplicates.

编辑：现在具有相同的重复等级。

var data = { a: [{ keyone: 'g', keytwo: 'tina', keythree: 21, keyfour: 15 }, { keyone: 'c', keytwo: 'anna', keythree: 21, keyfour: 15 }, { keyone: 'a', keytwo: 'anna', keythree: 22, keyfour: 15 }, { keyone: 's', keytwo: 'anna', keythree: 10, keyfour: 15 }, { keyone: 'v', keytwo: 'anna', keythree: 7, keyfour: 15 }], b: [{ keyone: 'f', keytwo: 'any', keythree: 45, keyfour: 100 }, { keyone: 'b', keytwo: 'any', keythree: 146, keyfour: 100 }, { keyone: 't', keytwo: 'any', keythree: 23, keyfour: 100 }, { keyone: 'h', keytwo: 'any', keythree: 11, keyfour: 100 }] },
    group = {};

Object.keys(data).reduce(function (r, k) {
    return r.concat(data[k].map(function (a) {
        return { obj: k, ref: a };
    }));
}, []).sort(function (a, b) {
    return a.ref.keythree - b.ref.keythree || a.ref.keyfour - b.ref.keyfour;
}).forEach(function (a, i, aa) {
    if (i && a.ref.keythree === aa[i - 1].ref.keythree && a.ref.keyfour === aa[i - 1].ref.keyfour) {
        a.ref.rankgroup = group[a.obj];
        a.ref.rankall = group.rankall;
    } else {
        group[a.obj] = (group[a.obj] || 0) + 1;
        group.rankall = (group.rankall || 0) + 1;
        a.ref.rankgroup = group[a.obj];
        a.ref.rankall = group.rankall;
    }
});

document.write('
' + JSON.stringify(data, 0, 4) + '
');