按对象数组中的值为对象分配排名

Question

我有一个像这样的对象数组:

var data = {
    a: [
        { 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 }
    ]
  };

我想根据组中的keythree和keyfour,以及整个数据集中的值为每个对象分配排名.我该怎么办？

更新:我在上面的代码中描述了排名.

结果对象:

var data = {
    a: [
        { keyone:'c', keytwo: 'anna', keythree: 21, keyfour: 15, rankgroup: 3, rankall: 4 },

        { keyone:'a', keytwo: 'anna', keythree: 22, keyfour: 15, rankgroup: 4, rankall: 5 },

        { keyone:'s', keytwo: 'anna', keythree: 22, keyfour: 15, rankgroup: 2, rankall: 2 },

        { keyone:'v', keytwo: 'anna', keythree: 7, keyfour: 15, rankgroup: 1, rankall: 1 }
    ],

    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 }
    ]
};

我在用lodash.我的想法是首先根据这些键对数组进行排序,然后遍历原始对象,通过比较另一个键来插入排序的索引.这是我尝试过的:

var keys = Object.keys(data);
var result = {};
var numkeys;
for(var i=0; i < keys.length; i++) {
    if(!numkeys) {
        var numkeys = _.keys(_.pick(data[keys[i]][0], _.isNumber));
  }

        for(var j=0;j<numkeys.length;j++) {
    var sorted = _.sortBy(data['a'], numkeys[j]);
        _.forEach(sorted, function(n, k) {

        //THIS FAILS
        var t = _.set(_.where(data[keys[i]], {keyone: n.keyone}), keys[i]+'rank', k);
        console.log(t);
        });

    }
  }

我该怎么办？我的逻辑似乎过于复杂,并且该set方法不会按键更新原始对象,而是在主对象之后添加新条目.

更新:注意22对象的重复出现a.这会在分配排名时出现问题,因为indexOf它将始终返回第一次出现的索引,因此第二次出现将永远不会为其分配索引,因此该值将是未定义的.

Answer 1

这就是我实现它的方法。

1. 将所有内容收集keythree到一个数组中并对它们进行排序（根据 进行分配rankall）index。

   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. 分配等级

   _.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;
       });
   });
   

检查这个小提琴

编辑

我正在为受骗者创建另一个数组

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

以及获得这些受骗物品的全球排名

    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);
}

看到项目是根据顺序中的所有属性排序的keythree，，， （如果您愿意，keyfour您可以更改内部顺序keytwo）keyone_.sortByAll

代码看起来比我想象的更难看。很快就会更新重构的代码

检查小提琴