我想通过迭代一个对象数组来创建具有Key和unique值的对象。
我的Json是
var data = [
{sex: 'male', beaconKey: '121', userKey: '01', key:'k1', date: 'Wed Jul 06 06:26:59 UTC 2016', freq: 20, preferenceList: ['Fashion','Business','Sports'], ageGroup: 'LESS_THAN_15'}
, {sex: 'male', beaconKey: '122', userKey: '01', key:'k2', date: 'Tue Jul 05 11:59:28 UTC 2016', freq: 22, preferenceList: ['Business'], ageGroup: 'BETWEEN_25_35'}
, {sex: 'female', beaconKey: '123', userKey: '01', key:'k3', date: 'Tue Jul 05 11:59:28 UTC 2016', freq: 26, preferenceList: ['Housing'], ageGroup: 'BETWEEN_35_45'}
, {sex: 'male', beaconKey: '121', userKey: '03', key:'k4', date: 'Wed Jul 06 06:26:59 UTC 2016', freq: 38, preferenceList: ['Business','Housing','Sports'], ageGroup: 'BETWEEN_25_35'}
, {sex: 'female', beaconKey: '124', userKey: '03', key:'k5', date: 'Tue Jul 05 11:59:28 UTC 2016', freq: 20, preferenceList: ['Business','Housing','Sports'], ageGroup: 'BETWEEN_45_55'}
, {sex: 'male', beaconKey: '125', userKey: '01', key:'k6', date: 'Wed Jul 06 06:26:59 UTC 2016', freq: 42, preferenceList: ['Jewels','Kids'], ageGroup: 'LESS_THAN_15'}
, {sex: 'female', beaconKey: '123', userKey: '02', key:'k7', date: 'Tue Jul 05 06:26:59 UTC 2016', freq: 34, preferenceList: ['Movies'], ageGroup: 'BETWEEN_55_65'}
, {sex: 'female', beaconKey: '121', userKey: '04', key:'k8', date: 'Wed Jul 06 06:26:59 UTC 2016', freq: 22, preferenceList: ['Culture'], ageGroup: 'BETWEEN_25_35'}
, {sex: 'female', beaconKey: '124', userKey: '01', key:'k9', date: 'Tue Jul 05 11:59:28 UTC 2016', freq: 115, preferenceList: ['Kids'], ageGroup: 'BETWEEN_35_45'}
, {sex: 'male', beaconKey: '122', userKey: '02', key:'k10', date: 'Wed Jul 06 06:26:59 UTC 2016', freq: 19, preferenceList: [], ageGroup: 'BETWEEN_45_55'}
, {sex: 'female', beaconKey: '121',userKey: '01', key:'k11', date: 'Wed Jul 06 06:26:59 UTC 2016', freq: 24, preferenceList: [], ageGroup: 'BETWEEN_35_45'}
];
我可以从上面的json中获得另一个对象
var filterObj = {
sex:[],
age:[],
category:[]
};
并且具有JSON中出现的唯一值,我也有preferenceList
的多个值,这些值在filterObj
中也应该是唯一的。
filterObj = {
sex:['male','female'],
age:['LESS_THAN_15','BETWEEN_25_35','BETWEEN_35_45','BETWEEN_45_55','BETWEEN_55_65'],
category:['Fashion','Business','Sports','Housing','Jewels','Kids','Movies','Kids']
};
答案 0 :(得分:3)
您基本上必须迭代数组data
。在数组中查找唯一值的最佳方法是使用key, value
数据结构(哈希表),如Javascript对象。
var sex = {};
var age = {};
data.forEach(function (obj, index) {
sex[obj.sex] = index;
age[obj.age] = index;
});
var filterObj = {};
filterObj['sex'] = Object.keys(sex);
filterObj['age'] = Object.keys(age);
答案 1 :(得分:3)
使用javascript中唯一的键和值对创建对象
function getMapObject(dataList, key) {
var lookup = {};
for ( var i in dataList) {
var tempArr = [];
if (lookup[dataList[i][key]] != undefined) {
tempArr = lookup[dataList[i][key]];
tempArr.push(dataList[i]);
lookup[dataList[i][key]] = tempArr;
} else {
tempArr.push(dataList[i]);
lookup[dataList[i][key]] = tempArr;
}
}
return lookup;}
答案 2 :(得分:1)
您可以将对象用作已插入项目的哈希表到过滤结果。
此提案使用所需属性的地图。它很容易扩展,只需添加另一个地图名称作为属性,将数据的原始属性添加为值。
{ "sex": "sex", "age": "ageGroup", "category": "preferenceList" }
var data = [{ sex: 'male', beaconKey: '121', userKey: '01', key: 'k1', date: 'Wed Jul 06 06:26:59 UTC 2016', freq: 20, preferenceList: ['Fashion', 'Business', 'Sports'], ageGroup: 'LESS_THAN_15' }, { sex: 'male', beaconKey: '122', userKey: '01', key: 'k2', date: 'Tue Jul 05 11:59:28 UTC 2016', freq: 22, preferenceList: ['Business'], ageGroup: 'BETWEEN_25_35' }, { sex: 'female', beaconKey: '123', userKey: '01', key: 'k3', date: 'Tue Jul 05 11:59:28 UTC 2016', freq: 26, preferenceList: ['Housing'], ageGroup: 'BETWEEN_35_45' }, { sex: 'male', beaconKey: '121', userKey: '03', key: 'k4', date: 'Wed Jul 06 06:26:59 UTC 2016', freq: 38, preferenceList: ['Business', 'Housing', 'Sports'], ageGroup: 'BETWEEN_25_35' }, { sex: 'female', beaconKey: '124', userKey: '03', key: 'k5', date: 'Tue Jul 05 11:59:28 UTC 2016', freq: 20, preferenceList: ['Business', 'Housing', 'Sports'], ageGroup: 'BETWEEN_45_55' }, { sex: 'male', beaconKey: '125', userKey: '01', key: 'k6', date: 'Wed Jul 06 06:26:59 UTC 2016', freq: 42, preferenceList: ['Jewels', 'Kids'], ageGroup: 'LESS_THAN_15' }, { sex: 'female', beaconKey: '123', userKey: '02', key: 'k7', date: 'Tue Jul 05 06:26:59 UTC 2016', freq: 34, preferenceList: ['Movies'], ageGroup: 'BETWEEN_55_65' }, { sex: 'female', beaconKey: '121', userKey: '04', key: 'k8', date: 'Wed Jul 06 06:26:59 UTC 2016', freq: 22, preferenceList: ['Culture'], ageGroup: 'BETWEEN_25_35' }, { sex: 'female', beaconKey: '124', userKey: '01', key: 'k9', date: 'Tue Jul 05 11:59:28 UTC 2016', freq: 115, preferenceList: ['Kids'], ageGroup: 'BETWEEN_35_45' }, { sex: 'male', beaconKey: '122', userKey: '02', key: 'k10', date: 'Wed Jul 06 06:26:59 UTC 2016', freq: 19, preferenceList: [], ageGroup: 'BETWEEN_45_55' }, { sex: 'female', beaconKey: '121', userKey: '01', key: 'k11', date: 'Wed Jul 06 06:26:59 UTC 2016', freq: 24, preferenceList: [], ageGroup: 'BETWEEN_35_45' }],
map = { sex: 'sex', age: 'ageGroup', category: 'preferenceList' },
filterObj = {};
data.forEach(function (a) {
Object.keys(map).forEach(function (k) {
filterObj[k] = filterObj[k] || [];
this[k] = this[k] || {};
(Array.isArray(a[map[k]]) ? a[map[k]] : [a[map[k]]]).forEach(function (b) {
if (!this[k][b]) {
this[k][b] = true;
filterObj[k].push(b);
}
}, this);
}, this);
}, {});
console.log(filterObj);
答案 3 :(得分:0)
试试这个
var data = [
{sex: 'male', beaconKey: '121', userKey: '01', key:'k1', date: 'Wed Jul 06 06:26:59 UTC 2016', freq: 20, preferenceList: ['Fashion','Business','Sports'], ageGroup: 'LESS_THAN_15'}
, {sex: 'male', beaconKey: '122', userKey: '01', key:'k2', date: 'Tue Jul 05 11:59:28 UTC 2016', freq: 22, preferenceList: ['Business'], ageGroup: 'BETWEEN_25_35'}
, {sex: 'female', beaconKey: '123', userKey: '01', key:'k3', date: 'Tue Jul 05 11:59:28 UTC 2016', freq: 26, preferenceList: ['Housing'], ageGroup: 'BETWEEN_35_45'}
, {sex: 'male', beaconKey: '121', userKey: '03', key:'k4', date: 'Wed Jul 06 06:26:59 UTC 2016', freq: 38, preferenceList: ['Business','Housing','Sports'], ageGroup: 'BETWEEN_25_35'}
, {sex: 'female', beaconKey: '124', userKey: '03', key:'k5', date: 'Tue Jul 05 11:59:28 UTC 2016', freq: 20, preferenceList: ['Business','Housing','Sports'], ageGroup: 'BETWEEN_45_55'}
, {sex: 'male', beaconKey: '125', userKey: '01', key:'k6', date: 'Wed Jul 06 06:26:59 UTC 2016', freq: 42, preferenceList: ['Jewels','Kids'], ageGroup: 'LESS_THAN_15'}
, {sex: 'female', beaconKey: '123', userKey: '02', key:'k7', date: 'Tue Jul 05 06:26:59 UTC 2016', freq: 34, preferenceList: ['Movies'], ageGroup: 'BETWEEN_55_65'}
, {sex: 'female', beaconKey: '121', userKey: '04', key:'k8', date: 'Wed Jul 06 06:26:59 UTC 2016', freq: 22, preferenceList: ['Culture'], ageGroup: 'BETWEEN_25_35'}
, {sex: 'female', beaconKey: '124', userKey: '01', key:'k9', date: 'Tue Jul 05 11:59:28 UTC 2016', freq: 115, preferenceList: ['Kids'], ageGroup: 'BETWEEN_35_45'}
, {sex: 'male', beaconKey: '122', userKey: '02', key:'k10', date: 'Wed Jul 06 06:26:59 UTC 2016', freq: 19, preferenceList: [], ageGroup: 'BETWEEN_45_55'}
, {sex: 'female', beaconKey: '121',userKey: '01', key:'k11', date: 'Wed Jul 06 06:26:59 UTC 2016', freq: 24, preferenceList: [], ageGroup: 'BETWEEN_35_45'}
];
var filterObj = { sex:[], age:[], category:[] };
//iterate all objects in data
data.forEach( function( obj ){
//check if value is already there in the array before putting the same in filterObj
filterObj.sex.indexOf( obj.sex ) == -1 && filterObj.sex.push( obj.sex );
filterObj.age.indexOf( obj.ageGroup ) == -1 && filterObj.age.push( obj.ageGroup );
filterObj.category = filterObj.category.concat( obj.preferenceList ) ;
});
//find unique categories
var map = {};
filterObj.category.forEach( function(cat){ map[cat] = true });
filterObj.category = Object.keys( map );
console.log( filterObj );
答案 4 :(得分:0)
我建议你使用lodash库:
var data = [
{sex: 'male', beaconKey: '121', userKey: '01', key:'k1', date: 'Wed Jul 06 06:26:59 UTC 2016', freq: 20, preferenceList: ['Fashion','Business','Sports'], ageGroup: 'LESS_THAN_15'}
, {sex: 'male', beaconKey: '122', userKey: '01', key:'k2', date: 'Tue Jul 05 11:59:28 UTC 2016', freq: 22, preferenceList: ['Business'], ageGroup: 'BETWEEN_25_35'}
, {sex: 'female', beaconKey: '123', userKey: '01', key:'k3', date: 'Tue Jul 05 11:59:28 UTC 2016', freq: 26, preferenceList: ['Housing'], ageGroup: 'BETWEEN_35_45'}
, {sex: 'male', beaconKey: '121', userKey: '03', key:'k4', date: 'Wed Jul 06 06:26:59 UTC 2016', freq: 38, preferenceList: ['Business','Housing','Sports'], ageGroup: 'BETWEEN_25_35'}
, {sex: 'female', beaconKey: '124', userKey: '03', key:'k5', date: 'Tue Jul 05 11:59:28 UTC 2016', freq: 20, preferenceList: ['Business','Housing','Sports'], ageGroup: 'BETWEEN_45_55'}
, {sex: 'male', beaconKey: '125', userKey: '01', key:'k6', date: 'Wed Jul 06 06:26:59 UTC 2016', freq: 42, preferenceList: ['Jewels','Kids'], ageGroup: 'LESS_THAN_15'}
, {sex: 'female', beaconKey: '123', userKey: '02', key:'k7', date: 'Tue Jul 05 06:26:59 UTC 2016', freq: 34, preferenceList: ['Movies'], ageGroup: 'BETWEEN_55_65'}
, {sex: 'female', beaconKey: '121', userKey: '04', key:'k8', date: 'Wed Jul 06 06:26:59 UTC 2016', freq: 22, preferenceList: ['Culture'], ageGroup: 'BETWEEN_25_35'}
, {sex: 'female', beaconKey: '124', userKey: '01', key:'k9', date: 'Tue Jul 05 11:59:28 UTC 2016', freq: 115, preferenceList: ['Kids'], ageGroup: 'BETWEEN_35_45'}
, {sex: 'male', beaconKey: '122', userKey: '02', key:'k10', date: 'Wed Jul 06 06:26:59 UTC 2016', freq: 19, preferenceList: [], ageGroup: 'BETWEEN_45_55'}
, {sex: 'female', beaconKey: '121',userKey: '01', key:'k11', date: 'Wed Jul 06 06:26:59 UTC 2016', freq: 24, preferenceList: [], ageGroup: 'BETWEEN_35_45'}
];
var filterObj = {
sex: _.uniq(_.map(data,'sex')),
age: _.uniq(_.map(data,'ageGroup')),
category: _.union.apply(_,_.map(data,'preferenceList'))
};
这里有plnkdr: