按数组中的对象分组最有效的方法是什么?

例如,给定此对象数组:

[ 
    { Phase: "Phase 1", Step: "Step 1", Task: "Task 1", Value: "5" },
    { Phase: "Phase 1", Step: "Step 1", Task: "Task 2", Value: "10" },
    { Phase: "Phase 1", Step: "Step 2", Task: "Task 1", Value: "15" },
    { Phase: "Phase 1", Step: "Step 2", Task: "Task 2", Value: "20" },
    { Phase: "Phase 2", Step: "Step 1", Task: "Task 1", Value: "25" },
    { Phase: "Phase 2", Step: "Step 1", Task: "Task 2", Value: "30" },
    { Phase: "Phase 2", Step: "Step 2", Task: "Task 1", Value: "35" },
    { Phase: "Phase 2", Step: "Step 2", Task: "Task 2", Value: "40" }
]

我正在表格中显示这些信息。我想通过不同的方法进行分组,但我想对值求和。

我将Undercore.js用于其groupby函数,这很有用,但并不能完成全部任务,因为我不希望它们“拆分”,而是“合并”,更像SQL groupby方法。

我要找的是能够合计特定值(如果需要)。

因此,如果我按阶段分组,我希望收到:

[
    { Phase: "Phase 1", Value: 50 },
    { Phase: "Phase 2", Value: 130 }
]

如果我组了阶段/步骤,我会收到:

[
    { Phase: "Phase 1", Step: "Step 1", Value: 15 },
    { Phase: "Phase 1", Step: "Step 2", Value: 35 },
    { Phase: "Phase 2", Step: "Step 1", Value: 55 },
    { Phase: "Phase 2", Step: "Step 2", Value: 75 }
]

是否有一个有用的脚本,或者我应该坚持使用Undercore.js,然后遍历生成的对象,自己计算总数?


当前回答

let x  = [
  {
    "id": "6",
    "name": "SMD L13",
    "equipmentType": {
      "id": "1",
      "name": "SMD"
    }
  },
  {
    "id": "7",
    "name": "SMD L15",
    "equipmentType": {
      "id": "1",
      "name": "SMD"
    }
  },
  {
    "id": "2",
    "name": "SMD L1",
    "equipmentType": {
      "id": "1",
      "name": "SMD"
    }
  }
];

function groupBy(array, property) {
  return array.reduce((accumulator, current) => {
    const object_property = current[property];
    delete current[property]

    let classified_element = accumulator.find(x => x.id === object_property.id);
    let other_elements = accumulator.filter(x => x.id !== object_property.id);

   if (classified_element) {
     classified_element.children.push(current)
   } else {
     classified_element = {
       ...object_property, 
       'children': [current]
     }
   }
   return [classified_element, ...other_elements];
 }, [])
}

console.log( groupBy(x, 'equipmentType') )

/* output 

[
  {
    "id": "1",
    "name": "SMD",
    "children": [
      {
        "id": "6",
        "name": "SMD L13"
      },
      {
        "id": "7",
        "name": "SMD L15"
      },
      {
        "id": "2",
        "name": "SMD L1"
      }
    ]
  }
]

*/

其他回答

常量动物=[{type:'狗',品种:“水坑”},{type:'狗',品种:“拉布拉多”},{类型:'cat',品种:“暹罗”},{type:'狗',品种:“法国斗牛犬”},{类型:'cat',品种:'泥'}];var groupBy=(arr,prop)=>{return arr.reduce((objs,obj)=>{const key=obj[prop];if(键){让fi=objs.findIndex(x=>x.key==key);如果(fi>=0){objs[fi].values=[…objs[fi].values,obj];}其他{对象.推送({key:键,值:[obj]})}}返回对象;}, []);}console.log(groupBy(动物,“类型”))

具有排序功能

export const groupBy = function groupByArray(xs, key, sortKey) {
      return xs.reduce(function(rv, x) {
        let v = key instanceof Function ? key(x) : x[key];
        let el = rv.find(r => r && r.key === v);

        if (el) {
          el.values.push(x);
          el.values.sort(function(a, b) {
            return a[sortKey].toLowerCase().localeCompare(b[sortKey].toLowerCase());
          });
        } else {
          rv.push({ key: v, values: [x] });
        }

        return rv;
      }, []);
    };

示例:

var state = [
    {
      name: "Arkansas",
      population: "2.978M",
      flag:
  "https://upload.wikimedia.org/wikipedia/commons/9/9d/Flag_of_Arkansas.svg",
      category: "city"
    },{
      name: "Crkansas",
      population: "2.978M",
      flag:
        "https://upload.wikimedia.org/wikipedia/commons/9/9d/Flag_of_Arkansas.svg",
      category: "city"
    },
    {
      name: "Balifornia",
      population: "39.14M",
      flag:
        "https://upload.wikimedia.org/wikipedia/commons/0/01/Flag_of_California.svg",
      category: "city"
    },
    {
      name: "Florida",
      population: "20.27M",
      flag:
        "https://upload.wikimedia.org/wikipedia/commons/f/f7/Flag_of_Florida.svg",
      category: "airport"
    },
    {
      name: "Texas",
      population: "27.47M",
      flag:
        "https://upload.wikimedia.org/wikipedia/commons/f/f7/Flag_of_Texas.svg",
      category: "landmark"
    }
  ];
console.log(JSON.stringify(groupBy(state,'category','name')));

从@mortb、@jmarceli的回答和这篇文章中,

我利用JSON.stringify()作为分组依据的PRIMITIVE VALUE多列的标识。

无第三方

function groupBy(list, keyGetter) {
    const map = new Map();
    list.forEach((item) => {
        const key = keyGetter(item);
        if (!map.has(key)) {
            map.set(key, [item]);
        } else {
            map.get(key).push(item);
        }
    });
    return map;
}

const pets = [
    {type:"Dog", age: 3, name:"Spot"},
    {type:"Cat", age: 3, name:"Tiger"},
    {type:"Dog", age: 4, name:"Rover"}, 
    {type:"Cat", age: 3, name:"Leo"}
];

const grouped = groupBy(pets,
pet => JSON.stringify({ type: pet.type, age: pet.age }));

console.log(grouped);

使用Lodash第三方

const pets = [
    {type:"Dog", age: 3, name:"Spot"},
    {type:"Cat", age: 3, name:"Tiger"},
    {type:"Dog", age: 4, name:"Rover"}, 
    {type:"Cat", age: 3, name:"Leo"}
];

let rslt = _.groupBy(pets, pet => JSON.stringify(
 { type: pet.type, age: pet.age }));

console.log(rslt);

想象一下,你有这样的东西:

〔{id:1,cat:'sedan'},{id:2,cat:'sport‘},{id:3,cat:'sport‘},{id:4,cat:'sadan‘}〕

通过这样做:const categories=[…new Set(cars.map((car)=>car.cat))]

你会得到这个:[“sadan”,“port”]

说明:1.首先,我们通过传递一个数组来创建一个新的Set。由于Set仅允许唯一值,因此将删除所有重复项。

现在重复项消失了,我们将使用扩展运算符将其转换回数组。。。

设置文档:https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Set排列运算符文档:https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Spread_syntax

使用ES6的简单解决方案:

该方法有一个返回模型,可以比较n个财产。

const compareKey = (item, key, compareItem) => {
    return item[key] === compareItem[key]
}

const handleCountingRelatedItems = (listItems, modelCallback, compareKeyCallback) => {
    return listItems.reduce((previousValue, currentValue) => {
        if (Array.isArray(previousValue)) {
        const foundIndex = previousValue.findIndex(item => compareKeyCallback(item, currentValue))

        if (foundIndex > -1) {
            const count = previousValue[foundIndex].count + 1

            previousValue[foundIndex] = modelCallback(currentValue, count)

            return previousValue
        }

        return [...previousValue, modelCallback(currentValue, 1)]
        }

        if (compareKeyCallback(previousValue, currentValue)) {
        return [modelCallback(currentValue, 2)]
        }

        return [modelCallback(previousValue, 1), modelCallback(currentValue, 1)]
    })
}

const itemList = [
    { type: 'production', human_readable: 'Production' },
    { type: 'test', human_readable: 'Testing' },
    { type: 'production', human_readable: 'Production' }
]

const model = (currentParam, count) => ({
    label: currentParam.human_readable,
    type: currentParam.type,
    count
})

const compareParameter = (item, compareValue) => {
    const isTypeEqual = compareKey(item, 'type', compareValue)
    return isTypeEqual
}

const result = handleCountingRelatedItems(itemList, model, compareParameter)

 console.log('Result: \n', result)
/** Result: 
    [
        { label: 'Production', type: 'production', count: 2 },
        { label: 'Testing', type: 'testing', count: 1 }
    ]
*/