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

此解决方案采用任意函数(而不是键),因此比上述解决方案更灵活,并允许箭头函数,这与LINQ中使用的lambda表达式类似:

Array.prototype.groupBy = function (funcProp) {
    return this.reduce(function (acc, val) {
        (acc[funcProp(val)] = acc[funcProp(val)] || []).push(val);
        return acc;
    }, {});
};

注意:是否要扩展Array的原型取决于您。

大多数浏览器支持的示例:

[{a:1,b:"b"},{a:1,c:"c"},{a:2,d:"d"}].groupBy(function(c){return c.a;})

使用箭头函数(ES6)的示例:

[{a:1,b:"b"},{a:1,c:"c"},{a:2,d:"d"}].groupBy(c=>c.a)

以上两个示例都返回:

{
  "1": [{"a": 1, "b": "b"}, {"a": 1, "c": "c"}],
  "2": [{"a": 2, "d": "d"}]
}

其他回答

_.groupBy([{tipo: 'A' },{tipo: 'A'}, {tipo: 'B'}], 'tipo');
>> Object {A: Array[2], B: Array[1]}

发件人:http://underscorejs.org/#groupBy

这里有一个ES6版本,不会在空成员上中断

function groupBy (arr, key) {
  return (arr || []).reduce((acc, x = {}) => ({
    ...acc,
    [x[key]]: [...acc[x[key]] || [], x]
  }), {})
}

使用ES6:

const groupBy = (items, key) => items.reduce(
  (result, item) => ({
    ...result,
    [item[key]]: [
      ...(result[item[key]] || []),
      item,
    ],
  }), 
  {},
);

我对公认的答案进行了扩展,包括按多个财产分组,然后再加上,使其完全起作用,没有变异。观看演示https://stackblitz.com/edit/typescript-ezydzv

export interface Group {
  key: any;
  items: any[];
}

export interface GroupBy {
  keys: string[];
  thenby?: GroupBy;
}

export const groupBy = (array: any[], grouping: GroupBy): Group[] => {
  const keys = grouping.keys;
  const groups = array.reduce((groups, item) => {
    const group = groups.find(g => keys.every(key => item[key] === g.key[key]));
    const data = Object.getOwnPropertyNames(item)
      .filter(prop => !keys.find(key => key === prop))
      .reduce((o, key) => ({ ...o, [key]: item[key] }), {});
    return group
      ? groups.map(g => (g === group ? { ...g, items: [...g.items, data] } : g))
      : [
          ...groups,
          {
            key: keys.reduce((o, key) => ({ ...o, [key]: item[key] }), {}),
            items: [data]
          }
        ];
  }, []);
  return grouping.thenby ? groups.map(g => ({ ...g, items: groupBy(g.items, grouping.thenby) })) : groups;
};

您可以在数组上使用forEach并构造一组新的项。以下是如何使用FlowType注释实现这一点

// @flow

export class Group<T> {
  tag: number
  items: Array<T>

  constructor() {
    this.items = []
  }
}

const groupBy = (items: Array<T>, map: (T) => number) => {
  const groups = []

  let currentGroup = null

  items.forEach((item) => {
    const tag = map(item)

    if (currentGroup && currentGroup.tag === tag) {
      currentGroup.items.push(item)
    } else {
      const group = new Group<T>()
      group.tag = tag
      group.items.push(item)
      groups.push(group)

      currentGroup = group
    }
  })

  return groups
}

export default groupBy

玩笑测试可以是这样的

// @flow

import groupBy from './groupBy'

test('groupBy', () => {
  const items = [
    { name: 'January', month: 0 },
    { name: 'February', month: 1 },
    { name: 'February 2', month: 1 }
  ]

  const groups = groupBy(items, (item) => {
    return item.month
  })

  expect(groups.length).toBe(2)
  expect(groups[1].items[1].name).toBe('February 2')
})