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

GroupBy one liner,ES2021解决方案

const groupBy = (x,f)=>x.reduce((a,b,i)=>((a[f(b,i,x)]||=[]).push(b),a),{});

TypeScript(类型脚本)

const groupBy = <T>(array: T[], predicate: (value: T, index: number, array: T[]) => string) =>
  array.reduce((acc, value, index, array) => {
    (acc[predicate(value, index, array)] ||= []).push(value);
    return acc;
  }, {} as { [key: string]: T[] });

示例

const groupBy = (x,f)=>x.reduce((a,b,i)=>((a[f(b,i,x)]||=[]).push(b),a),{});
// f -> should must return string/number because it will be use as key in object

// for demo

groupBy([1, 2, 3, 4, 5, 6, 7, 8, 9], v => (v % 2 ? "odd" : "even"));
// { odd: [1, 3, 5, 7, 9], even: [2, 4, 6, 8] };
const colors = [
  "Apricot",
  "Brown",
  "Burgundy",
  "Cerulean",
  "Peach",
  "Pear",
  "Red",
];

groupBy(colors, v => v[0]); // group by colors name first letter
// {
//   A: ["Apricot"],
//   B: ["Brown", "Burgundy"],
//   C: ["Cerulean"],
//   P: ["Peach", "Pear"],
//   R: ["Red"],
// };
groupBy(colors, v => v.length); // group by length of color names
// {
//   3: ["Red"],
//   4: ["Pear"],
//   5: ["Brown", "Peach"],
//   7: ["Apricot"],
//   8: ["Burgundy", "Cerulean"],
// }

const data = [
  { comment: "abc", forItem: 1, inModule: 1 },
  { comment: "pqr", forItem: 1, inModule: 1 },
  { comment: "klm", forItem: 1, inModule: 2 },
  { comment: "xyz", forItem: 1, inModule: 2 },
];

groupBy(data, v => v.inModule); // group by module
// {
//   1: [
//     { comment: "abc", forItem: 1, inModule: 1 },
//     { comment: "pqr", forItem: 1, inModule: 1 },
//   ],
//   2: [
//     { comment: "klm", forItem: 1, inModule: 2 },
//     { comment: "xyz", forItem: 1, inModule: 2 },
//   ],
// }

groupBy(data, x => x.forItem + "-" + x.inModule); // group by module with item
// {
//   "1-1": [
//     { comment: "abc", forItem: 1, inModule: 1 },
//     { comment: "pqr", forItem: 1, inModule: 1 },
//   ],
//   "1-2": [
//     { comment: "klm", forItem: 1, inModule: 2 },
//     { comment: "xyz", forItem: 1, inModule: 2 },
//   ],
// }

按映射分组

const groupByToMap = (x, f) =>
  x.reduce((a, b, i, x) => {
    const k = f(b, i, x);
    a.get(k)?.push(b) ?? a.set(k, [b]);
    return a;
  }, new Map());

TypeScript(类型脚本)

const groupByToMap = <T, Q>(array: T[], predicate: (value: T, index: number, array: T[]) => Q) =>
  array.reduce((map, value, index, array) => {
    const key = predicate(value, index, array);
    map.get(key)?.push(value) ?? map.set(key, [value]);
    return map;
  }, new Map<Q, T[]>());

其他回答

Array.prototype.groupBy = function(keyFunction) {
    var groups = {};
    this.forEach(function(el) {
        var key = keyFunction(el);
        if (key in groups == false) {
            groups[key] = [];
        }
        groups[key].push(el);
    });
    return Object.keys(groups).map(function(key) {
        return {
            key: key,
            values: groups[key]
        };
    });
};

为了补充Scott Sauyet的答案,一些人在评论中询问如何使用他的函数按值1、值2等分组,而不是仅对一个值分组。

只需编辑他的求和函数:

DataGrouper.register("sum", function(item) {
    return _.extend({}, item.key,
        {VALUE1: _.reduce(item.vals, function(memo, node) {
        return memo + Number(node.VALUE1);}, 0)},
        {VALUE2: _.reduce(item.vals, function(memo, node) {
        return memo + Number(node.VALUE2);}, 0)}
    );
});

保持主组(DataGrouper)不变:

var DataGrouper = (function() {
    var has = function(obj, target) {
        return _.any(obj, function(value) {
            return _.isEqual(value, target);
        });
    };

    var keys = function(data, names) {
        return _.reduce(data, function(memo, item) {
            var key = _.pick(item, names);
            if (!has(memo, key)) {
                memo.push(key);
            }
            return memo;
        }, []);
    };

    var group = function(data, names) {
        var stems = keys(data, names);
        return _.map(stems, function(stem) {
            return {
                key: stem,
                vals:_.map(_.where(data, stem), function(item) {
                    return _.omit(item, names);
                })
            };
        });
    };

    group.register = function(name, converter) {
        return group[name] = function(data, names) {
            return _.map(group(data, names), converter);
        };
    };

    return group;
}());

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”))

基于@Ceasar Bautista的原始想法,我修改了代码并使用typescript创建了一个groupBy函数。

static groupBy(data: any[], comparator: (v1: any, v2: any) => boolean, onDublicate: (uniqueRow: any, dublicateRow: any) => void) {
    return data.reduce(function (reducedRows, currentlyReducedRow) {
      let processedRow = reducedRows.find(searchedRow => comparator(searchedRow, currentlyReducedRow));

      if (processedRow) {
        // currentlyReducedRow is a dublicateRow when processedRow is not null.
        onDublicate(processedRow, currentlyReducedRow)
      } else {
        // currentlyReducedRow is unique and must be pushed in the reducedRows collection.
        reducedRows.push(currentlyReducedRow);
      }

      return reducedRows;
    }, []);
  };

此函数接受一个回调(比较器)和一个第二个回调(onDuplicate),该回调比较行并查找副本。

用法示例:

data = [
    { name: 'a', value: 10 },
    { name: 'a', value: 11 },
    { name: 'a', value: 12 },
    { name: 'b', value: 20 },
    { name: 'b', value: 1 }
  ]

  private static demoComparator = (v1: any, v2: any) => {
    return v1['name'] === v2['name'];
  }

  private static demoOnDublicate = (uniqueRow, dublicateRow) => {
    uniqueRow['value'] += dublicateRow['value'];    
  };

使命感

groupBy(data, demoComparator, demoOnDublicate) 

将执行计算值和的分组。

{name: "a", value: 33}
{name: "b", value: 21}

我们可以根据项目的需要创建任意多个回调函数,并根据需要聚合这些值。在一个例子中,我需要合并两个数组,而不是求和数据。

这是一个基于TS的功能,不是性能最好的,但很容易阅读和理解!

function groupBy<T>(array: T[], key: string): Record<string, T[]> {
const groupedObject = {}
for (const item of array) {
  const value = item[key]
    if (groupedObject[value] === undefined) {
  groupedObject[value] = []
  }
  groupedObject[value].push(item)
}
  return groupedObject
}

我们以->

const data = [
{ 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" },
];
console.log(groupBy(data, 'Step'))
{
'Step 1': [
    {
      Phase: 'Phase 1',
      Step: 'Step 1',
      Task: 'Task 1',
      Value: '5'
    },
    {
      Phase: 'Phase 1',
      Step: 'Step 1',
      Task: 'Task 2',
      Value: '10'
    }
  ],
  'Step 2': [
    {
      Phase: 'Phase 1',
      Step: 'Step 2',
      Task: 'Task 1',
      Value: '15'
    },
    {
      Phase: 'Phase 1',
      Step: 'Step 2',
      Task: 'Task 2',
      Value: '20'
    }
  ]
}