javascript - 对对象数组进行分组的最有效方法

Question

对数组中的对象进行分组的最有效方法是什么？

例如，给定这个对象数组：

[ 
    { 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" }
]

我在表格中显示此信息。我想对不同的方法进行分组，但我想对这些值求和。

我将 Underscore.js 用于它的 groupby 函数，这很有帮助，但并不能完全解决问题，因为我不希望它们“拆分”而是“合并”，更像 SQLgroup by方法。

我正在寻找的将能够总计特定值（如果需要）。

所以如果我做了 groupby Phase，我想收到：

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

如果我做了 groupy Phase/ Step，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 }
]

是否有对此有用的脚本，或者我应该坚持使用 Underscore.js，然后循环遍历生成的对象以自己进行总计？

Answer 1

如果你想避免使用外部库，你可以简洁地实现一个香草版本，groupBy()如下所示：

var groupBy = function(xs, key) {
  return xs.reduce(function(rv, x) {
    (rv[x[key]] = rv[x[key]] || []).push(x);
    return rv;
  }, {});
};

console.log(groupBy(['one', 'two', 'three'], 'length'));

// => {3: ["one", "two"], 5: ["three"]}

Answer 2

使用 ES6 Map 对象：

/**
 * @description
 * Takes an Array<V>, and a grouping function,
 * and returns a Map of the array grouped by the grouping function.
 *
 * @param list An array of type V.
 * @param keyGetter A Function that takes the the Array type V as an input, and returns a value of type K.
 *                  K is generally intended to be a property key of V.
 *
 * @returns Map of the array grouped by the grouping function.
 */
//export function groupBy<K, V>(list: Array<V>, keyGetter: (input: V) => K): Map<K, Array<V>> {
//    const map = new Map<K, Array<V>>();
function groupBy(list, keyGetter) {
    const map = new Map();
    list.forEach((item) => {
         const key = keyGetter(item);
         const collection = map.get(key);
         if (!collection) {
             map.set(key, [item]);
         } else {
             collection.push(item);
         }
    });
    return map;
}


// example usage

const pets = [
    {type:"Dog", name:"Spot"},
    {type:"Cat", name:"Tiger"},
    {type:"Dog", name:"Rover"}, 
    {type:"Cat", name:"Leo"}
];
    
const grouped = groupBy(pets, pet => pet.type);
    
console.log(grouped.get("Dog")); // -> [{type:"Dog", name:"Spot"}, {type:"Dog", name:"Rover"}]
console.log(grouped.get("Cat")); // -> [{type:"Cat", name:"Tiger"}, {type:"Cat", name:"Leo"}]

const odd = Symbol();
const even = Symbol();
const numbers = [1,2,3,4,5,6,7];

const oddEven = groupBy(numbers, x => (x % 2 === 1 ? odd : even));
    
console.log(oddEven.get(odd)); // -> [1,3,5,7]
console.log(oddEven.get(even)); // -> [2,4,6]

关于地图： https ://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Map

Answer 3

使用 ES6：

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

Answer 4

你可以Map从array.reduce().

const groupedMap = initialArray.reduce(
    (entryMap, e) => entryMap.set(e.id, [...entryMap.get(e.id)||[], e]),
    new Map()
);

与其他解决方案相比，这具有一些优势：

• 它不需要任何库（不像 eg _.groupBy()）
• 你得到的是一个 JavaScriptMap而不是一个对象（例如由返回的_.groupBy()）。这有很多好处，包括：
  • 它会记住第一次添加项目的顺序，
  • 键可以是任何类型，而不仅仅是字符串。
• AMap是比数组数组更有用的结果。但是，如果您确实想要一个数组数组，则可以调用Array.from(groupedMap.entries())（对于一[key, group array]对数组）或Array.from(groupedMap.values())（对于一个简单的数组数组）。
• 它非常灵活；通常，您计划接下来使用此地图执行的任何操作都可以作为缩减的一部分直接完成。

作为最后一点的示例，假设我有一个对象数组，我想通过 id 进行（浅）合并，如下所示：

const objsToMerge = [{id: 1, name: "Steve"}, {id: 2, name: "Alice"}, {id: 1, age: 20}];
// The following variable should be created automatically
const mergedArray = [{id: 1, name: "Steve", age: 20}, {id: 2, name: "Alice"}]

为此，我通常会先按 id 分组，然后合并每个结果数组。相反，您可以直接在以下位置进行合并reduce()：

const mergedArray = Array.from(
    objsToMerge.reduce(
        (entryMap, e) => entryMap.set(e.id, {...entryMap.get(e.id)||{}, ...e}),
        new Map()
    ).values()
);

Answer 5

我会检查lodash groupBy它似乎完全符合您的要求。它也非常轻巧且非常简单。

小提琴示例：https ://jsfiddle.net/r7szvt5k/

假设您的数组名称是arr带有 lodash 的 groupBy 只是：

import groupBy from 'lodash/groupBy';
// if you still use require:
// const groupBy = require('lodash/groupBy');

const a = groupBy(arr, function(n) {
  return n.Phase;
});
// a is your array grouped by Phase attribute

Answer 6

尽管linq的答案很有趣，但它也很重。我的方法有些不同：

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;
}());

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

您可以在 JSBin 上看到它的实际效果。

我没有在 Underscore 中看到任何has功能，尽管我可能会错过它。它与 非常相似_.contains，但用于比较_.isEqual而不是===用于比较。除此之外，其余的都是特定于问题的，尽管试图是通用的。

现在DataGrouper.sum(data, ["Phase"])返回

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

并DataGrouper.sum(data, ["Phase", "Step"])返回

[
    {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}
]

---

但这sum只是这里的一个潜在功能。您可以随意注册其他人：

DataGrouper.register("max", function(item) {
    return _.extend({}, item.key, {Max: _.reduce(item.vals, function(memo, node) {
        return Math.max(memo, Number(node.Value));
    }, Number.NEGATIVE_INFINITY)});
});

现在DataGrouper.max(data, ["Phase", "Step"])会回来

[
    {Phase: "Phase 1", Step: "Step 1", Max: 10},
    {Phase: "Phase 1", Step: "Step 2", Max: 20},
    {Phase: "Phase 2", Step: "Step 1", Max: 30},
    {Phase: "Phase 2", Step: "Step 2", Max: 40}
]

或者如果您注册了这个：

DataGrouper.register("tasks", function(item) {
    return _.extend({}, item.key, {Tasks: _.map(item.vals, function(item) {
      return item.Task + " (" + item.Value + ")";
    }).join(", ")});
});

然后打电话DataGrouper.tasks(data, ["Phase", "Step"])会给你

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

DataGrouper本身就是一个函数。您可以使用您的数据和要分组的属性列表来调用它。它返回一个数组，其元素是具有两个属性的对象：key是分组属性的集合，vals是包含不在键中的其余属性的对象数组。例如，DataGrouper(data, ["Phase", "Step"])将产生：

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

DataGrouper.register接受一个函数并创建一个新函数，该函数接受初始数据和要分组的属性。然后，这个新函数采用上述输出格式，并依次针对每个函数运行您的函数，返回一个新数组。生成的函数DataGrouper根据您提供的名称存储为 的属性，如果您只需要本地引用，也会返回该函数。

好吧，这是很多解释。代码相当简单，我希望！

Answer 7

这可能更容易完成linq.js，它旨在成为 JavaScript 中 LINQ 的真正实现（DEMO）：

var linq = Enumerable.From(data);
var result =
    linq.GroupBy(function(x){ return x.Phase; })
        .Select(function(x){
          return {
            Phase: x.Key(),
            Value: x.Sum(function(y){ return y.Value|0; })
          };
        }).ToArray();

结果：

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

或者，更简单地使用基于字符串的选择器（DEMO）：

linq.GroupBy("$.Phase", "",
    "k,e => { Phase:k, Value:e.Sum('$.Value|0') }").ToArray();

Answer 8

GroupBy one-liner，一种ES2021解决方案

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

打字稿

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

例子

const groupBy = (x, f) => x.reduce((a, b) => ((a[f(b)] ||= []).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 },
//   ],
// }

Answer 9

MDN在他们的文档中有这个例子。Array.reduce()

// Grouping objects by a property
// https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/Reduce#Grouping_objects_by_a_property#Grouping_objects_by_a_property

var people = [
  { name: 'Alice', age: 21 },
  { name: 'Max', age: 20 },
  { name: 'Jane', age: 20 }
];

function groupBy(objectArray, property) {
  return objectArray.reduce(function (acc, obj) {
    var key = obj[property];
    if (!acc[key]) {
      acc[key] = [];
    }
    acc[key].push(obj);
    return acc;
  }, {});
}

var groupedPeople = groupBy(people, 'age');
// groupedPeople is:
// { 
//   20: [
//     { name: 'Max', age: 20 }, 
//     { name: 'Jane', age: 20 }
//   ], 
//   21: [{ name: 'Alice', age: 21 }] 
// }

Answer 10

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

来自：http ://underscorejs.org/#groupBy

Answer 11

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]
        };
    });
};

Answer 12

你可以使用Alasql JavaScript 库来做到这一点：

var data = [ { Phase: "Phase 1", Step: "Step 1", Task: "Task 1", Value: "5" },
             { Phase: "Phase 1", Step: "Step 1", Task: "Task 2", Value: "10" }];

var res = alasql('SELECT Phase, Step, SUM(CAST([Value] AS INT)) AS [Value] \
                  FROM ? GROUP BY Phase, Step',[data]);

在 jsFiddle试试这个例子。

顺便说一句：在大型数组（100000 条记录及更多）上，Alasql 比 Linq 更快。请参阅jsPref 的测试。

评论：

• 这里我把Value放在方括号里，因为VALUE是SQL中的关键字
• 我必须使用 CAST() 函数将字符串值转换为数字类型。

Answer 13

一种较新的方法，其中包含一个用于分组的对象和另外两个函数来创建一个键并获取一个具有所需分组项的对象和另一个用于增加值的键。

const
    groupBy = (array, groups, valueKey) => {
        const
            getKey = o => groups.map(k => o[k]).join('|'),
            getObject = o => Object.fromEntries([...groups.map(k => [k, o[k]]), [valueKey, 0]]);

        groups = [].concat(groups);

        return Object.values(array.reduce((r, o) => {
            (r[getKey(o)] ??= getObject(o))[valueKey] += +o[valueKey];
            return r;
        }, {}));
    },
    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" }, { 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" }];

console.log(groupBy(data, 'Phase', 'Value'));
console.log(groupBy(data, ['Phase', 'Step'], 'Value'));

.as-console-wrapper { max-height: 100% !important; top: 0; }

旧方法：

虽然这个问题有一些答案，而且答案看起来有点复杂，但我建议使用 vanilla Javascript 进行 group-by 和嵌套（如有必要）Map。

function groupBy(array, groups, valueKey) {
    var map = new Map;
    groups = [].concat(groups);
    return array.reduce((r, o) => {
        groups.reduce((m, k, i, { length }) => {
            var child;
            if (m.has(o[k])) return m.get(o[k]);
            if (i + 1 === length) {
                child = Object
                    .assign(...groups.map(k => ({ [k]: o[k] })), { [valueKey]: 0 });
                r.push(child);
            } else {
                child = new Map;
            }
            m.set(o[k], child);
            return child;
        }, map)[valueKey] += +o[valueKey];
        return r;
    }, [])
};

var 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" }, { 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" }];

console.log(groupBy(data, 'Phase', 'Value'));
console.log(groupBy(data, ['Phase', 'Step'], 'Value'));

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 14

有点晚了，但也许有人喜欢这个。

ES6：

const users = [{
    name: "Jim",
    color: "blue"
  },
  {
    name: "Sam",
    color: "blue"
  },
  {
    name: "Eddie",
    color: "green"
  },
  {
    name: "Robert",
    color: "green"
  },
];
const groupBy = (arr, key) => {
  const initialValue = {};
  return arr.reduce((acc, cval) => {
    const myAttribute = cval[key];
    acc[myAttribute] = [...(acc[myAttribute] || []), cval]
    return acc;
  }, initialValue);
};

const res = groupBy(users, "color");
console.log("group by:", res);

Answer 15

检查答案 - 只是浅分组。理解减少非常好。问题还提供了额外聚合计算的问题。

这是一个 REAL GROUP BY 对象数组，由一些字段组成，1）计算的键名和 2）通过提供所需键的列表并将其唯一值转换为根键（如 SQL GROUP）来实现分组级联的完整解决方案BY 确实如此。

const inputArray = [ 
    { 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" }
];

var outObject = inputArray.reduce(function(a, e) {
  // GROUP BY estimated key (estKey), well, may be a just plain key
  // a -- Accumulator result object
  // e -- sequentally checked Element, the Element that is tested just at this itaration

  // new grouping name may be calculated, but must be based on real value of real field
  let estKey = (e['Phase']); 

  (a[estKey] ? a[estKey] : (a[estKey] = null || [])).push(e);
  return a;
}, {});

console.log(outObject);

玩estKey——您可以按多个字段进行分组，添加额外的聚合、计算或其他处理。

您也可以递归地对数据进行分组。例如最初按 分组Phase，然后按Step字段等。另外吹掉脂肪休息数据。

const inputArray = [
{ 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" }
  ];

/**
 * Small helper to get SHALLOW copy of obj WITHOUT prop
 */
const rmProp = (obj, prop) => ( (({[prop]:_, ...rest})=>rest)(obj) )

/**
 * Group Array by key. Root keys of a resulting array is value
 * of specified key.
 *
 * @param      {Array}   src     The source array
 * @param      {String}  key     The by key to group by
 * @return     {Object}          Object with grouped objects as values
 */
const grpBy = (src, key) => src.reduce((a, e) => (
  (a[e[key]] = a[e[key]] || []).push(rmProp(e, key)),  a
), {});

/**
 * Collapse array of object if it consists of only object with single value.
 * Replace it by the rest value.
 */
const blowObj = obj => Array.isArray(obj) && obj.length === 1 && Object.values(obj[0]).length === 1 ? Object.values(obj[0])[0] : obj;

/**
 * Recursive grouping with list of keys. `keyList` may be an array
 * of key names or comma separated list of key names whom UNIQUE values will
 * becomes the keys of the resulting object.
 */
const grpByReal = function (src, keyList) {
  const [key, ...rest] = Array.isArray(keyList) ? keyList : String(keyList).trim().split(/\s*,\s*/);
  const res = key ? grpBy(src, key) : [...src];
  if (rest.length) {
for (const k in res) {
  res[k] = grpByReal(res[k], rest)
}
  } else {
for (const k in res) {
  res[k] = blowObj(res[k])
}
  }
  return res;
}

console.log( JSON.stringify( grpByReal(inputArray, 'Phase, Step, Task'), null, 2 ) );

Answer 16

这是一个使用 ES6 的令人讨厌的、难以阅读的解决方案：

export default (arr, key) => 
  arr.reduce(
    (r, v, _, __, k = v[key]) => ((r[k] || (r[k] = [])).push(v), r),
    {}
  );

对于那些询问这甚至如何工作的人，这里有一个解释：

• 在这两者中=>，您都有免费的return

• 该Array.prototype.reduce函数最多需要 4 个参数。这就是为什么要添加第五个参数的原因，因此我们可以使用默认值在参数声明级别为组 (k) 进行廉价的变量声明。（是的，这是巫术）

• 如果我们当前的组在上一次迭代中不存在，我们创建一个新的空数组((r[k] || (r[k] = []))这将评估最左边的表达式，换句话说，一个现有的数组或一个空数组，这就是为什么push在那个表达式之后有一个立即，因为无论哪种方式，您都会得到一个数组。

• 当有 areturn时，逗号,运算符将丢弃最左边的值，返回此场景中调整后的前一个组。

一个更容易理解的版本是：

export default (array, key) => 
  array.reduce((previous, currentItem) => {
    const group = currentItem[key];
    if (!previous[group]) previous[group] = [];
    previous[group].push(currentItem);
    return previous;
  }, {});

编辑：

TS版本：

const groupBy = <T, K extends keyof any>(list: T[], getKey: (item: T) => K) =>
  list.reduce((previous, currentItem) => {
    const group = getKey(currentItem);
    if (!previous[group]) previous[group] = [];
    previous[group].push(currentItem);
    return previous;
  }, {} as Record<K, T[]>);

Answer 17

我想建议我的方法。首先，单独分组和聚合。让我们声明原型的“分组依据”功能。它需要另一个函数来为要分组的每个数组元素生成“哈希”字符串。

Array.prototype.groupBy = function(hash){
  var _hash = hash ? hash : function(o){return o;};

  var _map = {};
  var put = function(map, key, value){
    if (!map[_hash(key)]) {
        map[_hash(key)] = {};
        map[_hash(key)].group = [];
        map[_hash(key)].key = key;

    }
    map[_hash(key)].group.push(value); 
  }

  this.map(function(obj){
    put(_map, obj, obj);
  });

  return Object.keys(_map).map(function(key){
    return {key: _map[key].key, group: _map[key].group};
  });
}

分组完成后，您可以根据需要聚合数据

data.groupBy(function(o){return JSON.stringify({a: o.Phase, b: o.Step});})
    /* aggreagating */
    .map(function(el){ 
         var sum = el.group.reduce(
           function(l,c){
             return l + parseInt(c.Value);
           },
           0
         );
         el.key.Value = sum; 
         return el.key;
    });

共同点是有效的。我已经在 chrome 控制台中测试了这段代码。并随时改进和发现错误；）

Answer 18

没有突变：

const groupBy = (xs, key) => xs.reduce((acc, x) => Object.assign({}, acc, {
  [x[key]]: (acc[x[key]] || []).concat(x)
}), {})

console.log(groupBy(['one', 'two', 'three'], 'length'));
// => {3: ["one", "two"], 5: ["three"]}

Answer 19

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

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)

上面的两个例子都返回：

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

Answer 20

groupByArray(xs, key) {
    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);
        }
        else {
            rv.push({
                key: v,
                values: [x]
            });
        }
        return rv;
    }, []);
}

这个输出数组。

Answer 21

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:'sedan'}]

通过做这个： const categories = [...new Set(cars.map((car) => car.cat))]

你会得到这个： ['sedan','sport']

说明： 1. 首先，我们通过传递一个数组来创建一个新的 Set。因为 Set 只允许唯一值，所以所有重复项都将被删除。

2. 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/设置 传播运算符文档： https ://developer.mozilla.org/en-US/docs/Web/JavaScript /参考/运算符/Spread_syntax

Answer 22

根据以前的答案

const groupBy = (prop) => (xs) =>
  xs.reduce((rv, x) =>
    Object.assign(rv, {[x[prop]]: [...(rv[x[prop]] || []), x]}), {});

如果您的环境支持，使用对象扩展语法查看会更好一些。

const groupBy = (prop) => (xs) =>
  xs.reduce((acc, x) => ({
    ...acc,
    [ x[ prop ] ]: [...( acc[ x[ prop ] ] || []), x],
  }), {});

在这里，我们的 reducer 采用部分形成的返回值（从一个空对象开始），并返回一个由前一个返回值的展开成员组成的对象，以及一个新成员，其键是根据当前 iteree 的值计算得出的prop并且其值是该道具的所有值以及当前值的列表。

Answer 23

我不认为给出的答案是对问题的回应，我认为以下内容应该回答第一部分：

const arr = [ 
{ 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" }
]

const groupBy = (key) => arr.reduce((total, currentValue) => {
  const newTotal = total;
  if (
    total.length &&
    total[total.length - 1][key] === currentValue[key]
  )
    newTotal[total.length - 1] = {
      ...total[total.length - 1],
      ...currentValue,
      Value: parseInt(total[total.length - 1].Value) + parseInt(currentValue.Value),
    };
  else newTotal[total.length] = currentValue;
  return newTotal;
}, []);

console.log(groupBy('Phase'));

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

Answer 24

让我们在重用已经编写的代码（即下划线）的同时充分回答原始问题。如果您将其超过 100 个功能结合起来，您可以使用 Underscore 做更多的事情。以下解决方案证明了这一点。

第 1 步：通过属性的任意组合对数组中的对象进行分组。_.groupBy这使用了接受返回对象组的函数的事实。它还使用_.chain、_.pick、_.values和_.join。_.value请注意，_.value此处并非严格要求，因为链式值在用作属性名称时会自动展开。我将它包括在内是为了防止混淆，以防有人试图在不发生自动展开的上下文中编写类似的代码。

// Given an object, return a string naming the group it belongs to.
function category(obj) {
    return _.chain(obj).pick(propertyNames).values().join(' ').value();
}

// Perform the grouping.
const intermediate = _.groupBy(arrayOfObjects, category);

给定arrayOfObjects原始问题中的 并设置propertyNames为['Phase', 'Step']，intermediate将获得以下值：

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

第 2 步：将每个组缩减为单个平面对象，并将结果以数组的形式返回。除了我们之前看到的函数之外，下面的代码还使用了_.pluck、_.first、_.pick、和。在这种情况下保证返回一个对象，因为不会产生空组。在这种情况下是必要的。_.extend_.reduce_.map_.first_.groupBy_.value

// Sum two numbers, even if they are contained in strings.
const addNumeric = (a, b) => +a + +b;

// Given a `group` of objects, return a flat object with their common
// properties and the sum of the property with name `aggregateProperty`.
function summarize(group) {
    const valuesToSum = _.pluck(group, aggregateProperty);
    return _.chain(group).first().pick(propertyNames).extend({
        [aggregateProperty]: _.reduce(valuesToSum, addNumeric)
    }).value();
}

// Get an array with all the computed aggregates.
const result = _.map(intermediate, summarize);

鉴于intermediate我们之前获得的 并设置aggregateProperty为Value，我们得到result了提问者想要的：

[
    { 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 }
]

我们可以将所有这些放在一个以和作为参数的函数arrayOfObjects中。请注意，它实际上也可以是带有字符串键的普通对象，因为两者都可以接受。为此，我已重命名为.propertyNamesaggregatePropertyarrayOfObjects_.groupByarrayOfObjectscollection

function aggregate(collection, propertyNames, aggregateProperty) {
    function category(obj) {
        return _.chain(obj).pick(propertyNames).values().join(' ');
    }
    const addNumeric = (a, b) => +a + +b;
    function summarize(group) {
        const valuesToSum = _.pluck(group, aggregateProperty);
        return _.chain(group).first().pick(propertyNames).extend({
            [aggregateProperty]: _.reduce(valuesToSum, addNumeric)
        }).value();
    }
    return _.chain(collection).groupBy(category).map(summarize).value();
}

aggregate(arrayOfObjects, ['Phase', 'Step'], 'Value')现在将result再次给我们同样的。

我们可以更进一步，使调用者能够计算每个组中值的任何统计数据。我们可以这样做并且还允许调用者将任意属性添加到每个组的摘要中。我们可以在缩短代码的同时完成所有这些工作。我们用参数替换aggregateProperty参数iteratee并将其直接传递给_.reduce：

function aggregate(collection, propertyNames, iteratee) {
    function category(obj) {
        return _.chain(obj).pick(propertyNames).values().join(' ');
    }
    function summarize(group) {
        return _.chain(group).first().pick(propertyNames)
            .extend(_.reduce(group, iteratee)).value();
    }
    return _.chain(collection).groupBy(category).map(summarize).value();
}

实际上，我们将一些责任转移给了调用者；她必须提供一个iteratee可以传递给 的_.reduce，以便调用_.reduce将产生一个具有她想要添加的聚合属性的对象。例如，我们通过result以下表达式获得与之前相同的结果：

aggregate(arrayOfObjects, ['Phase', 'Step'], (memo, value) => ({
    Value: +memo.Value + +value.Value
}));

对于稍微复杂一点的例子iteratee，假设我们想要计算每个组的最大值 Value而不是总和，并且我们想要添加一个Tasks属性来列出Task组中出现的所有值。这是我们可以做到这一点的一种方法，使用aggregate上面的最后一个版本（和_.union）：

aggregate(arrayOfObjects, ['Phase', 'Step'], (memo, value) => ({
    Value: Math.max(memo.Value, value.Value),
    Tasks: _.union(memo.Tasks || [memo.Task], [value.Task])
}));

我们得到以下结果：

[
    { Phase: "Phase 1", Step: "Step 1", Value: 10, Tasks: [ "Task 1", "Task 2" ] },
    { Phase: "Phase 1", Step: "Step 2", Value: 20, Tasks: [ "Task 1", "Task 2" ] },
    { Phase: "Phase 2", Step: "Step 1", Value: 30, Tasks: [ "Task 1", "Task 2" ] },
    { Phase: "Phase 2", Step: "Step 2", Value: 40, Tasks: [ "Task 1", "Task 2" ] }
]

感谢@much2learn，他还发布了一个可以处理任意归约函数的答案。我又写了几个 SO 答案，展示了如何通过组合多个下划线函数来实现复杂的事情：

• https://stackoverflow.com/a/64938636/1166087
• https://stackoverflow.com/a/64094738/1166087
• https://stackoverflow.com/a/63625129/1166087
• https://stackoverflow.com/a/63088916/1166087

Answer 25

groupBy可以通过特定键或给定分组函数对数组进行分组的函数。打字。

groupBy = <T, K extends keyof T>(array: T[], groupOn: K | ((i: T) => string)): Record<string, T[]> => {
  const groupFn = typeof groupOn === 'function' ? groupOn : (o: T) => o[groupOn];

  return Object.fromEntries(
    array.reduce((acc, obj) => {
      const groupKey = groupFn(obj);
      return acc.set(groupKey, [...(acc.get(groupKey) || []), obj]);
    }, new Map())
  ) as Record<string, T[]>;
};

Answer 26

Array.prototype.groupBy = function (groupingKeyFn) {
    if (typeof groupingKeyFn !== 'function') {
        throw new Error("groupBy take a function as only parameter");
    }
    return this.reduce((result, item) => {
        let key = groupingKeyFn(item);
        if (!result[key])
            result[key] = [];
        result[key].push(item);
        return result;
    }, {});
}

var a = [
	{type: "video", name: "a"},
  {type: "image", name: "b"},
  {type: "video", name: "c"},
  {type: "blog", name: "d"},
  {type: "video", name: "e"},
]
console.log(a.groupBy((item) => item.type));

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

Answer 27

我会检查declarative-js groupBy它似乎完全符合您的要求。也是：

• 非常高性能（性能基准）
• 用打字稿写的，所以所有的打字都包括在内。
• 不强制使用第 3 方类似数组的对象。

import { Reducers } from 'declarative-js';
import groupBy = Reducers.groupBy;
import Map = Reducers.Map;

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" },
    { 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" }
];

data.reduce(groupBy(element=> element.Step), Map());
data.reduce(groupBy('Step'), Map());

Answer 28

让我们生成一个通用Array.prototype.groupBy()工具。只是为了多样化，让我们使用 ES6 奇特的扩展运算符在递归方法上进行一些 Haskellesque 模式匹配。另外，让我们Array.prototype.groupBy()接受一个回调，该回调将项目 ( e)、索引 ( i) 和应用的数组 ( a) 作为参数。

Array.prototype.groupBy = function(cb){
                            return function iterate([x,...xs], i = 0, r = [[],[]]){
                                     cb(x,i,[x,...xs]) ? (r[0].push(x), r)
                                                       : (r[1].push(x), r);
                                     return xs.length ? iterate(xs, ++i, r) : r;
                                   }(this);
                          };

var arr = [0,1,2,3,4,5,6,7,8,9],
    res = arr.groupBy(e => e < 5);
console.log(res);

Answer 29

Ceasar 的回答很好，但仅适用于数组内元素的内部属性（字符串的长度）。

这个实现更像：这个链接

const groupBy = function (arr, f) {
    return arr.reduce((out, val) => {
        let by = typeof f === 'function' ? '' + f(val) : val[f];
        (out[by] = out[by] || []).push(val);
        return out;
    }, {});
};

希望这可以帮助...

Answer 30

只是为了补充 Scott Sauyet 的回答，有些人在评论中询问如何使用他的函数对 value1、value2 等进行分组，而不是只对一个值进行分组。

只需编辑他的 sum 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)}
    );
});

保持主要（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;
}());

Answer 31

来自@mortb、@jmarceli 的回答和这篇文章，

我利用作为PRIMITIVE VALUE多列 group byJSON.stringify()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);

Answer 32

支持功能的基于ES6 reduce的版本iteratee。

iteratee如果未提供该功能，则按预期工作：

const data = [{id: 1, score: 2},{id: 1, score: 3},{id: 2, score: 2},{id: 2, score: 4}]

const group = (arr, k) => arr.reduce((r, c) => (r[c[k]] = [...r[c[k]] || [], c], r), {});

const groupBy = (arr, k, fn = () => true) => 
  arr.reduce((r, c) => (fn(c[k]) ? r[c[k]] = [...r[c[k]] || [], c] : null, r), {});

console.log(group(data, 'id'))     // grouping via `reduce`
console.log(groupBy(data, 'id'))   // same result if `fn` is omitted
console.log(groupBy(data, 'score', x => x > 2 )) // group with the iteratee

在OP问题的背景下：

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" }, { 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" } ]

const groupBy = (arr, k) => arr.reduce((r, c) => (r[c[k]] = [...r[c[k]] || [], c], r), {});
const groupWith = (arr, k, fn = () => true) => 
  arr.reduce((r, c) => (fn(c[k]) ? r[c[k]] = [...r[c[k]] || [], c] : null, r), {});

console.log(groupBy(data, 'Phase'))
console.log(groupWith(data, 'Value', x => x > 30 ))  // group by `Value` > 30

另一个ES6版本，它反转分组并使用valuesaskeys和keysas grouped values：

const data = [{A: "1"}, {B: "10"}, {C: "10"}]

const groupKeys = arr => 
  arr.reduce((r,c) => (Object.keys(c).map(x => r[c[x]] = [...r[c[x]] || [], x]),r),{});

console.log(groupKeys(data))

注意：为简洁起见，函数以简短形式（一行）发布并仅与想法相关。您可以扩展它们并添加额外的错误检查等。

Answer 33

发布是因为即使这个问题已有 7 年历史，我还没有看到满足原始标准的答案：

我不希望它们“分裂”而是“合并”，更像是按方法分组的 SQL。

我最初来到这篇文章是因为我想找到一种方法来减少对象数组（例如，当您从 csv 读取时创建的数据结构）并通过给定索引聚合以生成相同的数据结构。我正在寻找的返回值是另一个对象数组，而不是像我在这里看到的那样的嵌套对象或映射。

下面的函数接受一个数据集（对象数组）、一个索引列表（数组）和一个reducer函数，并将reducer函数应用于索引的结果作为对象数组返回。

function agg(data, indices, reducer) {

  // helper to create unique index as an array
  function getUniqueIndexHash(row, indices) {
    return indices.reduce((acc, curr) => acc + row[curr], "");
  }

  // reduce data to single object, whose values will be each of the new rows
  // structure is an object whose values are arrays
  // [{}] -> {{}}
  // no operation performed, simply grouping
  let groupedObj = data.reduce((acc, curr) => {
    let currIndex = getUniqueIndexHash(curr, indices);

    // if key does not exist, create array with current row
    if (!Object.keys(acc).includes(currIndex)) {
      acc = {...acc, [currIndex]: [curr]}
    // otherwise, extend the array at currIndex
    } else {
      acc = {...acc, [currIndex]: acc[currIndex].concat(curr)};
    }

    return acc;
  }, {})

  // reduce the array into a single object by applying the reducer
  let reduced = Object.values(groupedObj).map(arr => {
    // for each sub-array, reduce into single object using the reducer function
    let reduceValues = arr.reduce(reducer, {});

    // reducer returns simply the aggregates - add in the indices here
    // each of the objects in "arr" has the same indices, so we take the first
    let indexObj = indices.reduce((acc, curr) => {
      acc = {...acc, [curr]: arr[0][curr]};
      return acc;
    }, {});

    reduceValues = {...indexObj, ...reduceValues};


    return reduceValues;
  });


  return reduced;
}

我将创建一个返回 count(*) 和 sum(Value) 的 reducer：

reducer = (acc, curr) => {
  acc.count = 1 + (acc.count || 0);
  acc.value = +curr.Value + (acc.value|| 0);
  return acc;
}

最后，将 agg 函数与我们的 reducer 应用到原始数据集会产生一个对象数组，其中应用了适当的聚合：

agg(tasks, ["Phase"], reducer);
// yields:
Array(2) [
  0: Object {Phase: "Phase 1", count: 4, value: 50}
  1: Object {Phase: "Phase 2", count: 4, value: 130}
]

agg(tasks, ["Phase", "Step"], reducer);
// yields:
Array(4) [
  0: Object {Phase: "Phase 1", Step: "Step 1", count: 2, value: 15}
  1: Object {Phase: "Phase 1", Step: "Step 2", count: 2, value: 35}
  2: Object {Phase: "Phase 2", Step: "Step 1", count: 2, value: 55}
  3: Object {Phase: "Phase 2", Step: "Step 2", count: 2, value: 75}
]

Answer 34

let groupbyKeys = function(arr, ...keys) {
  let keysFieldName = keys.join();
  return arr.map(ele => {
    let keysField = {};
    keysField[keysFieldName] = keys.reduce((keyValue, key) => {
      return keyValue + ele[key]
    }, "");
    return Object.assign({}, ele, keysField);
  }).reduce((groups, ele) => {
    (groups[ele[keysFieldName]] = groups[ele[keysFieldName]] || [])
      .push([ele].map(e => {
        if (keys.length > 1) {
          delete e[keysFieldName];
        }
        return e;
    })[0]);
    return groups;
  }, {});
};

console.log(groupbyKeys(array, 'Phase'));
console.log(groupbyKeys(array, 'Phase', 'Step'));
console.log(groupbyKeys(array, 'Phase', 'Step', 'Task'));

Answer 35

这是一个不会破坏空成员的 ES6 版本

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

Answer 36

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;
      }, []);
    };

样本：

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')));

Answer 37

我已经扩展了已接受的答案，包括按多个属性进行分组，添加 thenby 并使其纯粹具有功能而没有突变。在https://stackblitz.com/edit/typescript-ezydzv上查看演示

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;
};

Answer 38

通常我使用带有预构建方法的Lodash JavaScript 实用程序库。groupBy()它非常易于使用，请在此处查看更多详细信息。

Answer 39

function groupBy(array, groupBy){
        return array.reduce((acc,curr,index,array) => {
           var  idx = curr[groupBy]; 
              if(!acc[idx]){
                    acc[idx] = array.filter(item => item[groupBy] === idx)
              } 
            return  acc; 

        },{})
    }

// call
groupBy(items,'Step')

Answer 40

我已经改进了答案。该函数采用组字段数组并返回分组对象，其键也是组字段的对象。

function(xs, groupFields) {
        groupFields = [].concat(groupFields);
        return xs.reduce(function(rv, x) {
            let groupKey = groupFields.reduce((keyObject, field) => {
                keyObject[field] = x[field];
                return keyObject;
            }, {});
            (rv[JSON.stringify(groupKey)] = rv[JSON.stringify(groupKey)] || []).push(x);
            return rv;
        }, {});
    }



let x = [
{
    "id":1,
    "multimedia":false,
    "language":["tr"]
},
{
    "id":2,
    "multimedia":false,
    "language":["fr"]
},
{
    "id":3,
    "multimedia":true,
    "language":["tr"]
},
{
    "id":4,
    "multimedia":false,
    "language":[]
},
{
    "id":5,
    "multimedia":false,
    "language":["tr"]
},
{
    "id":6,
    "multimedia":false,
    "language":["tr"]
},
{
    "id":7,
    "multimedia":false,
    "language":["tr","fr"]
}
]

groupBy(x, ['multimedia','language'])

//{
//{"multimedia":false,"language":["tr"]}: Array(3), 
//{"multimedia":false,"language":["fr"]}: Array(1), 
//{"multimedia":true,"language":["tr"]}: Array(1), 
//{"multimedia":false,"language":[]}: Array(1), 
//{"multimedia":false,"language":["tr","fr"]}: Array(1)
//}

Answer 41

按照 Joseph Nields 的回答，在https://github.com/padcom/array-prototype-functions#arrayprototypegroupbyfieldormapper中有一个用于对对象进行分组的 polyfill 。因此，与其一遍又一遍地写，你可能想要使用已经可用的东西。

Answer 42

在我的特定用例中，我需要按属性分组，然后删除分组属性。

无论如何，该属性仅出于分组目的而添加到记录中，并且对于向用户呈现没有意义。

    group (arr, key) {

        let prop;

        return arr.reduce(function(rv, x) {
            prop = x[key];
            delete x[key];
            (rv[prop] = (rv[prop] || [])).push(x);
            return rv;
        }, {});

    },

感谢@caesar-bautista 在最佳答案中的启动功能。

Answer 43

groupBy功能即将进入 JavaScript -目前处于第 3 阶段。

它可以在转译器配置中启用，我认为与 reduce 或第三方库（如 lodash 等）相比，解决方案要优雅得多。

const products = [
  { name: 'apples', category: 'fruits' },
  { name: 'oranges', category: 'fruits' },
  { name: 'potatoes', category: 'vegetables' }
];

const groupByCategory = products.groupBy(product => {
  return product.category;
});

console.log(groupByCategory);
// {
//   'fruits': [
//     { name: 'apples', category: 'fruits' }, 
//     { name: 'oranges', category: 'fruits' },
//   ],
//   'vegetables': [
//     { name: 'potatoes', category: 'vegetables' }
//   ]
// }

Answer 44

我从 underscore.js fiddler那里借用了这个方法

window.helpers=(function (){
    var lookupIterator = function(value) {
        if (value == null){
            return function(value) {
                return value;
            };
        }
        if (typeof value === 'function'){
                return value;
        }
        return function(obj) {
            return obj[value];
        };
    },
    each = function(obj, iterator, context) {
        var breaker = {};
        if (obj == null) return obj;
        if (Array.prototype.forEach && obj.forEach === Array.prototype.forEach) {
            obj.forEach(iterator, context);
        } else if (obj.length === +obj.length) {
            for (var i = 0, length = obj.length; i < length; i++) {
                if (iterator.call(context, obj[i], i, obj) === breaker) return;
            }
        } else {
            var keys = []
            for (var key in obj) if (Object.prototype.hasOwnProperty.call(obj, key)) keys.push(key)
            for (var i = 0, length = keys.length; i < length; i++) {
                if (iterator.call(context, obj[keys[i]], keys[i], obj) === breaker) return;
            }
        }
        return obj;
    },
    // An internal function used for aggregate "group by" operations.
    group = function(behavior) {
        return function(obj, iterator, context) {
            var result = {};
            iterator = lookupIterator(iterator);
            each(obj, function(value, index) {
                var key = iterator.call(context, value, index, obj);
                behavior(result, key, value);
            });
            return result;
        };
    };

    return {
      groupBy : group(function(result, key, value) {
        Object.prototype.hasOwnProperty.call(result, key) ? result[key].push(value) :              result[key] = [value];
        })
    };
})();

var arr=[{a:1,b:2},{a:1,b:3},{a:1,b:1},{a:1,b:2},{a:1,b:3}];
 console.dir(helpers.groupBy(arr,"b"));
 console.dir(helpers.groupBy(arr,function (el){
   return el.b>2;
 }));

Answer 45

var arr = [ 
    { Phase: "Phase 1", `enter code here`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" }
];

创建并清空对象。循环arr并添加 use Phase作为obj的唯一键。在循环arr时不断更新obj中的键总数。

const obj = {};
arr.forEach((item) => {
  obj[item.Phase] = obj[item.Phase] ? obj[item.Phase] + 
  parseInt(item.Value) : parseInt(item.Value);
});

结果将如下所示：

{ "Phase 1": 50, "Phase 2": 130 }

遍历obj以形成和resultArr。

const resultArr = [];
for (item in obj) {
  resultArr.push({ Phase: item, Value: obj[item] });
}
console.log(resultArr);

Answer 46

data = [{id:1, name:'BMW'}, {id:2, name:'AN'}, {id:3, name:'BMW'}, {id:1, name:'NNN'}]
key = 'id'//try by id or name
data.reduce((previous, current)=>{
    previous[current[key]] && previous[current[key]].length != 0 ? previous[current[key]].push(current) : previous[current[key]] = new Array(current)
    return previous;
}, {})

Answer 47

基于@Ceasar Bautista的最初想法，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）来比较行并找到重复项和聚合重复项的第二个回调（onDublicate）。

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'];    
  };

打电话

groupBy(data, demoComparator, demoOnDublicate) 

将执行一组计算值的总和。

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

我们可以根据项目的需要创建尽可能多的回调函数，并根据需要聚合这些值。例如，在一种情况下，我需要合并两个数组而不是对数据求和。

Answer 48

您可以forEach在数组上使用并构造一组新的项目。这是使用FlowType注释的方法

// @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

开玩笑的测试可以像

// @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')
})

Answer 49

下面的函数允许对任意字段进行分组（和求和值 - OP需要什么）。在解决方案中，我们定义cmp了根据分组比较两个对象的函数fields。在let w=...我们创建子集对象 x 字段的副本。在y[sumBy]=+y[sumBy]+(+x[sumBy])我们使用 '+' 将字符串转换为数字。

function groupBy(data, fields, sumBy='Value') {
  let r=[], cmp= (x,y) => fields.reduce((a,b)=> a && x[b]==y[b], true);
  data.forEach(x=> {
    let y=r.find(z=>cmp(x,z));
    let w= [...fields,sumBy].reduce((a,b) => (a[b]=x[b],a), {})
    y ? y[sumBy]=+y[sumBy]+(+x[sumBy]) : r.push(w);
  });
  return r;
}

const d = [ 
    { 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" }
];



function groupBy(data, fields, sumBy='Value') {
  let r=[], cmp= (x,y) => fields.reduce((a,b)=> a && x[b]==y[b], true);
  data.forEach(x=> {
    let y=r.find(z=>cmp(x,z));
    let w= [...fields,sumBy].reduce((a,b) => (a[b]=x[b],a), {})
    y ? y[sumBy]=+y[sumBy]+(+x[sumBy]) : r.push(w);
  });
  return r;
}


// TEST
let p=(t,o) => console.log(t, JSON.stringify(o));
console.log('GROUP BY:');

p('Phase', groupBy(d,['Phase']) );
p('Step', groupBy(d,['Step']) );
p('Phase-Step', groupBy(d,['Phase', 'Step']) );
p('Phase-Task', groupBy(d,['Phase', 'Task']) );
p('Step-Task', groupBy(d,['Step', 'Task']) );
p('Phase-Step-Task', groupBy(d,['Phase','Step', 'Task']) );

Answer 50

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"
      }
    ]
  }
]

*/

Answer 51

如果你使用 lodash 库就很简单

let temp = []
  _.map(yourCollectionData, (row) => {
    let index = _.findIndex(temp, { 'Phase': row.Phase })
    if (index > -1) {
      temp[index].Value += row.Value 
    } else {
      temp.push(row)
    }
  })

Answer 52

var newArr = data.reduce((acc, cur) => {
    const existType = acc.find(a => a.Phase === cur.Phase);
    if (existType) {
        existType.Value += +cur.Value;
        return acc;
    }

    acc.push({
        Phase: cur.Phase,
        Value: +cur.Value
    });
    return acc;
}, []);

Answer 53

使用 ES6 的简单解决方案：

该方法有一个返回模型并允许比较 n 个属性。

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 }
    ]
*/

Answer 54

如果您需要进行多分组：

    const populate = (entireObj, keys, item) => {
    let keysClone = [...keys],
        currentKey = keysClone.shift();

    if (keysClone.length > 0) {
        entireObj[item[currentKey]] = entireObj[item[currentKey]] || {}
        populate(entireObj[item[currentKey]], keysClone, item);
    } else {
        (entireObj[item[currentKey]] = entireObj[item[currentKey]] || []).push(item);
    }
}

export const groupBy = (list, key) => {
    return list.reduce(function (rv, x) {

        if (typeof key === 'string') (rv[x[key]] = rv[x[key]] || []).push(x);

        if (typeof key === 'object' && key.length) populate(rv, key, x);

        return rv;

    }, {});
}

const myPets = [
    {name: 'yaya', type: 'cat', color: 'gray'},
    {name: 'bingbang', type: 'cat', color: 'sliver'},
    {name: 'junior-bingbang', type: 'cat', color: 'sliver'},
    {name: 'jindou', type: 'cat', color: 'golden'},
    {name: 'dahuzi', type: 'dog', color: 'brown'},
];

// run 
groupBy(myPets, ['type', 'color']));

// you will get object like: 

const afterGroupBy = {
    "cat": {
        "gray": [
            {
                "name": "yaya",
                "type": "cat",
                "color": "gray"
            }
        ],
        "sliver": [
            {
                "name": "bingbang",
                "type": "cat",
                "color": "sliver"
            },
            {
                "name": "junior-bingbang",
                "type": "cat",
                "color": "sliver"
            }
        ],
        "golden": [
            {
                "name": "jindou",
                "type": "cat",
                "color": "golden"
            }
        ]
    },
    "dog": {
        "brown": [
            {
                "name": "dahuzi",
                "type": "dog",
                "color": "brown"
            }
        ]
    }
};

Answer 55

/**
 * array group by 
 * @category array
 * @function arrayGroupBy
 * @returns  {object} {"fieldName":[{...}],...}
 * @static
 * @author hht
 * @param {string}} key group key
 * @param {array} data array
 *
 * @example example 01 
 * --------------------------------------------------------------------------
 * import { arrayGroupBy } from "@xx/utils";
 * const array =  [
 *  {
 *    type: 'assets',
 *    name: 'zhangsan',
 *    age: '33',
 *  },
 *  {
 *    type: 'config',
 *    name: 'a',
 *    age: '13',
 *  },
 *  {
 *    type: 'run',
 *    name: 'lisi',
 *    age: '3',
 *  },
 *  {
 *    type: 'xx',
 *    name: 'timo',
 *    age: '4',
 *  },
 *];
 * arrayGroupBy(array,'type',);
 *
 * result:{
 *    assets: [{ age: '33', name: 'zhangsan', type: 'assets' }],
 *    config: [{ age: '13', name: 'a', type: 'config' }],
 *    run: [{ age: '3', name: 'lisi', type: 'run' }],
 *    xx: [{ age: '4', name: 'timo', type: 'xx' }],
 *  };
 *
 * @example example 02 null
 * --------------------------------------------------------------------------
 * const array = null;
 * arrayGroupBy(array,"type");
 *
 * result:{}
 *
 * @example example 03 key undefind
 * --------------------------------------------------------------------------
 * const array =  [
 *  {
 *    type: 'assets',
 *    name: 'zhangsan',
 *    age: '33',
 *  },
 *  {
 *    type: 'config',
 *    name: 'a',
 *    age: '13',
 *  },
 *  {
 *    type: 'run',
 *    name: 'lisi',
 *    age: '3',
 *  },
 *  {
 *    type: 'xx',
 *    name: 'timo',
 *    age: '4',
 *  },
 *];
 * arrayGroupBy(array,"xx");
 *
 * {}
 *
 */
  const arrayGroupBy = (data, key) => {
  if (!data || !Array.isArray(data)) return {};
  const groupObj = {};
  data.forEach((item) => {
    if (!item[key]) return;
    const fieldName = item[key];
    if (!groupObj[fieldName]) {
      groupObj[fieldName] = [item];
      return;
    }
    groupObj[fieldName].push(item);
  });
  return groupObj;
};

const array = [
    {
      type: 'assets',
      name: 'zhangsan',
      age: '33',
    },
    {
      type: 'config',
      name: 'a',
      age: '13',
    },
    {
      type: 'run',
      name: 'lisi',
      age: '3',
    },
    {
      type: 'run',
      name: 'wangmazi',
      age: '3',
    },
    {
      type: 'xx',
      name: 'timo',
      age: '4',
    },
  ];
console.dir(arrayGroupBy(array, 'type'))

<p>


describe('arrayGroupBy match', () => {
  const array = [
    {
      type: 'assets',
      name: 'zhangsan',
      age: '33',
    },
    {
      type: 'config',
      name: 'a',
      age: '13',
    },
    {
      type: 'run',
      name: 'lisi',
      age: '3',
    },
    {
      type: 'xx',
      name: 'timo',
      age: '4',
    },
  ];

  test('arrayGroupBy  ...', () => {
    const result = {
      assets: [{ age: '33', name: 'zhangsan', type: 'assets' }],
      config: [{ age: '13', name: 'a', type: 'config' }],
      run: [{ age: '3', name: 'lisi', type: 'run' }],
      xx: [{ age: '4', name: 'timo', type: 'xx' }],
    };

    expect(arrayGroupBy(array, 'type')).toEqual(result);
  });

  test('arrayGroupBy not match..', () => {
    // result
    expect(arrayGroupBy(array, 'xx')).toEqual({});
  });

  test('arrayGroupBy null', () => {
    let array = null;
    expect(arrayGroupBy(array, 'type')).toEqual({});
  });

  test('arrayGroupBy undefined', () => {
    let array = undefined;
    expect(arrayGroupBy(array, 'type')).toEqual({});
  });

  test('arrayGroupBy empty', () => {
    let array = [];
    expect(arrayGroupBy(array, 'type')).toEqual({});
  });
});

</p>