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？
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;
}));
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"
}
]
}
]
*/
这里有一个使用ES6的讨厌的、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),
{}
);
对于那些询问这是如何工作的人，这里有一个解释：
在这两个=>中，您可以获得免费回报Array.prototype.reduce函数最多包含4个参数。这就是为什么要添加第五个参数，这样我们就可以使用默认值在参数声明级别为组（k）创建一个廉价的变量声明。（是的，这是巫术）如果我们的当前组在上一次迭代中不存在，我们将创建一个新的空数组（（r[k]||（r[k]=[]））。这将计算到最左边的表达式，换句话说，一个现有数组或一个空数组，这就是为什么在该表达式之后会立即推送，因为无论哪种方式都会得到一个数组。当有一个返回时，逗号运算符将丢弃最左边的值，返回该场景中经过调整的前一组。
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[]>);
