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?
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。
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 }
]
*/