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基于现有对象数组的若干属性返回新的对象数组

object algorithm arrays javascript

kemotoe · 技术社区 · 6 年前

我正在绞尽脑汁想办法实现我所需要的。我的初始对象数组是70k个对象,每个对象有15个属性。通过过滤和映射,我将数据设置为我认为相关的部分。我需要根据每个对象的4个属性返回一个新的对象数组,它们是重复的 id's 不同的 levels 为每个人 code . 我的过滤数据看起来是这样的。

const arr = [{id: "12345", level: "1", current: "Y", code: "1"}, 
             {id: "12345", level: "0", current: "N", code: "1"},
             {id: "54321", level: "1", current: "N", code: "201"},
             {id: "54321", level: "2", current: "Y", code: "201"}, 
             {id: "54321", level: "3", current: "N", code: "201"},
             {id: "54321", level: "0", current: "Y", code: "401"},
             {id: "54321", level: "1", current: "N", code: "401"},
             {id: "54321", level: "3", current: "N", code: "201"},
             {id: "54321", level: "0", current: "N", code: "301"},
             {id: "121212", level: "0", current: "N", code: "3"},
             {id: "121212", level: "1", current: "N", code: "3"}]

我要做到的是 id 任何人 current 属于 Y 为每个人 代码 ,我需要找到最大的 level 为了这个 代码 . 因此,对于上面的示例,结果输出将是一个对象数组,如下所示:

const result = [{id: "12345", max: "1", code: "1"}, 
                {id: "54321", max: "3", code: "201"},
                {id: "54321", max: "1", code: "401"}]

我甚至不确定我想要的是可能的,还是我想错了。我一上午都在绞尽脑汁想办法。

4 回复 | 直到 6 年前

mickl 6 年前

你可以用 .filter() 和 .map() 根据 Y 然后 .reduce() 要获取最大值:

const arr = [{id: "12345", level: "1", current: "Y", code: "1"}, 
             {id: "12345", level: "0", current: "N", code: "1"},
             {id: "54321", level: "1", current: "N", code: "201"},
             {id: "54321", level: "2", current: "Y", code: "201"}, 
             {id: "54321", level: "3", current: "N", code: "201"},
             {id: "54321", level: "0", current: "Y", code: "401"},
             {id: "54321", level: "1", current: "N", code: "401"},
             {id: "54321", level: "3", current: "N", code: "201"},
             {id: "54321", level: "0", current: "N", code: "301"},
             {id: "121212", level: "0", current: "N", code: "3"},
             {id: "121212", level: "1", current: "N", code: "3"}]

let initialValues = arr.filter(x => x.current === "Y")
		       .map(x => ({ id: x.id, max: x.level, code: x.code }));

let result = arr.reduce((result,current) => {
	var value = result.find(x => x.id === current.id && x.code === current.code);
	if(!value) return result;
	if(value.max < current.level) {
		value.max = current.level;
	}
	return result;
}, initialValues);

console.log(result);

baao 6 年前

你可以简化成一张地图,取它的值

const arr = [{id: "12345", level: "1", current: "Y", code: "1"},
    {id: "12345", level: "0", current: "N", code: "1"},
    {id: "54321", level: "1", current: "N", code: "201"},
    {id: "54321", level: "2", current: "Y", code: "201"},
    {id: "54321", level: "3", current: "N", code: "201"},
    {id: "54321", level: "0", current: "Y", code: "401"},
    {id: "54321", level: "1", current: "N", code: "401"},
    {id: "54321", level: "3", current: "N", code: "201"},
    {id: "54321", level: "0", current: "N", code: "301"},
    {id: "121212", level: "0", current: "N", code: "3"},
    {id: "121212", level: "1", current: "N", code: "3"}];
    
    
const res = arr.reduce((a, b) =>
  b.current === 'Y' ?
  a.set(b.id + b.code, (a.get(b.id + b.code) || {
    level: Number.MIN_SAFE_INTEGER
  }).level < b.level ? b : a.get(b.id + b.code)) :
  a, new Map);
console.log([...res.values()]);

amrender singh 6 年前

你可以简单地实现 Array.reduce() 和 Object destructuring . 构造一个映射,每个键由 id 和 code . 简单使用 Object.values() 在地图上得到想要的结果:

const arr = [{id: "12345", level: "1", current: "Y", code: "1"}, {id: "12345", level: "0", current: "N", code: "1"}, {id: "54321", level: "1", current: "N", code: "201"}, {id: "54321", level: "2", current: "Y", code: "201"}, {id: "54321", level: "3", current: "N", code: "201"}, {id: "54321", level: "0", current: "Y", code: "401"}, {id: "54321", level: "1", current: "N", code: "401"}, {id: "54321", level: "3", current: "N", code: "201"}, {id: "54321", level: "0", current: "N", code: "301"}, {id: "121212", level: "0", current: "N", code: "3"}, {id: "121212", level: "1", current: "N", code: "3"}];
             
const result = Object.values(arr.reduce((a,{level,current, ...props})=>{
    a[props.id+"_"+props.code] = a[props.id+"_"+props.code] || props;
    a[props.id+"_"+props.code].max = Math.max((a[props.id+"_"+props.code].max || 0), level);
  return a;
},{}));
console.log(result);

Pavlo 6 年前

const resultObject = (arr, filter) => arr.reduce((sum, element) => {
  const currentInSum = sum[element.id];
  const filterSatisfied = element.current === filter;
  const isHighestLevel = !currentInSum  || currentInSum.level < element.level;
  if(filterSatisfied && isHighestLevel){
     sum[element.id] = element;
  }
  return sum;
}, {});
const result = Object.values(resultObject(arr, "Y"));

基本上你想使用一个对象,这样你就可以检查元素是否已经添加,如果当前的级别高于已经添加的级别,那么替换它,如果这是第一次出现,那么只需将它添加到对象中。

最后,只需从对象中获取所有值并将其转换为数组,就不必关心对象的id键。

这个算法需要O(n*2)时间来执行,因为必须运行一次arr,然后必须将对象中的所有值转换为一个数组