从数组中获取适当的对象[重复]答案

【问题标题】：Get appropriate objects from array [duplicate]从数组中获取适当的对象[重复]
【发布时间】：2020-04-05 18:53:34
【问题描述】：

我创建了一个如下所示的数组：

arrayObject = [{one: "1", two: "2", three: "3", four: "4", five: "5"}, 
               {one: "1", two: "2", three: "3", four: "4", five: "5"}];

如何获取/过滤该数组，使其仅包含“一”、“二”和“五”元素的 arrayObject，如下所示：

arrayObject = [{one: "1", two: "2", five: "5"}, 
               {one: "1", two: "2",  five: "5"}];

最好的方法是什么？

【问题讨论】：

您是要获取只有 3 个键的对象还是将对象更改为只有 3 个键？
您是要根据键基还是值基来获取数组对象？
我想把 arrya 改成 3 个键。
arrayObject.map(({ three, four, ...rest }) => rest)

标签： javascript arrays object

【解决方案1】：

您可以只映射想要的属性，例如解构并从变量中获取新对象。

var array = [{ one: "1", two: "2", three: "3", four: "4", five: "5" }, { one: "1", two: "2", three: "3", four: "4", five: "5" }],
    result = array.map(({ one, two, five }) => ({ one, two, five }));

console.log(result);

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【讨论】：

【解决方案2】：

如果您有一系列想要保留的项目，可以使用Object.fromEntries() 和.map()。通过将要保留的项目数组（这里我称之为keep）映射到键值对数组（即：[[key, value], ...]）。然后，您可以使用Object.fromEntries() 将给定数组转换为如下对象：

const keep = ["one", "two", "five"];
const arrayObject = [{ one: "1", two: "2", three: "3", four: "4", five: "5" }, { one: "1", two: "2", three: "3", four: "4", five: "5" } ];

const res = arrayObject.map(
  obj => Object.fromEntries(keep.map(key => [key, obj[key]])
));
console.log(res);

请注意，Object.fromEntries() 的浏览器支持有限，如果您需要更好的浏览器支持，您可以使用.reduce() 像这样：

const keep = ["one", "two", "five"];
const arrayObject = [{ one: "1", two: "2", three: "3", four: "4", five: "5" }, { one: "1", two: "2", three: "3", four: "4", five: "5" } ];
const res = arrayObject.map(
  obj => keep.map(key => [key, obj[key]]).reduce((acc, [k, v]) => ({...acc, [k]: v}), {})
);
console.log(res);

【讨论】：

【解决方案3】：

您可以像下面那样映射和使用传播

arrayObject = [{one: "1", two: "2", three: "3", four: "4", five: "5"}, 
               {one: "1", two: "2", three: "3", four: "4", five: "5"}];
               
let filtered=arrayObject.map(a=>{

  const {three,four,...rest}=a;
  return rest;
});
console.log(filtered);

【讨论】：

he 知道原因（他的答案在这里被否决，然后被删除）

【解决方案4】：

function extractObjectWithKeys(objects, keys) {
  return objects.reduce((acc, elem) => {
    let newObj = {};
    let filteredKeys = Object.keys(elem).filter(k => keys.includes(k));
    filteredKeys.forEach(key => {
      newObj[key] = elem[key];
    });
    return acc.concat(newObj);
  }, []);
}

const arrayObject = [
  {one: "1", two: "2", three: "3", four: "4", five: "5"}, 
  {one: "1", two: "2", three: "3", four: "4", five: "5"}
];

//Usage:
extractObjectWithKeys(arrayObject, ["one", "two", "five"])

【讨论】：