假设我有以下内容:
var array =
[
{"name":"Joe", "age":17},
{"name":"Bob", "age":17},
{"name":"Carl", "age": 35}
]
获得所有不同年龄的数组的最佳方法是什么，这样我就得到了一个结果数组:
[17, 35]
是否有一些方法，我可以选择结构数据或更好的方法，这样我就不必遍历每个数组检查“年龄”的值，并检查另一个数组是否存在，如果没有添加它?
如果有某种方法可以让我不用迭代就能得到不同的年龄……
目前效率低下的方式，我想改进…如果它的意思不是“数组”是一个对象的数组，而是一个对象的“映射”与一些唯一的键(即。"1,2,3")也可以。我只是在寻找最高效的方式。
以下是我目前的做法，但对我来说，迭代似乎只是为了提高效率，即使它确实有效……
var distinct = []
for (var i = 0; i < array.length; i++)
if (array[i].age not in distinct)
distinct.push(array[i].age)
如果你想从一个已知唯一对象属性的数组中过滤掉重复值，你可以使用下面的代码片段:
let arr = [
{ "name": "Joe", "age": 17 },
{ "name": "Bob", "age": 17 },
{ "name": "Carl", "age": 35 },
{ "name": "Carl", "age": 35 }
];
let uniqueValues = [...arr.reduce((map, val) => {
if (!map.has(val.name)) {
map.set(val.name, val);
}
return map;
}, new Map()).values()]
目前正在使用typescript库以orm方式查询js对象。你可以从下面的链接下载。这个答案解释了如何使用下面的库来解决。
https://www.npmjs.com/package/@krishnadaspc/jsonquery?activeTab=readme
var ageArray =
[
{"name":"Joe", "age":17},
{"name":"Bob", "age":17},
{"name":"Carl", "age": 35}
]
const ageArrayObj = new JSONQuery(ageArray)
console.log(ageArrayObj.distinct("age").get()) // outputs: [ { name: 'Bob', age: 17 }, { name: 'Carl', age: 35 } ]
console.log(ageArrayObj.distinct("age").fetchOnly("age")) // outputs: [ 17, 35 ]
Runkit live链接:https://runkit.com/pckrishnadas88/639b5b3f8ef36f0008b17512
从一组键中获取不同值的集合的方法。
您可以从这里获取给定的代码，并仅为所需的键添加映射，以获得唯一对象值的数组。
const
listOfTags = [{ id: 1, label: "Hello", color: "red", sorting: 0 }, { id: 2, label: "World", color: "green", sorting: 1 }, { id: 3, label: "Hello", color: "blue", sorting: 4 }, { id: 4, label: "Sunshine", color: "yellow", sorting: 5 }, { id: 5, label: "Hello", color: "red", sorting: 6 }],
keys = ['label', 'color'],
filtered = listOfTags.filter(
(s => o =>
(k => !s.has(k) && s.add(k))
(keys.map(k => o[k]).join('|'))
)(new Set)
)
result = filtered.map(o => Object.fromEntries(keys.map(k => [k, o[k]])));
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }
清洁解决方案
export abstract class Serializable<T> {
equalTo(t: Serializable<T>): boolean {
return this.hashCode() === t.hashCode();
}
hashCode(): string {
throw new Error('Not Implemented');
}
}
export interface UserFields {
firstName: string;
lastName: string;
}
export class User extends Serializable<User> {
constructor(private readonly fields: UserFields) {
super();
}
override hashCode(): string {
return `${this.fields.firstName},${this.fields.lastName}`;
}
}
const list: User[] = [
new User({ firstName: 'first', lastName: 'user' }),
new User({ firstName: 'first', lastName: 'user' }),
new User({ firstName: 'second', lastName: 'user' }),
new User({ firstName: 'second', lastName: 'user' }),
new User({ firstName: 'third', lastName: 'user' }),
new User({ firstName: 'third', lastName: 'user' }),
];
/**
* Let's create an map
*/
const userHashMap = new Map<string, User>();
/**
* We are adding each user into the map using user's hashCode value
*/
list.forEach((user) => userHashMap.set(user.hashCode(), user));
/**
* Then getting the list of users from the map,
*/
const uniqueUsers = [...userHashMap.values()];
/**
* Let's print and see we did right?
*/
console.log(uniqueUsers.map((e) => e.hashCode()));
Const数组= [
{" id ": " 93 ", "名称":" CVAM_NGP_KW "},
{" id ": " 94 ", "名称":" CVAM_NGP_PB "},
{" id ": " 93 ", "名称":" CVAM_NGP_KW "},
{" id ": " 94 ", "名称":" CVAM_NGP_PB "}
］
函数uniq(数组，字段){
返回数组中。Reduce((累加器，电流)=> {
如果(! accumulator.includes(当前(领域))){
accumulator.push(当前(领域))
｝
返回蓄电池;
}, []
）
｝
Const id = uniq(数组，'id');
console.log (ids)
/ *输出
(“93”,“94”)
* /
