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如何在JavaScript React中找到两个对象数组之间的所有交集？

javascriptarrays
6

我有两个对象数组。这些数组可以是任意长度。例如，我的第一个数组 "details" 包含三个对象。

let details = [
{a: 1, b: 200, name: "dad"}, 
{a:2, b: 250, name: "cat"}, 
{a:3, b: 212, name: "dog" } 
]

第二个数组由四个对象组成：

let certs = [
{id: 991, b: 250, dn: "qwerty", sign: "GOST"}, 
{id: 950, b: 251, dn: "how", sign: "RSA" }, 
{id: 100, b: 250, dn: "how are", sign: "twofish" }, 
{id: 957, b: 212, dn: "how you", sign: "black" }
]

如何获取由对象某个属性的所有交集组成的数组，例如由'b'？
我的意思是从“certs”数组中获取过滤后的数组，该数组仅包含三个对象而不是四个。
因为在certs数组中索引为1的元素具有属性“b”，其等于251。但是“details”数组中没有具有属性b等于251的对象。
因此，我的过滤后的certs数组应该只包含三个对象。如何实现？
我尝试了lodash方法，但它们都不适用。例如：
_.intersectionBy(certs, details, 'b')

只给了我这个：

0: {id: 991, b: 250, dn: "qwerty", sign: "GOST"}
1: {id: 957, b: 212, dn: "how you", sign: "black"}
length: 2

这个对象在最终数组中不存在：

{id: 100, b: 250, dn: "how are", sign: "twofish" }
4个回答
2
只需两次使用 filter() 方法： 

let details = [
  {a: 1, b: 200, name: "dad"}, 
  {a:2, b: 250, name: "cat"}, 
  {a:3, b: 212, name: "dog" } 
] 

let certs = [
  {id: 991, b: 250, dn: "qwerty", sign: "GOST"}, 
  {id: 950, b: 251, dn: "how", sign: "RSA" }, 
  {id: 100, b: 250, dn: "how are", sign: "twofish" }, 
  {id: 957, b: 212, dn: "how you", sign: "black" }
] 

const result = certs.filter(cert => {
 let arr = details.filter(detail => detail.b === cert.b)
 return !(arr.length === 0)
});

console.log(result)

0
你可以定义一个函数，接受一个表达式并对其进行评估，以便根据你的需要进行过滤。
下面的示例使用 filter 对你自己的 expression 进行过滤。 

let details = [
  {a: 1, b: 200, name: "dad"}, 
  {a:2, b: 250, name: "cat"}, 
  {a:3, b: 212, name: "dog" } 
];
let certs = [
  {id: 991, b: 250, dn: "qwerty", sign: "GOST"}, 
  {id: 950, b: 251, dn: "how", sign: "RSA" }, 
  {id: 100, b: 250, dn: "how are", sign: "twofish" }, 
  {id: 957, b: 212, dn: "how you", sign: "black" }
];

const intersectByExpression = (a, b, expression) => {
  return [...a,...b].filter(i => expression(i)); // <-- evaluates the expression.
  //     ^---------^-- concat items and filters.
}

// All elements where `b` exists and is exactly 250.
console.log(
  intersectByExpression(details, certs, o => o.b === 250)
  //                                    ^--------------^-- Expression to evaluate.
);

// Other samples.
// All elements where id is lower or equals 100.
console.log(
  intersectByExpression(details, certs, o => o.id <= 100)
);

// All elements where id is greater than 950 and b is greater or equals 250.
console.log(
  intersectByExpression(details, certs, o => o.id >= 950 && o.b >= 250)
);

0

有一个方便的库叫做lodash，它有一个名为intersectionBy的函数非常适合解决这个问题。

let details = [
{a: 1, b: 200, name: "dad"}, 
{a:2, b: 250, name: "cat"}, 
{a:3, b: 212, name: "dog" } 
] 
let certs = [
{id: 991, b: 250, dn: "qwerty", sign: "GOST"}, 
{id: 950, b: 251, dn: "how", sign: "RSA" }, 
{id: 100, b: 250, dn: "how are", sign: "twofish" }, 
{id: 957, b: 212, dn: "how you", sign: "black" }
] 
 _.intersectionBy(certs, details, 'b')
不行，lodash 不适用。我已经尝试过 lodash。如果我使用这个：_.intersectionBy(certs, details, 'b')，那么我只会得到一个仅包含两个对象的数组。它不包括具有重复 b 值的对象，只有唯一的对象。 - Dragon14
0
let details = [
{a: 1, b: 200, name: "dad"}, 
{a:2, b: 250, name: "cat"}, 
{a:3, b: 212, name: "dog" } 
];

let certs = [
{id: 991, b: 250, dn: "qwerty", sign: "GOST"}, 
{id: 950, b: 251, dn: "how", sign: "RSA" }, 
{id: 100, b: 250, dn: "how are", sign: "twofish" }, 
{id: 957, b: 212, dn: "how you", sign: "black" }
]

const result = certs.filter(cert => details.find(detail => detail.b === cert.b));

console.log(result);

如果您能提供更多的例子来解释您的问题，那就太好了，:)

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