如何在javascript中找到两个对象数组之间的所有交集？

Question

I have two arrays of objects. The Arrays can be of any length. For example my first array "details" consists of three objects,

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

second array consists of four objects:

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" }
] 

How to obtain an array that consists of all intersections by some property of object, for example by 'b' ?

I mean I want to obtain filtered array from "certs" array, that will contain only three objects not four.

Because element at index 1 in certs array has property "b" which equals to 251. But "details" array doesn't have object with property b equals to 251.

So my filtered certs array should consists of only three objects. How to implement this?

I tried lodash methods, but none of them fit. For example:

_.intersectionBy(certs, details, 'b')

gives me only this:

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

This object doesnt exist in final array:

{id: 100, b: 250, dn: "how are", sign: "twofish" }

Answer 1

There is a handy library called lodash and it has a function called intersectionBy which really fits this problem.

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')

Answer 2

You could define a function that will accept an expression and evaluate it in order to filter according to your need.

Below example uses filter against your own expression to filter the set of items.

 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 => ob === 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 && ob >= 250) ); 

Answer 3

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);

It would be great if you explain your problem with more examples, :)

Answer 4

Simply use the filter() method twice:

 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)