JavaScript中的两个数组内连接

Question

以下是我拥有的 2 个数组，

 var g= [ 
        { id: 36, name: 'AAA', goal: 'yes' },
        { id: 40, name: 'BBB', goal: 'yes' },
        { id: 39, name: 'JJJ', goal: 'yes' },
        { id: 27, name: 'CCC', goal: 'yes' }];

    var c= [ 
        { id: 36, color:"purple" },
        { id: 40, color:"purple" },
        { id: 100, color:"pink"} ];

期望的输出（左连接'id'）：

    res = [{ id: 36, name: 'AAA', goal: 'yes' , color:"purple"}, { id: 40, name: 'BBB', goal: 'yes', color:"purple" }]

这是一个合并的现有逻辑，但我正在寻找左连接的逻辑：

    function mergeBy(key, data) {
      return Array.from(data
          .reduce((m, o) => m.set(o[key], { ...m.get(o[key]), ...o }), new Map)
          .values()
      );
    }

Answer 1

这里有一些加入，你可以选择！

function* innerJoin(a, b, key) {
    let idx = new Map(b.map(x => [key(x), x]));
    for (let x of a) {
        let k = key(x);
        if (idx.has(k))
            yield {...x, ...idx.get(k)};
    }
}

function* leftJoin(a, b, key) {
    let idx = new Map(b.map(x => [key(x), x]));
    for (let x of a) {
        let k = key(x);
        yield idx.has(k) ? {...x, ...idx.get(k)} : x;
    }
}

function* rightJoin(a, b, key) {
    let idx = new Map(a.map(x => [key(x), x]));
    for (let x of b) {
        let k = key(x);
        yield idx.has(k) ? {...idx.get(k), ...x} : x;
    }
}


//

var A = [
    {id: 1, a: 'a1'},
    {id: 2, a: 'a2'},
    {id: 7, a: 'a3'},
    {id: 8, a: 'a4'}
];

var B = [
    {id: 1, b: 'b1'},
    {id: 2, b: 'b2'},
    {id: 9, b: 'b3'}
];


console.log('INNER:')
console.log(...innerJoin(A, B, x => x.id))
console.log('LEFT')
console.log(...leftJoin(A, B, x => x.id))
console.log('RIGHT')
console.log(...rightJoin(A, B, x => x.id))

Answer 2

您的预期结果要求的不是左连接。您要求使用id 对g 和c 进行内部连接并合并属性。下面应该这样做

var g= [ 
        { id: 36, name: 'AAA', goal: 'yes' },
        { id: 40, name: 'BBB', goal: 'yes' },
        { id: 39, name: 'JJJ', goal: 'yes' },
        { id: 27, name: 'CCC', goal: 'yes' }];

var c= [ 
        { id: 36, color:"purple" },
        { id: 40, color:"purple" },
        { id: 100, color:"pink"} ];
    
function mergeBy(key, dataL, dataR) {
  const rMap = dataR.reduce((m, o) => m.set(o[key], { ...m.get(o[key]), ...o }), new Map);
  
  return dataL.filter(x => rMap.get(x[key])).map(x => ({...x, ...rMap.get(x[key]) }));
}

console.log(mergeBy("id",g, c))

Answer 3

您可以获得不常用的键并过滤合并的结果。

const
    mergeCommon = (a, b, key) => {
        const aByKey = a.reduce((m, o) => m.set(o[key], o), new Map);

        return b.reduce((r, o) => {
            if (aByKey.has(o[key])) r.push({ ... aByKey.get(o[key]), ...o});
            return r;
        }, []);
    },
    g = [{ id: 36, name: 'AAA', goal: 'yes' , 'random': 27 }, { id: 40, name: 'BBB', goal: 'yes' }, { id: 39, name: 'JJJ', goal: 'yes' }, { id: 27, name: 'CCC', goal: 'yes' , lag: "23.3343" }],
    c = [{ id: 36, name: 'AAA', goal: 'yes', color:"purple" }, { id: 40, name: 'BBB', circle: 'yes', color:"purple" }, { id: 100, name: 'JJJ', circle: 'yes'}],
    result = mergeCommon(g, c, 'id');

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 4

您可以使用Maps 连接两个数组。三种连接Left Join、Right Join、Inner Join见代码sn-p。

所有三个连接都需要 O(N) 时间。

const g = [
  { id: 36, name: 'AAA', goal: 'yes', random: 27 },
  { id: 40, name: 'BBB', goal: 'yes' },
  { id: 39, name: 'JJJ', goal: 'yes' },
  { id: 27, name: 'CCC', goal: 'yes', lag: '23.3343' },
];

const c = [
  { id: 36, name: 'AAA', goal: 'yes', color: 'purple' },
  { id: 40, name: 'BBB', circle: 'yes', color: 'purple' },
  { id: 100, name: 'JJJ', circle: 'yes' },
];

const gMap = new Map(g.map(o => [o.id, o]));
const cMap = new Map(c.map(o => [o.id, o]));

const leftJoin = g.reduce(
  (a, o) => (cMap.has(o.id) ? [...a, { ...o, ...cMap.get(o.id) }] : [...a, o]),
  []
);

const rightJoin = c.reduce(
  (a, o) => (gMap.has(o.id) ? [...a, { ...o, ...gMap.get(o.id) }] : [...a, o]),
  []
);

const innerJoin = g.reduce(
  (a, o) => (cMap.has(o.id) ? [...a, { ...o, ...cMap.get(o.id) }] : a),
  []
);

console.log("LEFT JOIN\n", leftJoin)
console.log("RIGHT JOIN\n", rightJoin)
console.log("INNER JOIN\n", innerJoin)

Answer 5

var g = [ 
    { id: 36, name: 'AAA', goal: 'yes' , 'random':27},
    { id: 40, name: 'BBB', goal: 'yes' },
    { id: 39, name: 'JJJ', goal: 'yes' },
    { id: 27, name: 'CCC', goal: 'yes' , lag: "23.3343"}
];

var c = [ 
    { id: 36, name: 'AAA', goal: 'yes', color:"purple" },
    { id: 40, name: 'BBB', circle: 'yes', color:"purple" },
    { id: 100, name: 'JJJ', circle: 'yes'} 
];


const combinedArr = (arr1, arr2) => {
//Filter the arr2 and find the only matching elements from the first array
    const filteredArr = arr2.filter(({id}) => arr1.some(({id: arr1Id}) => arr1Id === id));
    //Loop through the filtered array and fetch the matching item from first and add obj from filtered array
    return filteredArr.map(obj => {
        return {
            ...arr1.find(({id}) => id === obj.id),
            ...obj
        }
    })
}    
console.log(combinedArr(g, c));

.as-console-wrapper { 
  max-height: 100% !important;
}

Answer 6

你去，一个班轮

 var g= [ 
        { id: 36, name: 'AAA', goal: 'yes' },
        { id: 40, name: 'BBB', goal: 'yes' },
        { id: 39, name: 'JJJ', goal: 'yes' },
        { id: 27, name: 'CCC', goal: 'yes' }];

    var c= [ 
        { id: 36, color:"purple" },
        { id: 40, color:"purple" },
        { id: 100, color:"pink"} ];



const result = g.map(i => !!c.find(e => e.id === i.id) && ({...i, ...c.find(e => e.id === i.id)})).filter(Boolean);
console.log(result);