我有两个数组,如下所示:
const persons = [
{
id: 1,
name: 'Peter',
job: 'Programmer'
},
{
id: 2,
name: 'Jeff',
job: 'Architect'
},
];
const salaries = [
{
id: 1,
salary: 3000,
departments: ['A', 'B']
},
{
id: 1,
salary: 4000,
departments: ['A', 'C']
},
{
id: 2,
salary: 4000,
departments: ['C', 'D']
}
];
现在,我需要以某种方式将此数组合并为一个,以便每个id仅存在一次。相同的键应该被替换,除了它是一个数组,然后我要它们添加/连接。因此,所需的结果应如下所示:
const result = [
{
id: 1,
name: 'Peter',
job: 'Programmer',
salary: 4000,
departments: ['A', 'B', 'C']
},
{
id: 2,
name: 'Jeff',
job: 'Architect',
salary: 4000,
departments: ['C', 'D']
}
];
我已经尝试过:
// double id's, arrays get replaced
Object.assign({}, persons, salaries)
// loadsh: double id's, arrays get concatenated
_.mergeWith(persons, salaries, (objValue, srcValue) => {
if (_.isArray(objValue)) {
return objValue.concat(srcValue);
}
});
// gives me a map but replaces arrays
new Map(salaries.map(x => [x.id, x])
有谁知道如何做到这一点?
最佳答案
您可以concat数组,然后使用id
和Map将所有具有相同Array.reduce()
的项组合在一起。
若要将具有相同id
的对象组合在一起,请从Map中获取对象。用Object.entries()
迭代新的Array.forEach()
。检查现有值是否为数组,如果没有分配该值。如果它是一个数组,请合并数组,并使用带有数组Set的spread使项目唯一。
要将Map转换回数组,可以传播Map.values()
迭代器。
const persons = [{"id":1,"name":"Peter","job":"Programmer"},{"id":2,"name":"Jeff","job":"Architect"}];
const salaries = [{"id":1,"salary":3000,"departments":["A","B"]},{"id":1,"salary":4000,"departments":["A","C"]},{"id":2,"salary":4000,"departments":["C","D"]}];
const result = [...persons.concat(salaries)
.reduce((r, o) => {
r.has(o.id) || r.set(o.id, {});
const item = r.get(o.id);
Object.entries(o).forEach(([k, v]) =>
item[k] = Array.isArray(item[k]) ?
[...new Set([...item[k], ...v])] : v
);
return r;
}, new Map()).values()];
console.log(result);
关于javascript - JS(ES6):基于id合并数组并连接子数组,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/49649360/