根据对象数组中的相似值对对象进行分组

Question

我有一个对象数组，我想将具有相似分支和环境的对象分组/合并，并同时连接它们的池。

const data = [
{
branch: "master",
environment: "dev",
pool: "6g",
service: "amex",
},
{
branch: "master",
environment: "dev",
pool: "6g",
service: "amex",
},
{
branch: "feature/rest",
environment: "dev",
pool: "2g",
service: "amex",
},
{
branch: "master",
environment: "dev",
pool: "4g",
service: "amex",
},
{
branch: "hotfix/23",
environment: "test",
pool: "9g",
service: "amex",
},
{
branch: "hotfix/23",
environment: "test",
pool: "1g",
service: "amex",
},
];

我想要的结果在下面的格式删除重复的对象，以及我试图减少它，但作为数组减少返回一个对象作为结果和其他对象被省略从响应什么数据结构或方式我可以用来实现结果?

const result = [
{
branch: "master",
environment: "dev",
pool: "6g, 4g",
service: "amex",
},
{
branch: "feature/rest",
environment: "dev",
pool: "2g",
service: "amex",
},
{
branch: "hotfix/23",
environment: "test",
pool: "9g,1g",
service: "amex",
},
];

Answer 1

const data = [{
branch: "master",
environment: "dev",
pool: "6g",
service: "amex",
},
{
branch: "feature/rest",
environment: "dev",
pool: "2g",
service: "amex",
},
{
branch: "master",
environment: "dev",
pool: "4g",
service: "amex",
},
{
branch: "hotfix/23",
environment: "test",
pool: "9g",
service: "amex",
},
{
branch: "hotfix/23",
environment: "test",
pool: "1g",
service: "amex",
},
];
var cacheMix = {};
for (var i = 0; i < data.length; i++) {
var item = data[i];
var compositeKey = item.environment + "~" + item.branch;
if (cacheMix[compositeKey]) {
cacheMix[compositeKey].pools[item.pool] = 1;
} else {
var pools = {}; pools[item.pool] = 1; //to avoid dublicate pools 
cacheMix[compositeKey] = {
branch: item.branch,
environment: item.environment,
service: item.service,
pools: pools 
}
}
}

var result = [];

for (var key in cacheMix) {
var item = cacheMix[key];
result.push({
branch: item.branch,
environment: item.environment,
service: item.service,
pool: Object.keys(item.pools).join(", ")
});
}
console.log(result);

Answer 2

用它们创建一个字典并填写值。

const data=[{branch:"master",environment:"dev",pool:"6g",service:"amex"},{branch:"feature/rest",environment:"dev",pool:"2g",service:"amex"},{branch:"master",environment:"dev",pool:"4g",service:"amex"},{branch:"hotfix/23",environment:"test",pool:"9g",service:"amex"},{branch:"hotfix/23",environment:"test",pool:"1g",service:"amex"},]

let x = {};
data.forEach(y => x[y.branch + "|" + y.environment] = y);
var res = Object.values(x).map(y => Object.assign({}, y)).map(y => 
{
y.pool = data.filter(d => d.branch == y.branch && d.environment == y.environment).map(x => x.pool).join(",");
return y;
})

如果您不关心不可变性(数据中的原始对象更改)，那么删除Object.assign map，它将产生相同的结果

Answer 3

我建议使用Array.reduce()，使用从branch和environment创建的键创建数据的映射。

一旦有了map对象，就可以使用Object.values()返回一个包含期望结果的数组。

const data = [ { branch: "master", environment: "dev", pool: "6g", service: "amex", }, { branch: "feature/rest", environment: "dev", pool: "2g", service: "amex", }, { branch: "master", environment: "dev", pool: "4g", service: "amex", }, { branch: "hotfix/23", environment: "test", pool: "9g", service: "amex", }, { branch: "hotfix/23", environment: "test", pool: "1g", service: "amex", }, ];
const result = Object.values(data.reduce((acc, { branch, environment, pool, service }) => {
// Our grouping key...
const key = `${branch}-${environment}`;
acc[key] = acc[key] || { branch, environment, pool: '', service };
acc[key].pool += ((acc[key].pool ? ", " : "" ) + pool);
return acc;
}, {}))
console.log('Result:', result)

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Answer 4

这是一种方法:

const data=[{branch:"master",environment:"dev",pool:"6g",service:"amex"},{branch:"feature/rest",environment:"dev",pool:"2g",service:"amex"},{branch:"master",environment:"dev",pool:"4g",service:"amex"},{branch:"hotfix/23",environment:"test",pool:"9g",service:"amex"},{branch:"hotfix/23",environment:"test",pool:"1g",service:"amex"},]
const res = data.reduce((acc, cur) =>
{
let isMatch = false
acc.forEach((el, idx) => {
if(el.branch === cur.branch && el.environment === cur.environment) {
if(acc[idx].service !== cur.service) {
acc[idx].service += `, ${cur.service}`
}
if(acc[idx].pool !== cur.pool) {
acc[idx].pool += `, ${cur.pool}`
}
isMatch = true
}
})
if(!isMatch) {
acc.push(cur)    
}

return acc
}, [])
console.log(res)

Answer 5

data.reduce((prev, curr) => {
const container = prev.find(
el => (el.branch === curr.branch) && (el.environment === curr.environment)
);
if (container) {
container.pool = container.pool + `,${curr.pool}`
return prev;
} else {
return prev.concat({...curr}) 
}
}, [])

，但我建议创建一个对象，keys等于分支名称。