删除多维数组中所有对象数组中不常见的对象

Question

我有一个数组对象的数组这是一个多维数组我需要删除所有数组中不常见的对象

这是多维数组:

const array111 = [
[{
name: 'test1',
parameters: {
employee_number: 1
}
},
{
name: 'test2',
parameters: {
employee_number: 2
}
}, {
name: 'test3',
parameters: {
employee_number: 4
}
},
{
name: 'test4',
parameters: {
employee_number: 5
}
},
{
name: 'test5',
parameters: {
employee_number: 6
}
}
],
[{
name: 'test1',
parameters: {
employee_number: 1
}
},
{
name: 'test4',
parameters: {
employee_number: 5
}
},
{
name: 'test5',
parameters: {
employee_number: 6
}
}
],
[{
name: 'test1',
parameters: {
employee_number: 1
}
},
{
name: 'test4',
parameters: {
employee_number: 4
}
},
{
name: 'test4',
parameters: {
employee_number: 5
}
},
{
name: 'test5',
parameters: {
employee_number: 6
}
}
],
[{
name: 'test1',
parameters: {
employee_number: 1
}
},
{
name: 'test4',
parameters: {
employee_number: 5
}
},
{
name: 'test5',
parameters: {
employee_number: 6
}
}
],
]
resultss = Object.values(array111.reduce((r, o) => {
key = []
key.push([...o.map(({
parameters
}) => parameters.employee_number)])
r.push(key)
return r;
}, []));
data = resultss.map(JSON.stringify).reverse().filter((e, i, a) => {
return a.indexOf(e, i) === 1;
}).reverse().map(JSON.parse)
filteredRes = array111.map((element) => {
element = element.filter(({
parameters
}) => data[0][0].some((val) => val === parameters.employee_number));
return element;
})
console.log(filteredRes)

在上面的数组employee_number "1,5,6"在所有对象数组和employee_number"2,4"应该被删除，这在所有数组中并不常见

我尝试的解决方案是给出正确的结果，但由于它是较长的方式，我想要较小的解决方案。

const array111 = [
[{
name: 'test1',
parameters: {
employee_number: 1
}
},
{
name: 'test4',
parameters: {
employee_number: 5
}
},
{
name: 'test5',
parameters: {
employee_number: 6
}
}
],
[{
name: 'test1',
parameters: {
employee_number: 1
}
},
{
name: 'test4',
parameters: {
employee_number: 5
}
},
{
name: 'test5',
parameters: {
employee_number: 6
}
}
],
[{
name: 'test1',
parameters: {
employee_number: 1
}
},
{
name: 'test4',
parameters: {
employee_number: 5
}
},
{
name: 'test5',
parameters: {
employee_number: 6
}
}
],
[{
name: 'test1',
parameters: {
employee_number: 1
}
},
{
name: 'test4',
parameters: {
employee_number: 5
}
},
{
name: 'test5',
parameters: {
employee_number: 6
}
}
],
]

以上是O/p

Answer 1

您可以使用Set和Map轻松实现结果

使用Set和Map

const array111 = [
[
{
name: "test1",
parameters: {
employee_number: 1,
},
},
{
name: "test2",
parameters: {
employee_number: 2,
},
},
{
name: "test3",
parameters: {
employee_number: 4,
},
},
{
name: "test4",
parameters: {
employee_number: 5,
},
},
{
name: "test5",
parameters: {
employee_number: 6,
},
},
],
[
{
name: "test1",
parameters: {
employee_number: 1,
},
},
{
name: "test4",
parameters: {
employee_number: 5,
},
},
{
name: "test5",
parameters: {
employee_number: 6,
},
},
],
[
{
name: "test1",
parameters: {
employee_number: 1,
},
},
{
name: "test4",
parameters: {
employee_number: 5,
},
},
{
name: "test5",
parameters: {
employee_number: 6,
},
},
],
[
{
name: "test1",
parameters: {
employee_number: 1,
},
},
{
name: "test4",
parameters: {
employee_number: 4,
},
},
{
name: "test4",
parameters: {
employee_number: 5,
},
},
{
name: "test5",
parameters: {
employee_number: 6,
},
},
],
];
const uniqueNamesFrequency = array111
.map((arr) => Array.from(arr.reduce((acc, curr) => acc.add(curr.parameters.employee_number),new Set())))
.reduce((acc, curr, i) => {
curr.forEach((id) => acc.set(id, (acc.get(id) ?? 0) + 1));
return acc;
}, new Map());
const result = array111.map((arr) =>arr.filter((o) => uniqueNamesFrequency.get(o.parameters.employee_number) === array111.length));
console.log(result);

/* This is not a part of answer. It is just to give the output fill height. So IGNORE IT */
.as-console-wrapper { max-height: 100% !important; top: 0; }

不使用Set

const array111 = [
[
{
name: "test1",
parameters: {
employee_number: 1,
},
},
{
name: "test2",
parameters: {
employee_number: 2,
},
},
{
name: "test3",
parameters: {
employee_number: 4,
},
},
{
name: "test4",
parameters: {
employee_number: 5,
},
},
{
name: "test5",
parameters: {
employee_number: 6,
},
},
],
[
{
name: "test1",
parameters: {
employee_number: 1,
},
},
{
name: "test4",
parameters: {
employee_number: 5,
},
},
{
name: "test5",
parameters: {
employee_number: 6,
},
},
],
[
{
name: "test1",
parameters: {
employee_number: 1,
},
},
{
name: "test4",
parameters: {
employee_number: 5,
},
},
{
name: "test5",
parameters: {
employee_number: 6,
},
},
],
[
{
name: "test1",
parameters: {
employee_number: 1,
},
},
{
name: "test4",
parameters: {
employee_number: 4,
},
},
{
name: "test4",
parameters: {
employee_number: 5,
},
},
{
name: "test5",
parameters: {
employee_number: 6,
},
},
],
];
const uniqueNamesFrequency = array111
.map((arr) => Array.from( arr.reduce((acc, curr) => {
if (!acc.includes(curr.parameters.employee_number))
acc.push(curr.parameters.employee_number);
return acc;
}, [])))
.reduce((acc, curr, i) => {
curr.forEach((id) => acc.set(id, (acc.get(id) ?? 0) + 1));
return acc;
}, new Map());
const result = array111.map((arr) => arr.filter( (o) => uniqueNamesFrequency.get(o.parameters.employee_number) === array111.length));
console.log(result);

/* This is not a part of answer. It is just to give the output fill height. So IGNORE IT */
.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 2

如果你不介意lodash图书馆:

const _ = require('lodash');
const uniqOnly = (arr) => {
const uniqPart = _.intersectionBy(...arr, (o) => o.parameters.employee_number);
return = arr.map(() => [...commomPart]);
}
console.log(uniqOnly(array111))
// output:
[
[
{ name: 'test1', parameters: [Object] },
{ name: 'test4', parameters: [Object] },
{ name: 'test5', parameters: [Object] }
],
[
{ name: 'test1', parameters: [Object] },
{ name: 'test4', parameters: [Object] },
{ name: 'test5', parameters: [Object] }
],
[
{ name: 'test1', parameters: [Object] },
{ name: 'test4', parameters: [Object] },
{ name: 'test5', parameters: [Object] }
],
[
{ name: 'test1', parameters: [Object] },
{ name: 'test4', parameters: [Object] },
{ name: 'test5', parameters: [Object] }
]
]