删除阵列中重复的对象，仅基于对象中的一个值

Question

我正在编写一些返回数组中唯一对象的代码。

我找到了一些代码，但它并不是我所需要的。

现在我们假设我有这个数组：

restuls= [
{
name: "name 1",
description: "",
group: "group 1"
},
{
name: "name 1",
description: "",
group: "group 1"
},
{
name: "name 1",
description: "",
group: "group 2"
},
]

通过下面的代码，我在这个数组中找到了所有唯一的对象，这意味着，如果对象中的任何值不同，它将是唯一的。因此，在这种情况下，它将返回2个对象(因为在第三个对象上-->group："group 2"(。

var array = results,
unique = Array.from(
new Set(array.map(o => JSON.stringify(o))),
s => JSON.parse(s)
);

我真正想要的是，它删除了所有重复的对象，但只查看"；name"；因此，在这种情况下，所有3个对象都有相同的名称，因此应该只显示一个。

有什么想法吗？

Answer 1

您可以使用for和Set-

const input =
[{name: "name 1",description: "",group: "group 1"},{name: "name 1",description: "",group: "group 1"},{name: "name 1",description: "",group: "group 2"},]
const result =
[]

const seen =
new Set


for (const v of input)
if (seen.has(v.name))
continue
else
(result.push(v), seen.add(v.name))

console.log(result)

输出-

[
{
"name": "name 1",
"description": "",
"group": "group 1"
}
]

---

或使用Map-

const input =
[{name: "name 1",description: "",group: "group 1"},{name: "name 1",description: "",group: "group 1"},{name: "name 1",description: "",group: "group 2"},]

const seen =
new Map

for (const v of input)
if (seen.has(v.name))
continue
else
seen.set(v.name, v)
const result =
Array.from(seen.values())

console.log(result)

输出-

[
{
"name": "name 1",
"description": "",
"group": "group 1"
}
]

Answer 2

results= [
{
name: "name 1",
description: "",
group: "group 1"
},
{
name: "name 1",
description: "",
group: "group 1"
},
{
name: "name 1",
description: "",
group: "group 2"
},
]
const map = new Map();
results.forEach(x => {
if(!map.has(x.name))
{ map.set(x.name, x) }
});
const arr = Array.from(map).map(x => x[1]);
console.log(arr)