如何在数组中删除具有相同键和值对的对象 [英] How to remove objects with the same key and value pair in arrays

问题描述

我查看了许多堆栈溢出问题，但似乎没有人完全回答我的问题。我有一个对象数组，我想通过删除键和值相同的所有对象来减少这些对象。

I've looked through many stack overflow questions, but none seem to quite answer my question. I have an array of objects, which I would like to reduce by deleting all objects where the key and value are the same.

所以我的对象数组是：

[{a:1},{a:2},{c:3},{b:1},{a:1},{c:3},{c:4},{a:1}]

最终结果应为：

[{a:1},{a:2},{c:3},{b:1},{c:4}]

我尝试过使用filer和map，但我只能获取数组中的第一个对象，而不是所有具有不同键/值对的对象阵列。我也尝试过使用filter和findIndex，但遇到同样的问题。

I've tried using filer and map, but I can only get the first object in the array, rather than all the objects that have different key/value pairs in the array. I've also tried using filter and findIndex, but with the same problem.

在将对象推入数组之前，我也无法过滤对象。

I also can't filter the objects before pushing them into the array.

有人能指出我正确的方向吗？

Can someone point me in the right direction?

推荐答案

你可以比较两个使用 JSON.stringify（）。然后，我们使用 reduce <添加到新阵列/ a>，如果它在数组中我们不添加它，否则我们会添加它。

You can compare the two items using JSON.stringify(). We then add to a new array using reduce, if it is in the array we don't add it otherwise we do add it.

const array = [{a:1},{a:2},{c:3},{b:1},{a:1},{c:3},{c:4},{a:1}]

let unique = array.reduce((res, itm) => {
  // Test if the item is already in the new array
  let result = res.find(item => JSON.stringify(item) == JSON.stringify(itm))
  // If not lets add it
  if(!result) return res.concat(itm)
  // If it is just return what we already have
  return res
}, [])

console.log(unique)

或者你可以使用设置（作为Fissure King metions）制作这样的项目的唯一列表：

Alternatively you could use a Set (as Fissure King metions) to make a unique list of items like this:

const array = [{a:1},{a:2},{c:3},{b:1},{a:1},{c:3},{c:4},{a:1}]

let unique = [...new Set(array.map(itm => JSON.stringify(itm)))].map(i => JSON.parse(i))

console.log(unique)

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