Question

我有两个对象数组。 localDataArray已经存储在我的应用程序中，remoteUpdateDataArray来自后端。

var localDataArray =   [
    { "date": "10/01/19", "category": "surf", "hash": "da39a3ee5e6b4b0d3255bfef95601890afd80709"}, 
    { "date": "10/01/19", "category": "skate", "hash": "a0f1490a20d0211c997b44bc357e1972deab8ae3"},
    { "date": "10/01/19", "category": "skate", "hash": "54fd1711209fb1c0781092374132c66e79e2241b"}
];


var remoteUpdateDataArray =   [
     { "date": "12/01/19", "category": "surf", "hash": "4a0a19218e082a343a1b17e5333409af9d98f0f5"}, 
     { "date": "11/01/19", "category": "surf", "hash": "54fd1711209fb1c0781092374132c66e79e2241b"}, 
     { "date": "10/01/19", "category": "surf", "hash": "da39a3ee5e6b4b0d3255bfef95601890afd80709"}, 
     { "date": "10/01/19", "category": "skate", "hash": "a0f1490a20d0211c997b44bc357e1972deab8ae3"},
     { "date": "10/01/19", "category": "skate", "hash": "54fd1711209fb1c0781092374132c66e79e2241b"}
];

我想从remoteUpdateDataArray中删除所有重复的对象。每个对象的唯一标识符是哈希。

到目前为止，我有以下代码：

let hashValue = "54fd1711209fb1c0781092374132c66e79e2241b"

var filteredResult = remoteUpdateDataArray.filter(x => x.hash !== hashValue);

结果：

var filteredResult =   [
         { "date": "12/01/19", "category": "surf", "hash": "4a0a19218e082a343a1b17e5333409af9d98f0f5"}, 
         { "date": "11/01/19", "category": "surf", "hash": "54fd1711209fb1c0781092374132c66e79e2241b"}, 
         { "date": "10/01/19", "category": "surf", "hash": "da39a3ee5e6b4b0d3255bfef95601890afd80709"}, 
         { "date": "10/01/19", "category": "skate", "hash": "a0f1490a20d0211c997b44bc357e1972deab8ae3"}
    ];

我如何还设法摆脱阵列内的另一个对象（在这种情况下为两个重复的对象）？请记住，这些数组可能会变得很大。

Answer 1

我会从您的第一个数组构建一个哈希列表（以保存迭代），然后只需使用include（）进行过滤

const inLocalData = localDataArray.map(({hash: e}) => e);
const result = remoteUpdateDataArray.filter(({hash: e}) => ! inLocalData.includes(e));
console.log(result);

<script>
var localDataArray = [{
    "date": "10/01/19",
    "category": "surf",
    "hash": "da39a3ee5e6b4b0d3255bfef95601890afd80709"
  },
  {
    "date": "10/01/19",
    "category": "skate",
    "hash": "a0f1490a20d0211c997b44bc357e1972deab8ae3"
  },
  {
    "date": "10/01/19",
    "category": "skate",
    "hash": "54fd1711209fb1c0781092374132c66e79e2241b"
  }
];


var remoteUpdateDataArray = [{
    "date": "12/01/19",
    "category": "surf",
    "hash": "4a0a19218e082a343a1b17e5333409af9d98f0f5"
  },
  {
    "date": "11/01/19",
    "category": "surf",
    "hash": "54fd1711209fb1c0781092374132c66e79e2241b"
  },
  {
    "date": "10/01/19",
    "category": "surf",
    "hash": "da39a3ee5e6b4b0d3255bfef95601890afd80709"
  },
  {
    "date": "10/01/19",
    "category": "skate",
    "hash": "a0f1490a20d0211c997b44bc357e1972deab8ae3"
  },
  {
    "date": "10/01/19",
    "category": "skate",
    "hash": "54fd1711209fb1c0781092374132c66e79e2241b"
  }
];
</script>

Answer 2

看来您需要缓存。将哈希存储到缓存中，并在U获取新数据时对其进行更新。

var localDataArray =   [
    { "date": "10/01/19", "category": "surf", "hash": "da39a3ee5e6b4b0d3255bfef95601890afd80709"}, 
    { "date": "10/01/19", "category": "skate", "hash": "a0f1490a20d0211c997b44bc357e1972deab8ae3"},
    { "date": "10/01/19", "category": "skate", "hash": "54fd1711209fb1c0781092374132c66e79e2241b"}
];

var cache = {}

// The result, make a copy of local data at start
var filtered = [].concat(localDataArray)

// initialize the cache
localDataArray.forEach(item => {
    cache[item.hash] = true
})

// -----------

var remoteUpdateDataArray =   [
     { "date": "12/01/19", "category": "surf", "hash": "4a0a19218e082a343a1b17e5333409af9d98f0f5"}, 
     { "date": "11/01/19", "category": "surf", "hash": "54fd1711209fb1c0781092374132c66e79e2241b"}, 
     { "date": "10/01/19", "category": "surf", "hash": "da39a3ee5e6b4b0d3255bfef95601890afd80709"}, 
     { "date": "10/01/19", "category": "skate", "hash": "a0f1490a20d0211c997b44bc357e1972deab8ae3"},
     { "date": "10/01/19", "category": "skate", "hash": "54fd1711209fb1c0781092374132c66e79e2241b"}
];

// filter duplicated items
remoteUpdateDataArray.forEach(item => {
    // item exists
    if(cache.hasOwnProperty(item.hash)) {
        return
    }
    // append hash and item
    cache[item.hash] = true
    // just append the new data items
    filtered.push(item)
})

Answer 3

为什么不使用lodash的uniqBy？

首先，使用spred opertaor将2个数组连接起来（您可以详细了解here）：

const newArray = [...localDataArray ,...remoteUpdateDataArray];

以及具有所有重复项的新数组

const filteredResult = _.uniqBy(newObject,'hash');

Answer 4

您也可以使用此

var localDataArray = [{
    "date": "10/01/19",
    "category": "surf",
    "hash": "da39a3ee5e6b4b0d3255bfef95601890afd80709"
  },
  {
    "date": "10/01/19",
    "category": "skate",
    "hash": "a0f1490a20d0211c997b44bc357e1972deab8ae3"
  },
  {
    "date": "10/01/19",
    "category": "skate",
    "hash": "54fd1711209fb1c0781092374132c66e79e2241b"
  }
];
var remoteUpdateDataArray = [{
    "date": "12/01/19",
    "category": "surf",
    "hash": "4a0a19218e082a343a1b17e5333409af9d98f0f5"
  },
  {
    "date": "11/01/19",
    "category": "surf",
    "hash": "54fd1711209fb1c0781092374132c66e79e2241b"
  },
  {
    "date": "10/01/19",
    "category": "surf",
    "hash": "da39a3ee5e6b4b0d3255bfef95601890afd80709"
  },
  {
    "date": "10/01/19",
    "category": "skate",
    "hash": "a0f1490a20d0211c997b44bc357e1972deab8ae3"
  },
  {
    "date": "10/01/19",
    "category": "skate",
    "hash": "54fd1711209fb1c0781092374132c66e79e2241b"
  }
];


let data =[];

localDataArray.map(item=>{
 data= remoteUpdateDataArray.find(el=>item.hash!=el.hash)
  
})

console.log(data)

用多个值过滤对象数组

4 个答案: