Question

我找不到类似的问题而且我有点卡住了。我有以下JSON数组：

[
    {
        "Name": "element1",
        "Attributes": ["1", "2"]
    },

    {
        "Name": "element2",
        "Attributes": ["1","3" ]
    },
    {
        "Name": "element3",
        "Attributes": []
    }
]

我正在尝试在“Attributes”属性中创建一个包含所有唯一元素的数组，但是我无法遍历每个对象，然后循环遍历数组元素以返回唯一值。我试图用filter（）或map（）来做这件事。

编辑：我想要一系列独特的元素，所以：[1,2,3]。

Answer 1

您可以使用 Array#reduce 和 Array#filter 方法

var data = [{
    "Name": "element1",
    "Attributes": ["1", "2"]
  },

  {
    "Name": "element2",
    "Attributes": ["1", "3"]
  }, {
    "Name": "element3",
    "Attributes": []
  }
]

console.log(
  // iterate over array elements
  data.reduce(function(arr, ele) {
    // push the unique values to array
    [].push.apply(arr,
      // filter out unique value
      ele.Attributes.filter(function(v) {
        // check element present in array
        return arr.indexOf(v) == -1;
      })
    );
    // return the unique array
    return arr;
    // set initial argument as an empty array
  }, [])
);

ES6 arrow function

 var data = [{
     "Name": "element1",
     "Attributes": ["1", "2"]
   },

   {
     "Name": "element2",
     "Attributes": ["1", "3"]
   }, {
     "Name": "element3",
     "Attributes": []
   }
 ]

 console.log(
   data.reduce((arr, ele) => ([].push.apply(arr, ele.Attributes.filter((v) => arr.indexOf(v) == -1)), arr), [])
 );

Answer 2

你可以用几种Array方法来做。例如：

var result = [
    {
        "Name": "element1",
        "Attributes": ["1", "2"]
    },

    {
        "Name": "element2",
        "Attributes": ["1","3" ]
    },
    {
        "Name": "element3",
        "Attributes": []
    }
]

// map to [ ["1", "2"], ["1", "3"], [] ]
.map(item => item.Attributes)

// flatten to [ "1", "2", "1", "3" ]
.reduce((prev, curr) => prev.concat(curr), [])

// filter unique [ "1", "2", "3" ]
.filter((item, i, arr) => arr.indexOf(item) === i)

console.log(result)

Answer 3

var uniqueArr = [];

var arr = [
    {
        "Name": "element1",
        "Attributes": ["1", "2"]
    },

    {
        "Name": "element2",
        "Attributes": ["1","3" ]
    },
    {
        "Name": "element3",
        "Attributes": []
    }
];

arr.forEach(function(obj) {
   var attr = obj.Attributes;
   attr.forEach(function(val){
       if (uniqueArray.indexOf(val) < 0) {
           uniqueArray.push(val)
       }
   });
})

Answer 4

如果lodash是一个选项，您可以轻松获得所需内容：

> _.chain(foo).map('Attributes').flatten().uniq().value()
["1", "2", "3"]

Answer 5

您有答案可供选择。只是为了好玩：这个使用es6

"use strict";
let uniqueAttr = [];
const obj = [
    {
        "Name": "element1",
        "Attributes": ["1", "2"]
    },

    {
        "Name": "element2",
        "Attributes": ["1","3" ]
    },
    {
        "Name": "element3",
        "Attributes": []
    }
];

obj.forEach( element => 
  element.Attributes.forEach( 
    attr => uniqueAttr.indexOf(attr) < 0  && uniqueAttr.push(attr)
  ) 
);

document.querySelector("#result").textContent = uniqueAttr;

<pre id="result"></pre>

Answer 6

试试这个，它有助于解决这个问题。

＆＃13;

  var data = [{
    "Name": "element1",
    "Attributes": ["1", "2"]
  }, {
    "Name": "element2",
    "Attributes": ["1", "3"]
  }, {
    "Name": "element3",
    "Attributes": []
  }];
  var Attributes = [];
  $.each(data, function(i, e) {
    $.each(e.Attributes, function(i, e) {
      Attributes.push(parseInt(e));
    });
  });
  Attributes = $.unique(Attributes);
  alert(Attributes);

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>

＆＃13;

Answer 7

在ES6 / ES2015中，您可以使用Set和spread operator：

const input = [
    {
        "Name": "element1",
        "Attributes": ["1", "2"]
    },

    {
        "Name": "element2",
        "Attributes": ["1","3" ]
    },
    {
        "Name": "element3",
        "Attributes": []
    }
];

const output = [...new Set([].concat(...input.map(item => item.Attributes)))];

console.log(output);

说明（由内而外）：

input.map(item => item.Attributes)产生一个由Attributes数组组成的数组
[].concat(...)使数组变平，即产生一个包含所有Attributes值（包括重复项）的数组
new Set()从数组中生成一个Set，即仅存储唯一的Attribute值
[...]根据集合的值生成一个数组，即生成所有唯一属性值的数组

Answer 8

以@dfsq答案为基础，您可以用单个map替换两个reduce和flatMap

var result = [
    {
        "Name": "element1",
        "Attributes": ["1", "2"]
    },

    {
        "Name": "element2",
        "Attributes": ["1","3" ]
    },
    {
        "Name": "element3",
        "Attributes": []
    }
]

// map & flatten to [ "1", "2", "1", "3" ]
.flatMap(item => item.Attributes)

// filter unique [ "1", "2", "3" ]
.filter((item, i, arr) => arr.indexOf(item) === i)

console.log(result)

从对象数组中的数组返回唯一的数组值

8 个答案: