将数组转换为对象数组

use*_*017 9 javascript arrays name-value

我如何将数组转换为JavaScript对象数组.

例如,我有一个数组作为

data = [
    ["fruits","frozen","fresh","rotten"],
    ["apples",884,494,494],
    ["oranges",4848,494,4949],
    ["kiwi",848,33,33]
]
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我想将其转换为名称值对.

例如,结果集合中的第一个对象将是

 {"fruits": "apple", "frozen": 884, "fresh": 494, "rotten": 494}
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等等剩下的数据.

cru*_*ush 8

DEMO

使用您提供的数据:

var data = [
    ["fruits","frozen","fresh","rotten"],
    ["apples",884,494,494],
    ["oranges",4848,494,4949],
    ["kiwi",848,33,33]
]
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以下函数将数组的第一个元素视为对象属性的键.然后它将遍历其余元素,并使用这些键将它们转换为对象.最后,它将返回这些新对象的数组.

function convertToArrayOfObjects(data) {
    var keys = data.shift(),
        i = 0, k = 0,
        obj = null,
        output = [];

    for (i = 0; i < data.length; i++) {
        obj = {};

        for (k = 0; k < keys.length; k++) {
            obj[keys[k]] = data[i][k];
        }

        output.push(obj);
    }

    return output;
}
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产量

[
    { fruits: 'apples', fresh: 494, frozen: 884, rotten: 494 },
    { fruits: 'oranges', fresh: 494, frozen: 4848, rotten: 4949 },
    { fruits: 'kiwi', fresh: 33, frozen: 848, rotten: 33 }
]
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Chr*_*ren 7

如果同时使用map和reduce,解决方案将变得非常优雅:

var collection = [
    ["fruits","frozen","fresh","rotten"],
    ["apples",884,494,494],
    ["oranges",4848,494,4949],
    ["kiwi",848,33,33]
];

var keys = collection.shift();
collection = collection.map(function (row) {
    return keys.reduce(function (obj, key, i) {
      obj[key] = row[i];
      return obj;
    }, {});
});
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输出:

[ 
  { fruits: 'apples', frozen: 884, fresh: 494, rotten: 494 },
  { fruits: 'oranges', frozen: 4848, fresh: 494, rotten: 4949 },
  { fruits: 'kiwi', frozen: 848, fresh: 33, rotten: 33 } 
]
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Pav*_*vlo 6

可以使用shift()+ map()+ 避免迭代两次forEach():

var data = [
    ["fruits","frozen","fresh","rotten"],
    ["apples",884,494,494],
    ["oranges",4848,494,4949],
    ["kiwi",848,33,33]
];

var collection = data.slice(); // make a copy
var keys = collection.shift();

collection = collection.map(function (e) {
    var obj = {};

    keys.forEach(function (key, i) {
        obj[key] = e[i];
    });

    return obj;
});
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演示:http://jsfiddle.net/d7W76/2/

  • 我同意,绝对更优雅.不幸的是,它[在Chrome 31.0.1650中慢了160倍](http://jsperf.com/testing-map-foreach-vs-2x-for-statements).(也许我设置测试错了?我没想到)另外,`map()`本质上是一个封装的迭代,所以你仍然迭代两次. (3认同)