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将对象数组转换为对象的最佳方法？

分类: 技术问答 • 2022-07-12 11:22:10
问题描述：

我需要将对象数组转换为对象。我已经完成了以下逻辑。有没有最好的方法来处理这个问题？将对象数组转换为对象的最佳方法？

Fiddler Version

var before = [{ 
 
    "x": ["2015-10-14T01:59:59.999+05:30", "2015-10-14T03:59:59.998+05:30", "2015-10-14T05:59:59.997+05:30", "2015-10-14T07:59:59.996+05:30", "2015-10-14T09:59:59.995+05:30", "2015-10-14T11:59:59.994+05:30", "2015-10-14T13:59:59.993+05:30", "2015-10-14T15:59:59.992+05:30", "2015-10-14T17:59:59.991+05:30", "2015-10-14T19:59:59.990+05:30", "2015-10-14T21:59:59.989+05:30", "2015-10-14T23:59:59.988+05:30"] 
 
}, { 
 
    "CleanCoal": ["0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00"] 
 
}, { 
 
    "Middelings": ["0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00"] 
 
}, { 
 
    "Prime": ["0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00"] 
 
}, { 
 
    "SpiralProd": ["0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00"] 
 
}]; 
 

 
var after = {}; 
 

 
for (var i = 0; i < before.length; i++) { 
 
    var keys = Object.keys(before[i]); 
 

 
    after[keys] = before[i][keys]; 
 
} 
 

 
console.log(after) 
 
document.writeln(JSON.stringify(after))

+0

你的版本是不正确的，如果'{A：1，B：4，...}'对象关键将不止一个。 –

+0

您的方法仅适用于内部数组每个只有一个键。多于一个就会失败。你需要第二个循环。 – Yoshi

答

您可以使用forEach遍历数组，你忘了，如果里面有多种元素迭代嵌套阵列上方的一个对象。

所以，有问题的代码不会为

var arr = [{'a': 'b', 'c': 'd'}];

演示

var before = [{ 
 
    "x": ["2015-10-14T01:59:59.999+05:30", "2015-10-14T03:59:59.998+05:30", "2015-10-14T05:59:59.997+05:30", "2015-10-14T07:59:59.996+05:30", "2015-10-14T09:59:59.995+05:30", "2015-10-14T11:59:59.994+05:30", "2015-10-14T13:59:59.993+05:30", "2015-10-14T15:59:59.992+05:30", "2015-10-14T17:59:59.991+05:30", "2015-10-14T19:59:59.990+05:30", "2015-10-14T21:59:59.989+05:30", "2015-10-14T23:59:59.988+05:30"] 
 
}, { 
 
    "CleanCoal": ["0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00"] 
 
}, { 
 
    "Middelings": ["0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00"] 
 
}, { 
 
    "Prime": ["0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00"] 
 
}, { 
 
    "SpiralProd": ["0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00"] 
 
}]; 
 

 
var after = {}; 
 

 
before.forEach(function(obj) { 
 
    // obj here is the element of the array, i.e. object 
 

 
    // Looping over all the keys of the object 
 
    Object.keys(obj).forEach(function(key) { 
 
    // key here is the key of the object 
 
    after[key] = obj[key]; 
 
    }); 
 
}); 
 

 
console.log(after); 
 
document.writeln('<pre>' + JSON.stringify(after, 0, 2) + '</pre>');

+0

谢谢你的回复。我还需要将日期格式从字符串更改为日期类型。我更新了代码片段。我可以如何有效地转换？ –

+1

@PremkumarJayaseelan请对日期格式更改提出不同的问题 – Tushar

+0

嗨，兄弟，感谢'JSON.stringify（after，0，2）'。我不知道。你怎么看我的答案。你能告诉我（这是否好方法）？ –

答

我们用最好的函数式编程的JavaScript中：

使用reduce我们不需要临时变量添加到外部范围。

for (var i in obj)一般快于Obj.keys(obj).forEach：https://jsperf.com/for-in-versus-object-keys-foreach

var before = [{ 
 
    "x": ["2015-10-14T01:59:59.999+05:30", "2015-10-14T03:59:59.998+05:30", "2015-10-14T05:59:59.997+05:30", "2015-10-14T07:59:59.996+05:30", "2015-10-14T09:59:59.995+05:30", "2015-10-14T11:59:59.994+05:30", "2015-10-14T13:59:59.993+05:30", "2015-10-14T15:59:59.992+05:30", "2015-10-14T17:59:59.991+05:30", "2015-10-14T19:59:59.990+05:30", "2015-10-14T21:59:59.989+05:30", "2015-10-14T23:59:59.988+05:30"] 
 
}, { 
 
    "CleanCoal": ["0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00"] 
 
}, { 
 
    "Middelings": ["0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00"] 
 
}, { 
 
    "Prime": ["0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00"] 
 
}, { 
 
    "SpiralProd": ["0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00"] 
 
}]; 
 
var after = before.reduce(function(dst, src){ 
 
    // import every key: val in src into dst 
 
    // and keep doing it for every element in the array. 
 
    for (var key in src) 
 
    if ({}.hasOwnProperty.call(src, key)) 
 
     dst[key] = src[key]; 
 
    return dst; 
 
}, {}); // start with an empty object to avoid modifying items in before array. 
 

 
document.writeln("<pre>" + JSON.stringify(after, null, 4) + "</pre>")

+0

仅有代码的答案不是最优的。你应该解释为什么你的代码符合OP的要求。 – RobG

答

如何Object.assign工作？

var before = [{ 
 
    "x": ["2015-10-14T01:59:59.999+05:30", "2015-10-14T03:59:59.998+05:30", "2015-10-14T05:59:59.997+05:30", "2015-10-14T07:59:59.996+05:30", "2015-10-14T09:59:59.995+05:30", "2015-10-14T11:59:59.994+05:30", "2015-10-14T13:59:59.993+05:30", "2015-10-14T15:59:59.992+05:30", "2015-10-14T17:59:59.991+05:30", "2015-10-14T19:59:59.990+05:30", "2015-10-14T21:59:59.989+05:30", "2015-10-14T23:59:59.988+05:30"] 
 
}, { 
 
    "CleanCoal": ["0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00"] 
 
}, { 
 
    "Middelings": ["0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00"] 
 
}, { 
 
    "Prime": ["0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00"] 
 
}, { 
 
    "SpiralProd": ["0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00", "0.00"] 
 
}]; 
 

 
var after = {}; 
 

 
before.map(function(i){ 
 
    return Object.assign(after,i); 
 
}) 
 

 

 
console.log(after) 
 
document.writeln(JSON.stringify(after))

答

在最新的浏览器[5的ECMAScript]，我们有reduce它返回一个对象下

var before = [{"x":["2015-10-14T01:59:59.999+05:30","2015-10-14T03:59:59.998+05:30","2015-10-14T05:59:59.997+05:30","2015-10-14T07:59:59.996+05:30","2015-10-14T09:59:59.995+05:30","2015-10-14T11:59:59.994+05:30","2015-10-14T13:59:59.993+05:30","2015-10-14T15:59:59.992+05:30","2015-10-14T17:59:59.991+05:30","2015-10-14T19:59:59.990+05:30","2015-10-14T21:59:59.989+05:30","2015-10-14T23:59:59.988+05:30"]},{"CleanCoal":["0.00","0.00","0.00","0.00","0.00","0.00","0.00","0.00","0.00","0.00","0.00","0.00"]},{"Middelings":["0.00","0.00","0.00","0.00","0.00","0.00","0.00","0.00","0.00","0.00","0.00","0.00"]},{"Prime":["0.00","0.00","0.00","0.00","0.00","0.00","0.00","0.00","0.00","0.00","0.00","0.00"]},{"SpiralProd":["0.00","0.00","0.00","0.00","0.00","0.00","0.00","0.00","0.00","0.00","0.00","0.00"]}]; 
 

 

 
var object = before.reduce(function(o, v, i) { 
 
    o[i] = v; 
 
    return o; 
 
}, {}); 
 

 
console.log(object);

试小提琴这里http://jsfiddle.net/pscytrgj/

+1

* reduce *不一定会返回一个对象，它将返回任何回调作为累加器返回的值（可以是任何值）。 MDN提供了[polyfill for * reduce *]（https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/Reduce#Polyfill），因此也适应了旧版浏览器。 – RobG

+0

@RobG是的，这是真的！有点为我学习以及:)谢谢！ –

答

你可以做到这一点通过JSON.stringify()

var before = [{ 
 
    "x": ["1", "2"], 
 
    y: { 
 
    a: 3 
 
    } 
 
}, { 
 
    "CleanCoal": ["0.00", "0.00"], 
 
    b: 4 
 
}, { 
 
    "Middelings": ["0.00", "0.00"] 
 
}, { 
 
    "Prime": ["0.00", "0.00", "0.00"] 
 
}, { 
 
    "SpiralProd": ["0.00", "0.00", "0.00"] 
 
}]; 
 

 
var after = {}; 
 
var b = "", 
 
    i = -1; 
 
while (before[++i]) { 
 
    var str = JSON.stringify(before[i]); 
 
    b += str.slice(1, str.length - 1); 
 
    if (before[i + 1]) b += ","; 
 
} 
 
after = JSON.parse("{" + b + "}"); 
 
console.log(after); 
 
document.write("<pre>" + JSON.stringify(after, 0, 3) + "</pre>")

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