Question

我有两个数组[14, 20 , 40 , 40]和["6:28 PM","7:28 PM","8:28 PM","9:28 PM"]。我想把它作为

[{x:14,y:"6:28 PM"},
{x:20,y:"7:28 PM"},
{x:40,y:"8:28 PM"},
{x:40,y:"9:28 PM"}]

我试图用另一个数组，例如a.push(x:values,y:time)推送数组，但这没有用。

Answer 1

const ids = [14, 20 , 40 , 40];
const times = ["6:28 PM","7:28 PM","8:28 PM","9:28 PM"];

const result = ids.map((id, i) => ({ x: id, y: times[i] }));

console.log(result);

与Angular没有关系。

Answer 2

使用任意数量的属性，您可以将名称为数组的键作为键，并使用简短的属性对象，并根据给定的数据构建新的数据集。

它与一个对象一起工作，该对象将具有更高键名的数组作为属性（short hand properties）。
 public int getCameraCount(){
    CameraManager manager = (CameraManager) context.getSystemService(Context.CAMERA_SERVICE);
    try {
        String[] strings = manager.getCameraIdList();
        return strings.length;
    } catch (CameraAccessException e) {
        e.printStackTrace();
        return 0;
    }
}
从该对象中获取键/值对（值是数组）的数组，并用Array#reduce进行缩减。
{ value, time }
这是通过将数组作为起始值并使用Array#forEach用值迭代数组来实现的。
result = Object
    .entries({ value, time })
    .reduce( ... )
在分配之前，将检查给定索引处的结果集，如果不存在，则将一个新对象作为默认值。
(r, [k, a]) => (a.forEach((v, i) => (r[i] = r[i] || {})[k] = v), r), // callback
[]                                                                   // start value
使用键r[i] = r[i] || {}，分配实际值。
k
结果是一个对象数组，其中包含来自命名变量的键和数组的值。

(r[i] = r[i] || {})[k] = v

var value = [14, 20, 40, 40],
    time = ["6:28 PM", "7:28 PM", "8:28 PM", "9:28 PM"],
    result = Object
        .entries({ value, time })
        .reduce((r, [k, a]) => (a.forEach((v, i) => (r[i] = r[i] || {})[k] = v), r), []);
    
console.log(result);

Answer 3

const values = [14, 20 , 40 , 40];
const times = ["6:28 PM", "7:28 PM", "8:28 PM", "9:28 PM"];

const myObj = values.map((value, index) => ({
 x: value,
 y: times[index]
}));

Answer 4

let a=[14, 20 , 40 , 40];
let b=["6:28 PM","7:28 PM","8:28 PM","9:28 PM"];
let result = a.map((t,index)=>Object.assign({x:t,y:b[index]}));

结果将如您所愿

Answer 5

let newArr = [];
a = [14, 20 , 40 , 40]
b = ["6:28 PM","7:28 PM","8:28 PM","9:28 PM"]
for(i=0;i<a.length;i++){
    let obj = {x: a[i], y: b[i]}
    newArr.push(obj)
}

Answer 6

受Pythons zip function的启发。像夹克上的拉链一样工作：

function zip(a, b) {
    var arr = [];
    for (var key in a) { arr.push([a[key], b[key]]); }
    return arr;
}

const ids = [14, 20, 40, 40];
const times = ["6:28 PM", "7:28 PM", "8:28 PM", "9:28 PM"];

const zipped = zip(ids, times)

Answer 7

function processor(values, times){
// Map function use two array value and return processed array of objects
   return values.map((value, index) => ({
     x: value,
     y: times[index]
   }));
}
// Pass two array as arguments for two which you want processed
processor([14, 20 , 40 , 40], ["6:28 PM", "7:28 PM", "8:28 PM", "9:28 PM"]);

Answer 8

get_index(s)

用键值存储数组

8 个答案: