我有两个数组,我想将它们合并到一个对象数组中...
第一个数组是日期(字符串):
let metrodates = [
"2008-01",
"2008-02",
"2008-03",..ect
];
第二个数组是数字:
let figures = [
0,
0.555,
0.293,..ect
]
我想合并它们以创建一个像这样的对象(因此数组项通过相似的索引进行匹配):
let metrodata = [
{data: 0, date: "2008-01"},
{data: 0.555, date: "2008-02"},
{data: 0.293, date: "2008-03"},..ect
];
到目前为止,我这样做:我创建一个空数组,然后循环遍历前两个数组之一以获取索引号(前两个数组的长度相同)...但是有没有更简单的方法(在 ES6 中)?
let metrodata = [];
for(let index in metrodates){
metrodata.push({data: figures[index], date: metrodates[index]});
}
最佳答案
最简单的方法可能是使用 map
和提供给回调的索引
let metrodates = [
"2008-01",
"2008-02",
"2008-03"
];
let figures = [
0,
0.555,
0.293
];
let output = metrodates.map((date,i) => ({date, data: figures[i]}));
console.log(output);
另一个选择是创建一个通用的 zip 函数,它将两个输入数组整理成一个数组。这通常称为“ zipper ”,因为它像 zipper 上的齿一样交错输入。
const zip = ([x,...xs], [y,...ys]) => {
if (x === undefined || y === undefined)
return [];
else
return [[x,y], ...zip(xs, ys)];
}
let metrodates = [
"2008-01",
"2008-02",
"2008-03"
];
let figures = [
0,
0.555,
0.293
];
let output = zip(metrodates, figures).map(([date, data]) => ({date, data}));
console.log(output);
另一种选择是创建一个通用的map函数,它接受多个源数组。映射函数将从每个源列表接收一个值。请参阅Racket's map procedure了解更多其使用示例。
这个答案可能看起来最复杂,但它也是最通用的,因为它接受任意数量的源数组输入。
const isEmpty = xs => xs.length === 0;
const head = ([x,...xs]) => x;
const tail = ([x,...xs]) => xs;
const map = (f, ...xxs) => {
let loop = (acc, xxs) => {
if (xxs.some(isEmpty))
return acc;
else
return loop([...acc, f(...xxs.map(head))], xxs.map(tail));
};
return loop([], xxs);
}
let metrodates = [
"2008-01",
"2008-02",
"2008-03"
];
let figures = [
0,
0.555,
0.293
];
let output = map(
(date, data) => ({date, data}),
metrodates,
figures
);
console.log(output);
关于arrays - ES6 : Merge two arrays into an array of objects,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/39711528/