需求

有这么两个数组

let metrodates = [
 "2008-01",
 "2008-02",
 "2008-03",..ect
];
let figures = [
 0,
 0.555,
 0.293,..ect
]

想要这样的结果

let result = [
   {data: 0, date: "2008-01"},
   {data: 0.555, date: "2008-02"},
   {data: 0.293, date: "2008-03"},..ect
];

方案一

let result = [];
for(let index in metrodates){
    result.push({data: figures[index], date: metrodates[index]});
}

此方案为最原始方法，简单，但过于low

方案二

let result = metrodates.map((date,i) => ({date, data: figures[i]}));

此方案使用了ES6中的map，简洁，但本质还是遍历，显得有些low

方案三

const zip = ([x,...xs], [y,...ys]) => {
  if (x === undefined || y === undefined)
    return [];
  else
    return [[x,y], ...zip(xs, ys)];
}
let result = zip(metrodates, figures).map(([date, data]) => ({date, data}));

此方案使用了ES6+递归，显得高大上起来了。

方案四

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 result = map((date, data) => ({date, data}), metrodates, figures);

此方案是方案三的加强版，它能接受多个数组映射成对象数组，威力无比！
原文链接：https://www.cnblogs.com/guanghe/p/11445426.html