javascript - 使用 es6 制作 uniqBy

Question

const uniqBy = (arr, fn) => {
  let res = []
  const hash = new Set()
  for (let i = 0; i < arr.length; i++) {
    const foo = typeof fn === 'function' ? fn(arr[i]) : arr[i][fn]
    if (!hash.has(foo)) {
      res.push(arr[i])
      hash.add(foo)
    }
  }
  return res
}

console.log(uniqBy([2.1, 1.2, 2.3], Math.floor)) // [2.1, 1.2]
console.log(uniqBy([{
  x: 1
}, {
  x: 2
}, {
  x: 1
}], 'x')) // [{x: 1 },{ x: 2 }]

这是我实现uniqBy的代码，但是代码太多了，我想用更少的代码得到更好的方法

Answer 1

 const uniqBy = (arr, fn, set = new Set) => arr.filter(el => (v => !set.has(v) && set.add(v))(typeof fn === "function" ? fn(el) : el[fn]));