Ramda：通过与另一个数组中的每个项目进行比较来从数组中获取对象

Question

我有一个像这样的数组：

ids = [1,3,5];

和另一个数组，如：

items: [
{id: 1, name: 'a'}, 
{id: 2, name: 'b'}, 
{id: 3, name: 'c'}, 
{id: 4, name: 'd'}, 
{id: 5, name: 'e'}, 
{id: 6, name: 'f'}
];

我想要的是另一个数组：

array = [{id: 1, name: 'a'}, {id: 3, name: 'c'}, {id: 5, name: 'e'}];

我无法理解它。到目前为止，我尝试过：

console.log(R.filter(R.propEq('id', <donnow what shud be here>), items);
console.log( R.pick(ids)(items))

Answer 1

您可以使用 .filter and .indexOf。请注意，这些是数组的 ECMA5 方法，在 IE8 中不起作用。

var ids = [1, 3, 5];
var items = [
  {id: 1, name: 'a'}, 
  {id: 2, name: 'b'}, 
  {id: 3, name: 'c'}, 
  {id: 4, name: 'd'}, 
  {id: 5, name: 'e'}, 
  {id: 6, name: 'f'}
];

var filtered = items.filter(function(obj) {
  return ids.indexOf(obj.id) > -1;
});
console.log(filtered); // [{id: 1, name: 'a'}, {id: 3, name: 'c'}, {id: 5, name: 'e'}];

Answer 2

我建议使用散列 table 来加快查找速度。

var ids = [1, 3, 5],
    items = [{id: 1, name: 'a'}, {id: 2, name: 'b'}, {id: 3, name: 'c'}, {id: 4, name: 'd'}, {id: 5, name: 'e'}, {id: 6, name: 'f'} ],
    filtered = items.filter(function(obj) {
        return this[obj.id];
    }, ids.reduce(function (r, a) {
        r[a] = true;
        return r;
    }, Object.create(null)));

document.write('<pre>' + JSON.stringify(filtered, 0, 4) + '</pre>');

Answer 3

或者可能是一个没有 Ramda 的 liner

items.filter(x=>ids.includes(x.id))

Answer 4

如果你还想使用 Ramda:

const ids = [1,3,5];

const items = [
{id: 1, name: 'a'}, 
{id: 2, name: 'b'}, 
{id: 3, name: 'c'}, 
{id: 4, name: 'd'}, 
{id: 5, name: 'e'}, 
{id: 6, name: 'f'}
];

console.log(

  R.filter(R.compose(R.flip(R.contains)(ids), R.prop('id')), items)

);

<script src="https://cdnjs.cloudflare.com/ajax/libs/ramda/0.26.1/ramda.min.js"></script>

Ramda：通过与另一个数组中的每个项目进行比较来从数组中获取对象

Ramda: get objects from array by comparing with each item in another array

javascript

arrays

ramda.js