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))
答案 0 :(得分:4)
您可以使用.filter
和.indexOf
。请注意,这些是用于阵列的ECMA5方法,并且在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'}];
答案 1 :(得分:4)
如果您仍想以ramda方式进行
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'}
];
R.filter(R.compose(R.flip(R.contains)(ids), R.prop('id')), items)
答案 2 :(得分:3)
或者可能是没有Ramda的一个班轮
items.filter(x=>ids.includes(x.id))
答案 3 :(得分:0)
我建议使用哈希表来加快查找速度。
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>');