Ramda：通過對每個項目在另一個數組進行比較得到數組對象

Question

我已經像一個數組：

ids = [1,3,5]; 

和另一個陣列，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'} 
]; 

我想要的是像另一個數組：

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

我無法理解它。到目前爲止，我嘗試過這樣的：

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

Answer 1

一個內膽如果您仍然想這樣做在ramda方式

Ramda REPL

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) 

Answer 2

您可以使用.filter和.indexOf。請注意，這些是數組的ECMA5方法，並且在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'}]; 

Answer 3

我建議使用散列表來加快查找速度。

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 4

或者可以沒有Ramda

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