Filter array of objects based on another array of

问题:

I have an array like this

arr1 =
[
   {A: 'red', B: 'blue'},
   {Q: 'green', R: 'blue'},
   {B: 'green', M: 'red'},
   {Q: 'white', R: 'blue'},
   ...
]

Each object has two key/value pairs, a letter and color. I have another array like this

filter=
[
   {A: 'val', B: 'someval'},
   {B: 'anothervalue', M: 'value'}
]

Is it possible to filter the first array such that the final result is an array that only has the objects that have the same keys as the second array. (without a for or while loop)

In this scenario it would be:

[
   {A: 'red', B: 'blue'},
   {B: 'green', M: 'red'}
]

So I want something like:

let filteredArr = arr1.filter(obj => 
    Object.keys(obj) == Object.keys(filter[someKey]));

But I'm not sure how to do this without looping over all the keys of filter.

回答1:

You could use a Set for the property names and filter the array.

var array = [{ A: 'red', B: 'blue' }, { Q: 'green', R: 'blue' }, { B: 'green', M: 'red' }, { Q: 'white', R: 'blue' }],
    filter = [{ A: 'val', B: 'someval' }, { B: 'anothervalue', M: 'value' }],
    getKey = o => Object.keys(o).sort().join('|'),
    result = array.filter((s => o => s.has(getKey(o)))(new Set(filter.map(getKey))));
   
console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

回答2:

You can do this with fiter(), some() and every()

var arr1 = [{"A":"red","B":"blue"},{"Q":"green","R":"blue"},{"B":"green","M":"red"},{"Q":"white","R":"blue"}]
var filter = [{"A":"val","B":"someval"},{"B":"anothervalue","M":"value"}]

var result = arr1.filter(function(e) {
  return filter.some(function(a) {
    var keyE = Object.keys(e);
    var keyA = Object.keys(a);

    return keyE.length == keyA.length && keyE.every(function(k) {
      return keyA.includes(k)
    })
  })
})

console.log(result)