从特定的数组中删除数组中的元素

Question

I have two arrays. First one is an array of indexes and second one is an array of objects. They look like this:

var nums = [0, 2];
var obj = [Object_1, Object_2, Object_3];

In this particular case I need to remove all " obj " elements except obj[0] and obj[2] . So the result will look like this:

obj = [Object_2]

There are also may be cases when nums = [0, 1, 2] and obj = [Object_1, Object_2, Object_3] ; In that case I dont need to remove any elements.

The " obj " length is always greater than "nums" length.

So I started with finding only the elements that I need to save:

nums.forEach(function(key) {
    obj.forEach(function(o, o_key) {
        if (key === o_key) {
            console.log(key, o);
            // deleting remaining elements
        }
    });
});

The question: How can I remove elements that dont meets my condition? I dont need the new array, I want to modify the existing "obj" array. How can I achieve this functionality? Or should I use some another techniques?

Answer 1

You coudld check if the length of the indices is the same length of the object array and return or delete the objects at the given indices.

It needs a sorted array for indices, because Array#splice changes the length of the array. (With an array with descending sorted indices, you could use Array#forEach instead of Array#reduceRight .)

 function mutate(objects, indices) { if (objects.length === indices.length) { return; } indices.reduceRight(function (_, i) { objects.splice(i, 1); }, null); } var objects = [{ o: 1 }, { o: 2 }, { o: 3 }]; mutate(objects, [0, 1, 2]); // keep all items console.log(objects); objects = [{ o: 1 }, { o: 2 }, { o: 3 }]; // delete items at index 0 and 2 mutate(objects, [0, 2]); console.log(objects); 

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Answer 2

您可以使用过滤器执行此操作，假设nums是要保留的元素索引上的数组。

obj = obj.filter((o, i) => nums.indexOf(i) > -1); 

Answer 3

If you want to keep the same array object, you need to use splice eg in a simple reverse for in order to not mess the indices:

for (var i=obj.length-1; i>=0; i--) {
    if (nums.indexOf(i) < 0) {
        obj.splice(i, 1);
    }
}

This is assuming the list of indices (nums) is ordered. If not, we first need to sort it:

var sortedNums = nums.sort(function (a, b) {  return a - b;  });

And then use the sortedNums to check the indexOf