开发者

How can I filter a Javascript array and maintain it's indices?

开发者 https://www.devze.com 2023-04-08 15:32 出处:网络
Here is what I have currently: var myArray = []; myArray[15] = 2; myArray[6323] = 1; myArray[349022] = 3;

Here is what I have currently:

var myArray = [];
myArray[15] = 2;
myArray[6323] = 1;
myArray[349022] = 3;
myArray = myArray.filter(function(i,v,a){return ! (typeof v == 'undefined')});
myArray = myArray.sort(function(a,b){return (a - b)});

When I console.log() allocationOrder after the "filter" function, I get the values that I expect, but the indices are not maintained. How can I maintain the indices but also remove the undefined values (due to having spread out indices)? Keep in mind that I need to sort the array by value afterwards (开发者_运维技巧which is why I avoided objects as a solution).


If you're working with filter, you could supply it with an extra callback for the index and assign it in the returned object:

source = source.filter((instance, index) => {
  instance.originalIndex = index;
}); 

My usage was in working with adding options to a car, there was a filterable list of options, and each one could only be applied to the car once (meaning it needed to be stripped from the available options array as well as the filtered array). Hopefully that helps!


You can't have your cake and eat it too.

You can't have a sparse array (only some values filled in) and have no undefined values in between. That's just not how javascript arrays work. Further, you can remove the elements that have undefined values and expect the other values to stay at the same index. By defintion, removing those other values, collapses the array (changing indexes).

Perhaps if you can describe what you're really trying to accomplish we can figure out a better way to approach the problem.

I'm thinking that what you want is an array of objects where each object in the array has an index and a value. The indexes on the objects in the array will never change, but the container array can be sorted by those indexes.

var mySortedArray = [
    {index:15, value: 2},
    {index:6323, value: 1},
    {index:349022, value: 3}
];

mySortedArray = myArray.sort(function(a, b) {a.value - b.value});

Now you have a sorted array that's three elements long and it's sorted by value and the original index is preserved and available.

The only drawback is that you can't easily access it by the index. If you wanted to be able to have both ordered, sorted access and access by the index key, you could make a parallel data structure that make key access quick. In Javascript, there is no single data structure that supports both key access and sorted order access.

If you want to build a parallel data structure that would support fast access by key (and would include the value and the sortOrder value, you could build such an object like this:

// build a parallel data structure for keyed access
var myKeyedObject = {}, temp, item;
for (var i = 0; i < mySortedArray.length; i++) {
    item = mySortedArray[i];
    temp = {};
    temp.value = item.value;
    temp.index = item.index;
    temp.sortOrder = i;
    myKeyedObject[item.index] = temp;
}


You could map the original index before filtering. When filtering and eventually iterating over the items just destructure to get hold of the item and its original index.

var fruit = ['apple', 'banana', 'mango'];

fruit
  .map((item, index) => ({ item, index }))
  .filter(({ item }) => item == 'mango')
  .forEach(({ item, index}) => console.log(`${index} - ${item}`));

This will log 2 - mango.


While JavaScript arrays behave rather nicely if you delete their entries, there is sadly no built-in function that does this. However, it isn't too hard to write one:

Array.prototype.filterAssoc = function(callback) {
    var a = this.slice();
    a.map(function(v, k, a) {
        if(!callback(v, k, a)) delete a[k];
    });
    return a;
};

With this added to the array prototype, myArray.filterAssoc(...).map(...) will map over only the key-value pairs that were filtered for, just as expected.

That answers the question in the title. As for your side note about wanting to sort the array afterward... sort does not behave nicely on sparse arrays. The easiest way to do that really is to sort an array of objects.

0

精彩评论

暂无评论...
验证码 换一张
取 消