如何获得“未过滤”的数组项？

Question

Let's say I have an array which I filter by calling myItems.filter(filterFunction1) and get some items from it.

Then I want to run another filtering function filterFunction2 against the remaining items which were not selected by filterFunction1 .

Is that possible to get the remaining items that were left out after calling a filtering function?

Answer 1

You'd have to rerun the filter with an inverted predicate, which seems wasteful. You should reduce the items instead and bin them into one of two bins:

const result = arr.reduce((res, item) => {
    res[predicate(item) ? 'a' : 'b'].push(item);
    return res;
}, { a: [], b: [] });

predicate here is the callback you'd give to filter .

Answer 2

Unfortunately, there is no one-step solution based on filter . Still the solution is a simple one-liner:

Here's an example

 const arr = [ 1,2,3,4,5,6,7,8 ]; const filtered = arr.filter(x=>!!(x%2)) const remaining = arr.filter(x=>!filtered.includes(x)) console.log(filtered, remaining);

Answer 3

You could map an array of flags and then filter by the flags values.

 const cond = v => !(v % 2); var array = [1, 2, 3, 4, 5], flags = array.map(cond), result1 = array.filter((_, i) => flags[i]), result2 = array.filter((_, i) => !flags[i]); console.log(result1); console.log(result2);

Answer 4

You can achieve that using Array.reduce .

const myItems = [...];

const { result1, result2 } = myItems.reduce(
  (result, item) => {
    if (filterFunc1(item)) {
      result.result1.push(item);
    } else if (filterFunc2(item)) {
      result.result2.push(item);
    }
    return result;
  },
  { result1: [], result2: [] },
);

Answer 5

If you don't want to use reduce , you may want to iterate the array once and acquire the filtered and unfiltered items in a single shot, using a plain efficient for..of loop:

 function filterAndDiscriminate(arr, filterCallback) { const res = [[],[]]; for (var item of arr) { res[~~filterCallback(item)].push(item); } return res; } const [unfiltered, filtered] = filterAndDiscriminate([1,2,3,4,5], i => i <= 3); console.log(filtered, unfiltered);

Answer 6

There's a way more simple and readable way to do this:

const array1 = []
const array2 = []
itemsToFilter.forEach(item => item.condition === met ? array1.push(challenge) : array2.push(challenge))