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如何在使用javascript的过滤器时获取排除的元素

filter arrays javascript

Rewind · 技术社区 · 3 年前

Javascript的 filter 返回包含所有通过测试的元素的数组。

如何在不重新运行测试的情况下轻松获取所有未通过测试的元素,但反过来呢?即使你必须再次运行测试,最好的方法是如何做到这一点。

let arr; // this is the array on which the filter will be run [SET ELSEWHERE]
let fn; // The filter function [SET ELSEWHERE]
let goodElements; // This will be the new array of the good elements passing the test
let badElements; // This will be the new array of the elements failing the test

goodElements = arr.filter(fn);

// SO HOW IS badElements set????

怎么样 badElements 设置

5 回复 | 直到 3 年前

Spectric amamoto 3 年前

如果不想进行两次迭代,可以使用 for 循环和三元运算符:

let arr = [1, 2, 3];
let fn = (e) => e % 2 == 0;
let goodElements = [];
let badElements = [];

for(const e of arr) (fn(e) ? goodElements : badElements).push(e);
console.log(goodElements);
console.log(badElements);

否则,只需使用反转条件 ! 操作人员

let arr = [1, 2, 3];
let fn = (e) => e % 2 == 0;
let goodElements;
let badElements;

goodElements = arr.filter(fn);
badElements = arr.filter(e => !fn(e));
console.log(goodElements);
console.log(badElements);

Barmar 3 年前

不使用 filter() 。如果你想把数据分成两个数组,那就自己做吧。

function partition(array, fn) {
  let goodArray = [],
    badArray = [];
  array.forEach(el => {
    if (fn(el)) {
      goodArray.push(el);
    } else {
      badArray.push(el);
    }
  });
  return [goodArray, badArray];
}

let [goodElemements, badElements] = partition(arr, fn);

您也可以使用 reduce()

function partition(array, fn) {
  return array.reduce(acc, el => {
    if (fn(el)) {
      acc[0].push(el);
    } else {
      acc[1].push(el);
    }
  }, [[],[]]);
}

let [goodElemements, badElements] = partition(arr, fn);

tarkh 3 年前

如果你想严格遵守 Array.filter 只有一个循环,然后考虑这样的事情:

UPD:基于 @AlvaroFlaÃ±oLarrondo 评论,为当前方法添加了外部条件函数。

// Array of elements
const words = ['spray', 'limit', 'elite', 'exuberant', 'destruction', 'present'];

// External filter method
const fn = e => e.length > 6;

// Define empty bad array
const bad = [];

// Define good array as result from filter function
const good = words.filter(word => {
  // In filter condition return good values
  if(fn(word)) return word;
  // And skip else values by just pushing them
  // to bad array, without returning
  else bad.push(word);
});

// Results
console.log(good);
console.log(bad);

Guerric P 3 年前

你可以使用 reduce 以便根据谓词将数组拆分为两个数组,如本例所示:

const arr = [1, 0, true, false, "", "foo"];
const fn = element => !element;

const [goodElements, badElements] = arr.reduce(
  ([truthies, falsies], cur) =>
    fn(cur) ? [truthies, [...falsies, cur]] : [[...truthies, cur], falsies],
  [[], []]
);

console.log(goodElements, badElements);

Snowmonkey 3 年前

我看到了两条可能的路线:

// in this case, if it's not in `goodElements`, it's a bad 'un
badElements = arr.filter( el => !goodElements.includes(el) );

或者这样:

// we don't **know** if fn needs the optional parameters, so we will
//  simply pass them. If it doesn't need 'em, they'll be ignored.
badElements = arr.filter( (el, idx, arr) => !fn(el, idx, arr) );

ManuelMB 3 年前

const arr = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
const fn = (x) => x % 2 === 0

const removeItems = (array, itemsToRemove) => {
    return array.filter(v => {
      return !itemsToRemove.includes(v);
    });
}
  
const goodElements = arr.filter(fn)
console.log(goodElements) // [ 0, 2, 4, 6, 8 ]

const badElements = removeItems(arr, goodElements)
console.log(badElements) // [ 1, 3, 5, 7, 9 ]