删除在索引i 上的元素, 不突变原始数组 :

/**
* removeElement
* @param {Array} array
* @param {Number} index
*/
function removeElement(array, index) {
   return Array.from(array).splice(index, 1);
}

// Another way is
function removeElement(array, index) {
   return array.slice(0).splice(index, 1);
}

根据所有主要正确的答复并考虑到建议的最佳做法(特别是不直接使用Array.prototype),我提出了以下代码:

function arrayWithout(arr, values) {
  var isArray = function(canBeArray) {
    if (Array.isArray) {
      return Array.isArray(canBeArray);
    }
    return Object.prototype.toString.call(canBeArray) === '[object Array]';
  };

  var excludedValues = (isArray(values)) ? values : [].slice.call(arguments, 1);
  var arrCopy = arr.slice(0);

  for (var i = arrCopy.length - 1; i >= 0; i--) {
    if (excludedValues.indexOf(arrCopy[i]) > -1) {
      arrCopy.splice(i, 1);
    }
  }

  return arrCopy;
}

在审查上述功能时,尽管运作良好,但我意识到业绩可能有所改进。 使用ES6而不是ES5是一种更好的方法。

const arrayWithoutFastest = (() => {
  const isArray = canBeArray => ('isArray' in Array) 
    ? Array.isArray(canBeArray) 
    : Object.prototype.toString.call(canBeArray) === '[object Array]';

  let mapIncludes = (map, key) => map.has(key);
  let objectIncludes = (obj, key) => key in obj;
  let includes;

  function arrayWithoutFastest(arr, ...thisArgs) {
    let withoutValues = isArray(thisArgs[0]) ? thisArgs[0] : thisArgs;

    if (typeof Map !== 'undefined') {
      withoutValues = withoutValues.reduce((map, value) => map.set(value, value), new Map());
      includes = mapIncludes;
    } else {
      withoutValues = withoutValues.reduce((map, value) => { map[value] = value; return map; } , {}); 
      includes = objectIncludes;
    }

    const arrCopy = [];
    const length = arr.length;

    for (let i = 0; i < length; i++) {
      // If value is not in exclude list
      if (!includes(withoutValues, arr[i])) {
        arrCopy.push(arr[i]);
      }
    }

    return arrCopy;
  }

  return arrayWithoutFastest;  
})();

如何使用 :

const arr = [1,2,3,4,5,"name", false];

arrayWithoutFastest(arr, 1); // will return array [2,3,4,5,"name", false]
arrayWithoutFastest(arr, 'name'); // will return [2,3,4,5, false]
arrayWithoutFastest(arr, false); // will return [2,3,4,5]
arrayWithoutFastest(arr,[1,2]); // will return [3,4,5,"name", false];
arrayWithoutFastest(arr, {bar: "foo"}); // will return the same array (new copy)

我目前正在写博客文章, 其中我已设定数个无问题的阵列解决方案基准, 并比较运行时间。 一旦我完成此文章, 我将更新此答案, 并用链接更新。 仅供参考, 我比较了上述与没有 Lodash 的比较, 以防浏览器支持Map注意我没有用Array.prototype.indexOf或Array.prototype.includes将exlcude Values 包装在Map或Object让查询更快!

我还有一个从阵列中移除的好办法:

var words = ['spray', 'limit', 'elite', 'exuberant', 'destruction', 'present'];

const result = words.filter(word => word.length > 6);

console.log(result);
// expected output: Array ["exuberant", "destruction", "present"]

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/filter

let array = [5,5,4,4,2,3,4]    
let newArray = array.join(',').replace('5','').split(',')

如果您想要删除一个当前项目, 此示例将有效 。

我喜欢这个组合的版本, 以其值来去除元素, 使用$.inArray:

$(document).ready(function(){
    var arr = ["C#","Ruby","PHP","C","C++"];
    var itemtoRemove = "PHP";
    arr.splice($.inArray(itemtoRemove, arr),1);
});

    Array.prototype.remove = function(start, end) {
        var n = this.slice((end || start) + 1 || this.length);
        return this.length = start < 0 ? this.length + start : start,
        this.push.apply(this, n)
    }

start和end可以是负数。在这种情况下,它们会从数组的末尾计数。

如果只有start中指定,只删除一个元素。

函数返回新数组长度。

z = [0,1,2,3,4,5,6,7,8,9];

newlength = z.remove(2,6);

(8) [0, 1, 7, 8, 9]

z=[0,1,2,3,4,5,6,7,8,9];

newlength = z.remove(-4,-2);

(7) [0, 1, 2, 3, 4, 5, 9]

z=[0,1,2,3,4,5,6,7,8,9];

newlength = z.remove(3,-2);

(4) [0, 1, 2, 9]