比较和过滤两个数组

Question

我一直在尝试实现一个给出两个数组的函数,

array1的元素用作过滤掉array2中元素的条件.

例如:

array1= [apple, grapes, oranges]

array2= [potato, pears, grapes, berries, apples, oranges]

在加入函数之后,array2应该有这样的元素:

filter_twoArrays(array1,array2)

array2= [grapes, apples, oranges]

我尝试了下面的代码,使用for循环和array.splice(),但我看到的问题是,当我使用splice方法时,似乎它改变了for循环中array2的长度:

function filter_twoArrays(filter,result){

  for(i=0; i< filter.length; i++){
    for(j=0; j< result.length; j++){
      if(filter[i] !== result[j]){
        result.splice(j,1)
      }
    }
  }

如何优化过滤功能将非常感谢任何输入

干杯!

Answer 1

您可以使用过滤器如下

var array1 = ['apples', 'grapes', 'oranges', 'banana'],
  array2 = ['potato', 'pears', 'grapes', 'berries', 'apples', 'oranges'];

var intersection = array1.filter(function(e) {
  return array2.indexOf(e) > -1;
});

console.log(intersection);

您还可以在Array原型上添加此方法,并直接在数组上调用它

Array.prototype.intersection = function(arr) {
  return this.filter(function(e) {
    return arr.indexOf(e) > -1;
  });
};

var array1 = ['apples', 'grapes', 'oranges', 'banana'],
  array2 = ['potato', 'pears', 'grapes', 'berries', 'apples', 'oranges'];

var intersection = array1.intersection(array2);
console.log(intersection);

Answer 2

您好，这是函数 array_intersect php 的移植。应该对你有好处 http://phpjs.org/functions/array_intersect/

function array_intersect(arr1) {
  //  discuss at: http://phpjs.org/functions/array_intersect/
  // original by: Brett Zamir (http://brett-zamir.me)
  //        note: These only output associative arrays (would need to be
  //        note: all numeric and counting from zero to be numeric)
  //   example 1: $array1 = {'a' : 'green', 0:'red', 1: 'blue'};
  //   example 1: $array2 = {'b' : 'green', 0:'yellow', 1:'red'};
  //   example 1: $array3 = ['green', 'red'];
  //   example 1: $result = array_intersect($array1, $array2, $array3);
  //   returns 1: {0: 'red', a: 'green'}

  var retArr = {},
    argl = arguments.length,
    arglm1 = argl - 1,
    k1 = '',
    arr = {},
    i = 0,
    k = '';

  arr1keys: for (k1 in arr1) {
    arrs: for (i = 1; i < argl; i++) {
      arr = arguments[i];
      for (k in arr) {
        if (arr[k] === arr1[k1]) {
          if (i === arglm1) {
            retArr[k1] = arr1[k1];
          }
          // If the innermost loop always leads at least once to an equal value, continue the loop until done
          continue arrs;
        }
      }
      // If it reaches here, it wasn't found in at least one array, so try next value
      continue arr1keys;
    }
  }

  return retArr;
}