Javascript算法查找数组中不在另一个数组中的元素

I'm looking for a good algorithm to get all the elements in one array that are not elements in another array. So given these arrays:

我正在寻找一种好的算法来获取一个数组中不是另一个数组中的元素的所有元素。所以给定这些数组：

var x = ["a","b","c","t"];
var y = ["d","a","t","e","g"];

I want to end up with this array:

我想最终得到这个数组：

var z = ["d","e","g"];

I'm using jquery, so I can take advantage of $.each() and $.inArray(). Here's the solution I've come up with, but it seems like there should be a better way.

我正在使用jquery，所以我可以利用$ .each（）和$ .inArray（）。这是我提出的解决方案，但似乎应该有更好的方法。

// goal is to get rid of values in y if they exist in x
var x = ["a","b","c","t"];
var y = ["d","a","t","e","g"];

var z = [];
$.each(y, function(idx, value){
  if ($.inArray(value,x) == -1) {
    z.push(value);
  }
});
alert(z);  // should be ["d","e","g"]

Here is the code in action. Any ideas?

这是代码的实际应用。有任何想法吗？

7 个解决方案

#1

var z = $.grep(y, function(el){return $.inArray(el, x) == -1});

Also, that method name is too short for its own good. I would expect it to mean isElementInArray, not indexOf.

此外，该方法名称太短，不适合自己的好处。我希望它的意思是isElementInArray，而不是indexOf。

For a demo with objects, see http://jsfiddle.net/xBDz3/6/

有关对象的演示，请参阅http://jsfiddle.net/xBDz3/6/

#2

Late answer with the new ECMA5 javascript:

使用新的ECMA5 javascript后期回答：

var x = ["a","b","c","t"];
var y = ["d","a","t","e","g"];

myArray = y.filter( function( el ) {
  return x.indexOf( el ) < 0;
});

#3

Here's an alternative using underscore.js:

这是使用underscore.js的替代方法：

function inAButNotInB(A, B) {
  return _.filter(A, function (a) {
    return !_.contains(B, a);
  });
}

#4

in ES6 simply

在ES6中简单

var x = ["a", "b", "c", "t"];
var y = ["d", "a", "t", "e", "g"];

console.log( y.filter(e => !x.includes(e)) );

(another option is y.filter(e => x.indexOf(e)===-1) )

（另一种选择是y.filter（e => x.indexOf（e）=== - 1））

#5

Maybe jLinq can help you?

也许jLinq可以帮到你？

It lets you run queries like this against javascript objects.

它允许您针对javascript对象运行这样的查询。

For example:

例如：

var users = [ { name: "jacob", age: 25 },  { name: "bob" , age: 30 }]
var additionalusers = [ { name: "jacob", age: 25 },  { name: "bill" , age: 25 }]

var newusers = jLinq.from(users).except(additionalusers).select();

>>> newusers = [ { name: "bob" , age: 30 } ]

It's a bit overkill for you at the moment, but it's a robust solution that I was glad to learn about.

这对你来说有点矫枉过正，但这是一个我很乐意了解的强大解决方案。

It can do intersects, unions, handle boolean logic and all kinds of great linq style goodness.

它可以做交叉，联合，处理布尔逻辑和各种伟大的linq风格的善良。

#6

This is a late answer, but it uses no libraries so some may find it helpful.

这是一个迟到的答案，但它没有使用库，所以有些人可能会发现它有用。

/**
 * Returns a non-destructive Array of elements that are not found in
 * any of the parameter arrays.
 *
 * @param {...Array} var_args   Arrays to compare.
 */
Array.prototype.uniqueFrom = function() {
  if (!arguments.length)
    return [];
  var a1 = this.slice(0); // Start with a copy

  for (var n=0; n < arguments.length; n++) {
    var a2 = arguments[n];
    if (!(a2 instanceof Array))
      throw new TypeError( 'argument ['+n+'] must be Array' );

    for(var i=0; i<a2.length; i++) {
      var index = a1.indexOf(a2[i]);
      if (index > -1) {
        a1.splice(index, 1);
      } 
    }
  }
  return a1;
}

Example:

例：

var sheetUsers = ['joe@example.com','fred@example.com','sam@example.com'];
var siteViewers = ['joe@example.com','fred@example.com','lucy@example.com'];
var viewersToAdd = sheetUsers.uniqueFrom(siteViewers);  // [sam@example.com]
var viewersToRemove = siteViewers.uniqueFrom(sheetUsers);  // [lucy@example.com]

#7

Make sorted copies of the arrays first. If the top elements are equal, remove them both. Otherwise remove the element that is less and add it to your result array. If one array is empty, then add the rest of the other array to the result and finish. You can iterate through the sorted arrays instead of removing elements.

首先制作数组的排序副本。如果顶部元素相等，则将它们都删除。否则，删除较少的元素并将其添加到结果数组中。如果一个数组为空，则将另一个数组的其余部分添加到结果中并完成。您可以遍历已排序的数组而不是删除元素。

// assume x and y are sorted
xi = 0; yi = 0; xc = x.length; yc = y.length;
while ( xi < xc && yi < yc ) {
  if ( x[xi] == y[yi] ) {
    xi += 1;
    yi += 1;
  } else if ( x[xi] < y[yi] ) {
    z.push( x[xi] );
    xi += 1;
  } else {
    z.push( y[yi] );
    yi += 1;
  }
}
// add remainder of x and y to z.  one or both will be empty.

#1

var z = $.grep(y, function(el){return $.inArray(el, x) == -1});