比较在JavaScript 2多维数组 [英] Compare two multidimensional arrays in javascript

问题描述

我有两个数组：

var array_old = [{id:"5436", title:"I Like you boy"}, {id:"5437", title:"Hello how are you"}];
var array_new = [{id:"5436", title:"I Like you boy"}, {id:"1132", title:"I'm fine"}];

$.each(array_old, function(id, array)
{
    if(!$.inArray(array['id'], array_new, 1)>-1){
        alert(array['id'] + " does not exist in array_new");
    }
});

我要检查如果array_new存在array_old的ID，所以我在这个例子中期待code输出5437中array_new不存在。

I want to check if the IDs of array_old exist in array_new, so I'm expecting the code to output "5437 does not exist in array_new" in this example.

我找不到可以让我做的任何功能，所以我应该怎么办呢？

I can't find any function that would allow me to do that, so how should I do it?

推荐答案

取决于如何大的阵列 - 你可能想使用更高性能的解决方案。

Depends on how big your arrays are - you might want to use a more performant solution.

• 最简单的方法（这你和@Tebb找到）具有Θ（N * M）


• 如果您将优化这一点（打破了，如果你没[不]研究发现元素 - 见@gonchuki），你仍然在 O（N * M）


• 您可以假设两个数组都是以相同的顺序，并且只运行一个循环： O（分钟（N，M））。如果你需要在这之前给他们排序，你会得到 O（N *登录N + M *日志M）。


• 最好将使用一个哈希表 O（1）查找，致使 O（N + M）。您可以轻松地使用JS对象为：

• The easiest solution (which both you and @Tebb found) has Θ(n*m)
• If you would optimize this a bit (breaking out if you did [not] found the element - see @gonchuki), you are still at O(n*m)
• You could assume that both arrays are in the same order, and run only one loop: O(min(n,m)). If you'd need to sort them before that, you'd get O(n*log n+m*log m).
• Best would be using a hash table for O(1) lookup, resulting in O(n+m). You can easily use a JS object for that:

var counts = {};
for (var i=0; i<array_new.length; i++)
    counts[array_new[i].id] = (counts[array_new[i].id] || 0) + 1;

return array_old.every(function(item) {
    return item.id in counts && counts[item.id]--;
});

（演示）

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