是否有一种方法可以在JavaScript中返回两个数组之间的差异?
例如:
var a1 = ['a', 'b'];
var a2 = ['a', 'b', 'c', 'd'];
// need ["c", "d"]
是否有一种方法可以在JavaScript中返回两个数组之间的差异?
例如:
var a1 = ['a', 'b'];
var a2 = ['a', 'b', 'c', 'd'];
// need ["c", "d"]
当前回答
使用http://phrogz.net/JS/ArraySetMath.js你可以:
var array1 = ["test1", "test2","test3", "test4"];
var array2 = ["test1", "test2","test3","test4", "test5", "test6"];
var array3 = array2.subtract( array1 );
// ["test5", "test6"]
var array4 = array1.exclusion( array2 );
// ["test5", "test6"]
其他回答
数据:
var new_storage = JSON.parse('[{"id_order":"0003"},{"id_order":"0004"},{"id_order":"0006"}]');
var old_storage = JSON.parse('[{"id_order":"0001"},{"id_order":"0002"},{"id_order":"0003"},{"id_order":"0004"},{"id_order":"0005"}]');
使用过滤器:
var diff = new_storage
.filter(x => {if(!(old_storage.filter(y => y.id_order==x.id_order)).length){return x}})
.concat(old_storage
.filter(x => {if(!(new_storage.filter(y => y.id_order==x.id_order)).length){return x}})
)
console.log(JSON.stringify(diff))
两个数组的结果不同
[{"id_order":"0006"},{"id_order":"0001"},{"id_order":"0002"},{"id_order":"0005"}]
我一直在寻找一个不涉及使用不同库的简单答案,我想出了我自己的答案,我想这里没有提到过。 我不知道它的效率如何,但它确实有效;
function find_diff(arr1, arr2) {
diff = [];
joined = arr1.concat(arr2);
for( i = 0; i <= joined.length; i++ ) {
current = joined[i];
if( joined.indexOf(current) == joined.lastIndexOf(current) ) {
diff.push(current);
}
}
return diff;
}
对于我的代码,我也需要删除副本,但我想这并不总是可取的。
我想主要的缺点是它可能会比较许多已经被拒绝的选择。
我在这里读到的答案有很多问题,使得它们在实际编程应用中价值有限。
First and foremost, you're going to want to have a way to control what it means for two items in the array to be "equal". The === comparison is not going to cut it if you're trying to figure out whether to update an array of objects based on an ID or something like that, which frankly is probably one of the most likely scenarios in which you will want a diff function. It also limits you to arrays of things that can be compared with the === operator, i.e. strings, ints, etc, and that's pretty much unacceptable for grown-ups.
其次,diff操作有三种状态结果:
在第一个数组中但不在第二个数组中的元素 两个数组共用的元素 在第二个数组中但不在第一个数组中的元素
我认为这意味着你需要不少于2个循环,但我愿意接受肮脏的技巧,如果有人知道如何将其减少到一个。
这里是我拼凑的一些东西,我想强调的是,我绝对不在乎它在旧版本的Microshaft浏览器中不起作用。如果您在IE这样的较差的编码环境中工作,那么您就可以自行修改它,使其在您无法满意的限制范围内工作。
Array.defaultValueComparison = function(a, b) {
return (a === b);
};
Array.prototype.diff = function(arr, fnCompare) {
// validate params
if (!(arr instanceof Array))
arr = [arr];
fnCompare = fnCompare || Array.defaultValueComparison;
var original = this, exists, storage,
result = { common: [], removed: [], inserted: [] };
original.forEach(function(existingItem) {
// Finds common elements and elements that
// do not exist in the original array
exists = arr.some(function(newItem) {
return fnCompare(existingItem, newItem);
});
storage = (exists) ? result.common : result.removed;
storage.push(existingItem);
});
arr.forEach(function(newItem) {
exists = original.some(function(existingItem) {
return fnCompare(existingItem, newItem);
});
if (!exists)
result.inserted.push(newItem);
});
return result;
};
如果你的数组包含对象,如果你想比较一个属性,就会变得有点困难。
幸运的是,lodash使用_contains和_.pluck使这非常简单:
var list1 = [{id: 1},{id: 2}];
var list1 = [{id: 1},{id: 2}, {id: 3}];
//es6
var results = list2.filter(item => {
return !_.contains(_.pluck(list1, 'id'), item.id);
});
//es5
var results = list2.filter(function(item){
return !_.contains(_.pluck(list1, 'id'), item.id);
});
//results contains [{id: 3}]
这是目前为止最简单的方法来得到你正在寻找的结果,使用jQuery:
var diff = $(old_array).not(new_array).get();
Diff现在包含了old_array中不在new_array中的内容