这应该能奏效:

var mergedList = _.map(a1, function(item){
    return _.extend(item, _.findWhere(a2, { id: item.id }));
});

这假设a1中第二个对象的id应该是2而不是“2”

假设ID是字符串，并且顺序无关紧要，您可以

1. 创建哈希表。
2. 迭代这两个数组，并将数据存储在哈希表中，由ID索引。 如果已经有一些具有该ID的数据，请使用object.assign(ES6，可以是多填充的）更新它。
3. 获取包含哈希映射值的数组。

var hash = Object.create(null);
a1.concat(a2).forEach(function(obj) {
    hash[obj.id] = Object.assign(hash[obj.id] || {}, obj);
});
var a3 = Object.keys(hash).map(function(key) {
    return hash[key];
});

在ECMAScript6中，如果ID不一定是字符串，则可以使用map:

var hash = new Map();
a1.concat(a2).forEach(function(obj) {
    hash.set(obj.id, Object.assign(hash.get(obj.id) || {}, obj))
});
var a3 = Array.from(hash.values());

ES6简化了这一点:

let merge = (obj1, obj2) => ({...obj1, ...obj2});

注意重复的键将被合并，并且以第二个对象的值为准，忽略第一个对象的重复值。

示例:

let obj1 = {id: 1, uniqueObj1Key: "uniqueKeyValueObj1", repeatedKey: "obj1Val"};
let obj2 = {id: 1, uniqueObj2Key: "uniqueKeyValueObj2", repeatedKey: "obj2Val"};

merge(obj1, obj2)
// {id: 1, uniqueObj1Key: "uniqueKeyValueObj1", repeatedKey: "obj2Val", uniqueObj2Key: "uniqueKeyValueObj2"}
merge(obj2, obj1)
// {id: 1, uniqueObj2Key: "uniqueKeyValueObj2", repeatedKey: "obj1Val", uniqueObj1Key: "uniqueKeyValueObj1"}

var a1 = [{ id : 1, name : "test"}, { id : 2, name : "test2"}]
var a2 = [{ id : 1, count : "1"}, {id : 2, count : "2"}]
var merge = (obj1, obj2) => ({...obj1, ...obj2});
_.zipWith(a1, a2, merge)
(2) [{…}, {…}]
   0: {id: 1, name: "test", count: "1"}
   1: {id: 2, name: "test2", count: "2"}

如果您有一组数组要合并，您可以这样做:

var arrayOfArraysToMerge = [a1, a2, a3, a4]; //a3 and a4 are arrays like a1 and a2 but with different properties and same IDs.
_.zipWith(...arrayOfArraysToMerge, merge)
(2) [{…}, {…}]
   0: {id: 1, name: "test", count: "1", extra1: "val1", extra2: 1}
   1: {id: 2, name: "test2", count: "2", extra1: "val2", extra2: 2}