检测数组中的对象是否具有一个相同的属性，但与其他对象具有不同的其他属性

Question

拿着这个数组：

[
  {"color": "blue","type": 1},
  {"color": "red","type": 1},
  {"color": "blue","type": 1},
  {"color": "green","type": 1},
  {"color": "green","type": 1},
  {"color": "red","type": 2},
  {"color": "red","type": 1},
  {"color": "green","type": 2},
  {"color": "red","type": 3},
];

如何查找数组中哪个“颜色”具有不同的“类型”(与所有其他具有相同“名称”的对象相比)？

我希望能够遍历这个数组并创建第二个数组，如下所示：

{red, green}

请注意，蓝色是通用的，因为所有具有“颜色”的对象都有相同的“类型”。

我得到的最接近的是：https://jsfiddle.net/0wgjs5zh/，但是它将所有颜色添加到添加了不同类型的数组中：

arr.forEach(function(item){
  if(newArr.hasOwnProperty(item.color+ '-' +item.type)) {
   // newArr[item.color+ '-' +item.type].push(item);
  }
  else {
    newArr[item.color+ '-' +item.type] = item;
  }
});

// RESULT
{blue-1, green-1, green-2, red-1, red-2, red-3}

Answer 1

您可以使用两次传递，一次用于集合，一次用于生成结果数组。

var array = [{ "color": "blue", "type": 1 }, { "color": "red", "type": 1 }, { "color": "blue", "type": 1 }, { "color": "green", "type": 1 }, { "color": "green", "type": 1 }, { "color": "red", "type": 2 }, { "color": "red", "type": 1 }, { "color": "green", "type": 2 }, { "color": "red", "type": 3 }, ],
    result,
    object = Object.create(null);

array.forEach(function (a) {
    object[a.color] = object[a.color] || {};
    object[a.color][a.type] = true;
});

result = Object.keys(object).filter(function (k) {
    return Object.keys(object[k]).length > 1;
});

console.log(result);

​

Answer 2

我的解决方案如下所示

​

var cls = [
  {"color": "blue","type": 1},
  {"color": "red","type": 1},
  {"color": "blue","type": 1},
  {"color": "green","type": 1},
  {"color": "green","type": 1},
  {"color": "red","type": 2},
  {"color": "red","type": 1},
  {"color": "green","type": 2},
  {"color": "red","type": 3},
],

    map = cls.reduce((p,c) => (p[c.color] ? !~p[c.color].indexOf(c.type) && p[c.color].push(c.type)
                                          : p[c.color] = [c.type], p),{}),
 result = Object.keys(map).reduce((p,c) => map[c].length > 1 && p.concat(c) || p,[]);

console.log(result);

​