根据对象属性将多个数组组合为一个答案

【问题标题】：Group many arrays into one based on object properties根据对象属性将多个数组组合为一个
【发布时间】：2019-09-09 00:17:51
【问题描述】：

我正在开发一个电子商务网站。我必须管理产品变化。 REST API 响应如下所示：

"variations": [
  {
    "attributeCombinations": [
      {
        "name": "Color",
        "valueName": "GRIS"
      },
      {
        "name": "Talle",
        "valueName": "L"
      }
    ]
  },
  {
    "attributeCombinations": [
      {
        "name": "Color",
        "valueName": "GRIS"
      },
      {
        "name": "Talle",
        "valueName": "XXL"
      }
    ]
  },
  {
    "attributeCombinations": [
      {
        "name": "Color",
        "valueName": "Azul"
      },
      {
        "name": "Talle",
        "valueName": "M"
      }
    ]
  },
  {
    "attributeCombinations": [
      {
        "name": "Color",
        "valueName": "GRIS"
      },
      {
        "name": "Talle",
        "valueName": "S"
      }
    ]
  },
  {
    "attributeCombinations": [
      {
        "name": "Color",
        "valueName": "GRIS"
      },
      {
        "name": "Talle",
        "valueName": "XL"
      }
    ]
  },
  {
    "attributeCombinations": [
      {
        "name": "Color",
        "valueName": "Azul"
      },
      {
        "name": "Talle",
        "valueName": "S"
      }
    ]
  },
  {
    "attributeCombinations": [
      {
        "name": "Color",
        "valueName": "GRIS"
      },
      {
        "name": "Talle",
        "valueName": "M"
      }
    ]
  }
]

这个 JSON 对我来说在前端工作效率很低。很难根据其类型（名称）和值（值名称）对变体进行分组，因为它们位于不同的“数组数组”上。

最终的结果应该是这样的：

variations: [
  {
    name: 'Color',
    values: ['Gris', 'Azul']
  },
  {
    name: 'Talle',
    values: ['S', 'M', 'L', 'XL', 'XXL']
  }
]

到目前为止，我已经达到了这个结果：

[
  { name: "Color", values: ["GRIS"] },
  { name: "Talle", values: ["L"] },
  { name: "Color", values: ["GRIS"] },
  { name: "Talle", values: ["XXL"] },
  { name: "Color", values: ["Azul"] },
  { name: "Talle", values: ["M"] },
  { name: "Color", values: ["GRIS"] },
  { name: "Talle", values: ["S"] },
  { name: "Color", values: ["GRIS"] },
  { name: "Talle", values: ["XL"] },
  { name: "Color", values: ["Azul"] },
  { name: "Talle", values: ["S"] },
  { name: "Color", values: ["GRIS"] },
  { name: "Talle", values: ["M"] }
]

基于此代码：

let variations = this.$page.product.variations;
let newArr = [];

variations.forEach(variation => {
        newArr.push(
          Object.values(
            variation.attributeCombinations.reduce(
              (result, { name, valueName }) => {
                // Create new group
                if (!result[name]) {
                  result[name] = {
                    name,
                    values: []
                  };
                }
                // Append to group
                result[name].values.push(valueName);

                return result;
              },
              {}
            )
          )
        );
      });
return newArr.reduce((acc, cur) => acc.concat(cur));

【问题讨论】：

您想要的输出与原始结构相比丢失了信息。例如，原始结构没有 Azul-XL 组合。此信息无法从您想要的输出中得出。你确定吗？
@trincot 是的！我注意到了这一点。但是，就目前而言，以及我将为此功能提供的特定用例，这并不是真正的问题。不过谢谢！
那么任何答案都符合您的需求吗？还是您还在寻找答案？
Naga 的回答正是我所需要的。非常感谢

标签： javascript arrays object ecmascript-2016

【解决方案1】：

为了达到预期的效果，请使用以下使用 reduce ， forEach 和每个循环的索引的选项

循环遍历变量数组
对于位置 0 的第一个索引，推送到累加器 (acc)
使用第二个循环，使用 .forEach 循环遍历属性组合
将 valueNames 推送到相应的名称 - Color 和 Talle

var arr = {"variations": [
  {
    "attributeCombinations": [
      {
        "name": "Color",
        "valueName": "GRIS"
      },
      {
        "name": "Talle",
        "valueName": "L"
      }
    ]
  },
  {
    "attributeCombinations": [
      {
        "name": "Color",
        "valueName": "GRIS"
      },
      {
        "name": "Talle",
        "valueName": "XXL"
      }
    ]
  },
  {
    "attributeCombinations": [
      {
        "name": "Color",
        "valueName": "Azul"
      },
      {
        "name": "Talle",
        "valueName": "M"
      }
    ]
  },
  {
    "attributeCombinations": [
      {
        "name": "Color",
        "valueName": "GRIS"
      },
      {
        "name": "Talle",
        "valueName": "S"
      }
    ]
  },
  {
    "attributeCombinations": [
      {
        "name": "Color",
        "valueName": "GRIS"
      },
      {
        "name": "Talle",
        "valueName": "XL"
      }
    ]
  },
  {
    "attributeCombinations": [
      {
        "name": "Color",
        "valueName": "Azul"
      },
      {
        "name": "Talle",
        "valueName": "S"
      }
    ]
  },
  {
    "attributeCombinations": [
      {
        "name": "Color",
        "valueName": "GRIS"
      },
      {
        "name": "Talle",
        "valueName": "M"
      }
    ]
  }
]
          }


console.log(arr.variations.reduce((acc, val, idx) => {
  val.attributeCombinations.forEach((v, i) => {
  if(idx === 0) {
    acc.push({name: v.name , valueName:  Array.of(v.valueName)})
  }else{
    i === 0 ? acc[0].valueName.push(v.valueName) : acc[1].valueName.push(v.valueName) 
  }
})
  return acc
},[]).map(v => v = {name :v.name,  valueName: [...new Set(v.valueName)]}));

codepen - https://codepen.io/nagasai/pen/ZZxVzp?editors=1010

【讨论】：

【解决方案2】：

您可以使用函数reduce 进行分组，使用函数Object.values 提取值。

Set 对象提供了一种删除重复值的方法。

let variations = [   {     "attributeCombinations": [       {         "name": "Color",         "valueName": "GRIS"       },       {         "name": "Talle",         "valueName": "L"       }     ]   },   {     "attributeCombinations": [       {         "name": "Color",         "valueName": "GRIS"       },       {         "name": "Talle",         "valueName": "XXL"       }     ]   },   {     "attributeCombinations": [       {         "name": "Color",         "valueName": "Azul"       },       {         "name": "Talle",         "valueName": "M"       }     ]   },   {     "attributeCombinations": [       {         "name": "Color",         "valueName": "GRIS"       },       {         "name": "Talle",         "valueName": "S"       }     ]   },   {     "attributeCombinations": [       {         "name": "Color",         "valueName": "GRIS"       },       {         "name": "Talle",         "valueName": "XL"       }     ]   },   {     "attributeCombinations": [       {         "name": "Color",         "valueName": "Azul"       },       {         "name": "Talle",         "valueName": "S"       }     ]   },   {     "attributeCombinations": [       {         "name": "Color",         "valueName": "GRIS"       },       {         "name": "Talle",         "valueName": "M"       }     ]   } ];
let result = Object.values(variations.reduce((a, {attributeCombinations: combinations}) => {
  combinations.forEach(({name, valueName}) => {
    (a[name] || (a[name] = {name, values: new Set()})).values.add(valueName);
  });
  
  return a;
}, {}));

console.log(Array.from(result[0].values));// for understanding the object set
console.log(result);

.as-console-wrapper { min-height: 100%; }

【讨论】：

我不知道 Set 对象。你的代码真的很干净！结果并不完全符合我的需要，但非常感谢您的帮助。谢谢！

【解决方案3】：

您可以使用reduce 循环遍历数组并汇总为一个对象。使用forEach 循环通过attributeCombinations 并构造所需的输出。使用Object.values将对象转化为数组。

var variations = [{"attributeCombinations":[{"name":"Color","valueName":"GRIS"},{"name":"Talle","valueName":"L"}]},{"attributeCombinations":[{"name":"Color","valueName":"GRIS"},{"name":"Talle","valueName":"XXL"}]},{"attributeCombinations":[{"name":"Color","valueName":"Azul"},{"name":"Talle","valueName":"M"}]},{"attributeCombinations":[{"name":"Color","valueName":"GRIS"},{"name":"Talle","valueName":"S"}]},{"attributeCombinations":[{"name":"Color","valueName":"GRIS"},{"name":"Talle","valueName":"XL"}]},{"attributeCombinations":[{"name":"Color","valueName":"Azul"},{"name":"Talle","valueName":"S"}]},{"attributeCombinations":[{"name":"Color","valueName":"GRIS"},{"name":"Talle","valueName":"M"}]}];

var result = Object.values(variations.reduce((c, v) => {
  v.attributeCombinations.forEach(({name,valueName}) => {
    c[name] = c[name] || {name,values: []}
    c[name].values.push(valueName);
  });
  return c;
}, {}));

console.log(result);

如果您想要唯一值，可以使用Set 并添加另一个循环将Set 对象转换为数组。

var variations = [{"attributeCombinations":[{"name":"Color","valueName":"GRIS"},{"name":"Talle","valueName":"L"}]},{"attributeCombinations":[{"name":"Color","valueName":"GRIS"},{"name":"Talle","valueName":"XXL"}]},{"attributeCombinations":[{"name":"Color","valueName":"Azul"},{"name":"Talle","valueName":"M"}]},{"attributeCombinations":[{"name":"Color","valueName":"GRIS"},{"name":"Talle","valueName":"S"}]},{"attributeCombinations":[{"name":"Color","valueName":"GRIS"},{"name":"Talle","valueName":"XL"}]},{"attributeCombinations":[{"name":"Color","valueName":"Azul"},{"name":"Talle","valueName":"S"}]},{"attributeCombinations":[{"name":"Color","valueName":"GRIS"},{"name":"Talle","valueName":"M"}]}]

var result = Object.values(variations.reduce((c, v) => {
  v.attributeCombinations.forEach(({name,valueName}) => {
    c[name] = c[name] || {name,values: new Set}
    c[name].values.add(valueName);
  });
  return c;
}, {})).map(o => {
  o.values = [...o.values.values()]
  return o;
});

console.log(result);

【讨论】：