Question

我有以下对象数组，

var data = [
    {
        label: "Book1",
        data: "US edition"
    },
    {
        label: "Book1",
        data: "UK edition"
    },
    {
        label: "Book2",
        data: "CAN edition"
    }
];

我想根据属性'label'合并重复的对象所以最终输出将如下所示，

var data = [
    {
        label: "Book1",
        data: ["US edition", "UK edition"] //data attribute is merged
    },
    {
        label: "Book2",
        data: "CAN edition"
    }
];

有人可以帮我确定方法吗？

Answer 1

我可能会使用filter进行循环，跟踪我之前看到的对象地图，沿着这些行（编辑以反映您同意是的，有意义的是(entry).data始终是一个数组）：

var seen = {};
data = data.filter(function(entry) {
    var previous;

    // Have we seen this label before?
    if (seen.hasOwnProperty(entry.label)) {
        // Yes, grab it and add this data to it
        previous = seen[entry.label];
        previous.data.push(entry.data);

        // Don't keep this entry, we've merged it into the previous one
        return false;
    }

    // entry.data probably isn't an array; make it one for consistency
    if (!Array.isArray(entry.data)) {
        entry.data = [entry.data];
    }

    // Remember that we've seen it
    seen[entry.label] = entry;

    // Keep this one, we'll merge any others that match into it
    return true;
});

在ES6环境中，我使用seen = new Map()而不是seen = {}。

注意：Array.isArray是由ES5定义的，因此像IE8这样的一些相当老的浏览器都没有。它可以很容易地进行填充/填充，但是：

if (!Array.isArray) {
    Array.isArray = (function() {
        var toString = Object.prototype.toString;
        return function(a) {
            return toString.call(a) === "[object Array]";
        };
    })();
}

旁注：即使我没有看到两个值，我也可能始终使entry.data成为数组，因为数据一致结构更容易处理。我上面没有这样做，因为当只有一个匹配的条目时，你的最终结果显示data只是一个字符串。 （我们现在已经完成了这个。）

实例（ES5版）：

＆＃13;

var data = [
    {
        label: "Book1",
        data: "US edition"
    },
    {
        label: "Book1",
        data: "UK edition"
    },
    {
        label: "Book2",
        data: "CAN edition"
    }
];
snippet.log("Before:");
snippet.log(JSON.stringify(data, null, 2), "pre");
var seen = {};
data = data.filter(function(entry) {
    var previous;

    // Have we seen this label before?
    if (seen.hasOwnProperty(entry.label)) {
        // Yes, grab it and add this data to it
        previous = seen[entry.label];
        previous.data.push(entry.data);

        // Don't keep this entry, we've merged it into the previous one
        return false;
    }

    // entry.data probably isn't an array; make it one for consistency
    if (!Array.isArray(entry.data)) {
        entry.data = [entry.data];
    }

    // Remember that we've seen it
    seen[entry.label] = entry;

    // Keep this one, we'll merge any others that match into it
    return true;
});
snippet.log("After:");
snippet.log(JSON.stringify(data, null, 2), "pre");

＆＃13;

<!-- Script provides the `snippet` object, see http://meta.stackexchange.com/a/242144/134069 -->
<script src="http://tjcrowder.github.io/simple-snippets-console/snippet.js"></script>

＆＃13;

Answer 2

尝试使用以下方法，它可以找到并且很短

 var data = [
    {
        label: "Book1",
        data: "US edition"
    },
    {
        label: "Book1",
        data: "UK edition"
    },
    {
        label: "Book2",
        data: "CAN edition"
    }
];

const result = Array.from(new Set(data.map(s => s.label)))
    .map(lab => {
      return {
        label: lab,
        data: data.filter(s => s.label === lab).map(edition => edition.data)
      }
    })

console.log(result);

Answer 3

此代码在最新版本的Firefox上测试。要在其他浏览器上工作，请将库的Array.isArray更改为lodash或您喜欢的任何内容。

var data = [
    {
        label: "Book1",
        data: "US edition"
    },
    {
        label: "Book1",
        data: "UK edition"
    },
    {
        label: "Book2",
        data: "CAN edition"
    }
],
i = 0,
j = data.length - 1,
current;

for (;i < data.length; i++) {
  current = data[i];
  for (;j > i; j--) {
     if (current.label === data[j].label) {
       if (Array.isArray(current.data)) {
         current.data = current.data.concat([data[j].data]);
       } else {
         current.data = [].concat([data[j].data, current.data]);
       }
       data.splice(j, 1);
     }

  }
} 

console.log(data);

Answer 4

我遇到了同样的情况，我希望在这里使用Set代替我的

const data = [
  {
    label: "Book1",
    data: "US edition"
  },
  {
    label: "Book1",
    data: "UK edition"
  },
  {
    label: "Book2",
    data: "CAN edition"
  },
  {
    label: "Book3",
    data: "CAN edition"
  },
  {
    label: "Book3",
    data: "CANII edition"
  }
];

const filteredArr = data.reduce((acc, current) => {
  const x = acc.find(item => item.label === current.label);
  if (!x) {
    const newCurr = {
      label: current.label,
      data: [current.data]
    }
    return acc.concat([newCurr]);
  } else {
    const currData = x.data.filter(d => d === current.data);
    if (!currData.length) {
      const newData = x.data.push(current.data);
      const newCurr = {
        label: current.label,
        data: newData
      }
      return acc;
    } else {
      return acc;
    }
    
  }
}, []);

console.log(filteredArr);

Answer 5

var data = [
    {
        label: "Book1",
        data: "US edition"
    },
    {
        label: "Book1",
        data: "UK edition"
    },
    {
        label: "Book2",
        data: "CAN edition"
    }
].filter(function(e, index ,b) {
    k = b.map(z => z.label)
    if(k.includes(e.label, index+1)){
        indexPosition = k.indexOf(e.label, index+1);
        b[indexPosition].data = (e.data+"||"+b[indexPosition].data);
        
    }else{
        e.data = e.data.split("||").length > 1 ? e.data.split("||") : e.data;
        return e;
    }
});

console.log(data)

合并对象数组中的重复对象

5 个答案: