Question

所以我有一个对象数组，我创建了两个字典来映射名称。我想找到所有具有相同键的对象并将其合并。同时连接该值。这是我做过的，但是卡住了：

const wordcloudData = {
  'pretty cool': [3, 1, ['161', '329']],
  'pretty damn': [2, 1, ['111', '131']],
  'pretty nice': [1, 1, ['211', '499']],
  'great': [4, 1, ['18', '19']],
};

const dict = {
  'pretty cool': 1, 
  'pretty damn': 1,
  'pretty nice': 1,
};

const dictNames = {
  1: 'nice',
}


const formattedArray = _
  .chain(wordcloudData)
  .keys()
  .map(item => {
    const [weight, color, reviews] = wordcloudData[item];

    return {
      name: dictNames[dict[item]] || item,
      weight: weight,
      color: color,
      reviews: reviews,
    }
  })
  /* Here i'm getting that type of array:
   [ 
    { name: 'nice', weight: 1, color: 1, reviews: [ '211', '499' ] },
    { name: 'nice', weight: 2, color: 1, reviews: [ '111', '131' ] },
    { name: 'nice', weight: 3, color: 1, reviews: [ '161', '329' ] },
    { name: 'great', weight: 4, color: 1, reviews: [ '18', '19' ] } 
   ]
  */
  .groupBy('name')
  .map(_.spread(_.assign)) // Here i'm trying to get rid of objects with same key, but something goes wrong
  .value();

我认为删除重复项时做错了吗？我下一步该怎么办？

要解释我想要的内容，将显示我想要的数组：

对象的初始数组：

{
 'pretty cool': [3, 1, ['161', '329']],
 'pretty damn': [2, 1, ['111', '131']],
 'pretty nice': [1, 1, ['211', '499']],
 'great': [4, 1, ['18', '19']],
};

对象的结果数组：

{ name: 'nice', weight: 6, color: 1, reviews: [ '161', '329', '111', '131', '211', '499'] },
{ name: 'great', weight: 4, color: 1, reviews: [ '18', '19' ] }

Answer 1

使用_.mergeWith()并根据键选择如何组合值：

const wordcloudData = {
  'pretty cool': [3, 1, ['161', '329']],
  'pretty damn': [2, 1, ['111', '131']],
  'pretty nice': [1, 1, ['211', '499']],
  'great': [4, 1, ['18', '19']],
};

const dict = {
  'pretty cool': 1, 
  'pretty damn': 1,
  'pretty nice': 1,
};

const dictNames = {
  1: 'nice',
}

const merger = _.partial(_.invoke, {
  weight: _.add,
  reviews: _.concat,
});

const formattedArray = _(wordcloudData)
  .map(([weight, color, reviews], item) => ({
    name: _.get(dictNames, dict[item], item),
    weight,
    color,
    reviews
  }))
  .groupBy('name')
  .map(v => _.mergeWith(...v, (o, s, k) => merger(k, o, s)))
  .value();
  
console.log(formattedArray);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>

Answer 2

这里有一个没有lodash并使用reduce()的版本：

const wordcloudData = {
  'pretty cool': [3, 1, ['161', '329']],
  'pretty damn': [2, 1, ['111', '131']],
  'pretty nice': [1, 1, ['211', '499']],
  'great': [4, 1, ['18', '19']],
};

const dict = {
  'pretty cool': 1, 
  'pretty damn': 1,
  'pretty nice': 1,
};

const dictNames = {
  1: 'nice',
}

let merged = Object.keys(wordcloudData).reduce((res, curr) =>
{
    let new_name = dictNames[dict[curr]] || curr;
    let [weigth, color, review] = wordcloudData[curr];
    let found = res.findIndex(x => x.name === new_name);
    
    if (found >= 0)
    {        
        res[found].weigth += weigth;
        res[found].color = color;
        res[found].review.push(...review);
    }
    else
    {
        res.push({
            name: new_name,
            weigth: weigth,
            color: color,
            review: review
        });
    }

    return res;

}, []);

console.log(merged);

Answer 3

这是一个使用纯ES6的简短解决方案，它将合并逻辑分为其自己的功能：https://jsbin.com/qevohugene/edit?js,console

const wordcloudData = {
  'pretty cool': [3, 1, ['161', '329']],
  'pretty damn': [2, 1, ['111', '131']],
  'pretty nice': [1, 1, ['211', '499']],
  'great': [4, 1, ['18', '19']],
};

const dict = {
  'pretty cool': 1, 
  'pretty damn': 1,
  'pretty nice': 1,
};

const dictNames = {
  1: 'nice',
}

const mergeItems = (oldItem, newItem) => ({
  name: newItem.name,
  weight: newItem.weight + oldItem.weight,
  color: newItem.color,
  reviews: [...oldItem.reviews, ...newItem.reviews]
});

const formattedArray = Object.entries(wordcloudData)
  .map(([key, [weight, color, reviews]]) => ({
      name: dictNames[dict[key]] || key,
      weight: weight,
      color: color,
      reviews: reviews,
    }))
  .reduce((accum, item) => {
    return accum.some(x => x.name === item.name)
      ? accum.map(x => x.name === item.name ? mergeItems(x, item) : x)
      : [...accum, item];
  }, []);
console.log(formattedArray)

希望有帮助。

javascript通过相同的键查找对象并将其合并

3 个答案: