Question

我有一个集合A和数组B，其结构如下：

A：

{
    "_id" : ObjectId("5160757496cc6207a37ff778"),
    "name" : "Pomegranate Yogurt Bowl",
    "description" : "A simple breakfast bowl made with Greek yogurt, fresh pomegranate juice, puffed quinoa cereal, toasted sunflower seeds, and honey."
},
{
  "_id": ObjectId("5160757596cc62079cc2db18"),
  "name": "Krispy Easter Eggs",
  "description": "Imagine the Easter Bunny laying an egg.     Wait. That’s not anatomically possible.     And anyway, the Easter Bunny is a b..."
}

B：

var names = ["egg", "garlic", "cucumber", "kale", "pomegranate", "sunflower", "fish", "pork", "apple", "sunflower", "strawberry", "banana"]

我的目标是从A返回一个文档，该文档在数组B中存在最多的单词。在这种情况下，它应该返回第一个"_id" : ObjectId("5160757496cc6207a37ff778")。

我不确定如何解决这个问题：

这不起作用：

db.A.find({
    "description": {
      "$in": names
    }
  }, function(err, data) {
    if (err) console.log(err);
    console.log(data);
  });

Answer 1

这取决于您想要投入的“单词”类型，以及它们是否被视为“停止单词”，例如"a"，"the"，"with"等或如果这些事情的数量真的不重要。

如果它们无关紧要，请考虑$text索引并进行搜索。

第一个指数：

db.A.createIndex({ "name": "text", "description": "text" })

然后构建搜索：

var words = [
  "egg", "garlic", "cucumber", "kale", "pomegranate",
  "sunflower", "fish", "pork", "apple", "sunflower",
  "strawberry", "banana"
];

var search = words.join(" ")

db.A.find(
    { "$text": { "$search": search } },
    { "score": { "$meta": "textScore" } }
).sort({ "score": { "$meta": "textScore" }}).limit(1)

返回第一个文档：

{
    "_id" : ObjectId("5160757496cc6207a37ff778"),
    "name" : "Pomegranate Yogurt Bowl",
    "description" : "A simple breakfast bowl made with Greek yogurt, fresh pomegranate juice, puffed quinoa cereal, toasted sunflower seeds, and honey.",
    "score" : 1.7291666666666665
}

另一方面，如果您需要计算“停用词”，那么mapReduce可以为您找到结果：

db.A.mapReduce(
  function() {
    var words = [
      "egg", "garlic", "cucumber", "kale", "pomegranate",
      "sunflower", "fish", "pork", "apple", "sunflower",
      "strawberry", "banana"
    ];

    var count = 0;

    var fulltext = this.name.toLowerCase() + " " + this.description.toLowerCase();

    // Increment count by number of matches
    words.forEach(function(word) {
      count += ( fulltext.match(new RegExp(word,"ig")) || [] ).length;
    });

    emit(null,{ count: count, doc: this });

  },
  function(key,values) {
    // Sort largest first, return first
    return values.sort(function(a,b) {
      return a.count < b.count;
    })[0];
  },
  { "out": { "inline": 1 } }
)

结果：

{
    "_id" : null,
    "value" : {
        "count" : 4,
        "doc" : {
            "_id" : ObjectId("5160757496cc6207a37ff778"),
            "name" : "Pomegranate Yogurt Bowl",
            "description" : "A simple breakfast bowl made with Greek yogurt, fresh pomegranate juice, puffed quinoa cereal, toasted sunflower seeds, and honey."
        }
    }
}

所以“文本”索引方法按匹配数“加权”，然后只返回最大加权匹配。

mapReduce操作遍历每个文档并编制得分。然后“减速器”对结果进行分类，并保持得分最高的那个。

请注意，“reducer”可以多次调用，因此“不会”尝试立即对集合中的所有文档进行排序。但它仍然是真正的“蛮力”。

找到字段Mongodb中存在的最多出现的单词

1 个答案: