如何使用 PyMongo 在 MongoDB 中获取不同的名称和计数 [英] How to get distinct name and count in MongoDB using PyMongo

问题描述

我有以下集合，如下所示.我想要的只是不同的名称"和计数.例如 Betty 出现了 2 次，所以我想要的输出是 Betty:2、Vic:1、Veronica:2.我可以通过发出命令db.Car.find().distinct('Name')"来获取不同的名称，但不确定如何获取计数.

<代码>{"姓名": "贝蒂","汽车": "吉普",}{"姓名": "贝蒂","车": "货车",}{"姓名": "维克",汽车":法拉利"，}{"名称": "维罗妮卡","汽车": "公共汽车",}{"名称": "维罗妮卡","车": "货车",}

解决方案

你可以只使用 $group 按 Name 字段分组并使用 $sum 运算符以获取 Count 字段.

类似于以下内容:

db.collection.aggregate([{$组":{"_id": "$Name",数数": {$sum":1}}},{$项目":{"名称": "$_id",计数":1，"_id":0}}])

以上将产生以下输出:

<预><代码>[{计数":2，姓名":贝蒂"},{计数":1，姓名":维克"},{计数":2，名称":维罗妮卡"}]

I have the below collection as shown below. All I want is the distinct "Name" and the count. For example Betty appears 2 times, so the output I want is Betty:2, Vic:1, Veronica:2. I am able to get the distinct Name by issuing the command "db.Car.find().distinct('Name')" but not sure how to get the count.

{
    "Name": "Betty",
    "Car": "Jeep",
}
{
    "Name": "Betty",
    "Car": "Van",
}
{
    "Name": "Vic",
    "Car": "Ferrari",
}
{
    "Name": "Veronica",
    "Car": "Bus",
}
{
    "Name": "Veronica",
    "Car": "Van",
}

解决方案

You can just use $group to group by Name field and use $sum operator in it to get the Count field.

Something like below:

db.collection.aggregate([
  {
    "$group": {
      "_id": "$Name",
      "Count": {
        "$sum": 1
      }
    }
  },
  {
    "$project": {
      "Name": "$_id",
      "Count": 1,
      "_id":0
    }
  }
])

The above will produce the following output:

[
  {
    "Count": 2,
    "Name": "Betty"
  },
  {
    "Count": 1,
    "Name": "Vic"
  },
  {
    "Count": 2,
    "Name": "Veronica"
  }
]

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