Question

我需要使用PyMongo构建一个查询，它从MongoDB数据库中的两个相关集合中获取数据。

集合X包含字段UserId，Name和EmailId：

[
  {
    "UserId" :    "941AB",
    "Name" :      "Alex Andresson",
    "EmailId" :   "alex@example.com"
  },
  {
    "UserId" :    "768CD",
    "Name" :      "Bryan Barnes",
    "EmailId" :   "bryan@example.com"
  }
]

集合Y包含字段UserId1，UserID2和Rating：

[
  {
    "UserId1" :  "941AB",
    "UserId2" :  "768CD",
    "Rating" :   0.8
   }
]

我需要打印UserId1和UserId2的名称和电子邮件ID以及评级，如下所示：

[
  {
    "UserId1" :    "941AB",
    "UserName1" :  "Alex Andresson"
    "UserEmail1" : "alex@example.com",
    "UserId2" :    "768CD",
    "UserName2" :  "Bryan Barnes"
    "UserEmail2" : "bryan@example.com",
    "Rating":      0.8
  }
]

这意味着我需要从集合Y和X中获取数据。我现在正在与PyMongo合作，但我找不到它的解决方案。有人甚至可以给我一个关于这个概念或方法的伪代码如何推进它。

Answer 1

您需要手动加入或使用一些可以为您完成此操作的库 - 也许mongoengine。

基本上，您需要找到您感兴趣的评分，然后找到与这些评分相关的用户。

示例：

#!/usr/bin/env python3

import pymongo
from random import randrange

client = pymongo.MongoClient()
db = client['test']

# clean collections
db['users'].drop()
db['ratings'].drop()

# insert data
user_count = 100
rating_count = 20

db['users'].insert_many([
    {'UserId': i, 'Name': 'John', 'EmailId': i}
    for i in range(user_count)])

db['ratings'].insert_many([
    {'UserId1': randrange(user_count), 'UserId2': randrange(user_count), 'Rating': i}
    for i in range(rating_count)])

# don't forget the indexes
db['users'].create_index('UserId')
# but it would be better if we used _id as the UserId

# if you want to make queries based on Rating value, then add also this index:
db['ratings'].create_index('Rating')

# now print ratings with users that have value 10+

# simple approach:
ratings = db['ratings'].find({'Rating': {'$gte': 10}})
for rating in ratings:
    u1 = db['users'].find_one({'UserId': rating['UserId1']})
    u2 = db['users'].find_one({'UserId': rating['UserId2']})
    print('Rating between {} (UserId {:2}) and {} (UserId {:2}) is {:2}'.format(
        u1['Name'], u1['UserId'], u2['Name'], u2['UserId'], rating['Rating']))

print('---')

# optimized approach:
ratings = list(db['ratings'].find({'Rating': {'$gte': 10}}))
user_ids = {r['UserId1'] for r in ratings}
user_ids |= {r['UserId2'] for r in ratings}
users = db['users'].find({'UserId': {'$in': list(user_ids)}})
users_by_id = {u['UserId']: u for u in users}
for rating in ratings:
    u1 = users_by_id.get(rating['UserId1'])
    u2 = users_by_id.get(rating['UserId2'])
    print('Rating between {} (UserId {:2}) and {} (UserId {:2}) is {:2}'.format(
        u1['Name'], u1['UserId'], u2['Name'], u2['UserId'], rating['Rating']))

请注意，第一种方法为评级调用一个find，每个评级调用两个find，但第二种方法总共调用三个find。如果您通过网络访问MongoDB，这将导致巨大的性能差异。

如果可能，我建议使用_id代替UserId。

当然，使用SQL数据库时，这个特定的用例会更容易。如果您使用MongoDB获得性能并且读取的内容比写入多得多，那么请考虑将相关用户缓存到评级文档中。

使用PyMongo，我需要获取另一个集合的字段

1 个答案: