学校数据库中有200个文件。我必须删除每个包含"类型":"" homework"和最低分。
{
"_id" : 0,
"name" : "aimee Zank",
"scores" :
[
{
"type" : "exam",
"score" : 1.463179736705023
},
{
"type" : "quiz",
"score" : 11.78273309957772
},
{
"type" : "homework",
"score" : 6.676176060654615
},
{
"type" : "homework",
"score" : 35.8740349954354
}
]
}
例如,这里
{
"type" : "homework",
"score" : 6.676176060654615
}
必须删除,因为得分= 6.6< 35.8
我对所有文件进行了分类:
db.students.find({"scores.type":"homework"}).sort({"scores.score":1})
但我不知道如何删除分数最低的文档并输入:homework ??? 注意:如何通过不使用聚合方法来解决它?例如,通过排序然后更新。
答案 0 :(得分:1)
这可以通过几个步骤完成。第一步是使用汇总框架和$match
,$unwind
和$group
运算符来获取具有最低分数的文档列表,这些运算符可简化您的文档以查找每个文档的最低分数文件:
lowest_scores_docs = db.school.aggregate([
{ "$match": {"scores.type": "homework"} },
{ "$unwind": "$scores" }, { "$match": {"scores.type": "homework"} },
{ "$group": { "_id":"$_id", "lowest_score": {"$min": "$scores.score" } } } ] )
第二步是遍历上面的字典并在更新查询中使用$pull
运算符从数组中删除元素,如下所示:
for result in lowest_scores_docs["result"]:
db.school.update({ "_id": result["_id"] },
{ "$pull": { "scores": { "score": result["lowest_score"] } } } )
答案 1 :(得分:0)
import pymongo
import sys
# connnecto to the db on standard port
connection = pymongo.MongoClient("mongodb://localhost")
db = connection.school # attach to db
students = db.students # specify the colllection
try:
cursor = students.find({})
print(type(cursor))
for doc in cursor:
hw_scores = []
for item in doc["scores"]:
if item["type"] == "homework":
hw_scores.append(item["score"])
hw_scores.sort()
hw_min = hw_scores[0]
#students.update({"_id": doc["_id"]},
# {"$pull":{"scores":{"score":hw_min}}})
except:
print ("Error trying to read collection:" + sys.exc_info()[0])