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我的数据库中有一个名为“item”的对象，我想写一些python，它会从每组student_id中选择最低分……我是一个python菜鸟……我该怎么做?

{u'student_id': 197, u'_id': ObjectId('50906d7fa3c412bb040eb88d'), u'type': u'homework', u'score': 10.90872422518918}
{u'student_id': 197, u'_id': ObjectId('50906d7fa3c412bb040eb88e'), u'type': u'homework', u'score': 88.3871242475841}
{u'student_id': 198, u'_id': ObjectId('50906d7fa3c412bb040eb892'), u'type': u'homework', u'score': 17.46279901047208}
{u'student_id': 198, u'_id': ObjectId('50906d7fa3c412bb040eb891'), u'type': u'homework', u'score': 76.18366499496366}
{u'student_id': 199, u'_id': ObjectId('50906d7fa3c412bb040eb895'), u'type': u'homework', u'score': 49.34223066136407}
{u'student_id': 199, u'_id': ObjectId('50906d7fa3c412bb040eb896'), u'type': u'homework', u'score': 58.09608083191365}

更新：

运行我的代码时遇到问题...我可以完成一些健全性检查吗？我收到这个错误...

NameError: name 'getResultFromDatabase' is not defined

谢谢你的麻烦......这是我的代码......

import pymongo

from itertools import groupby
from pymongo import Connection

data = getResultFromDatabase()
connection = Connection('localhost', 27017)

db = connection.students

item = db.grades.find({'type' : 'homework'}).sort([('student_id',pymongo.ASCENDING),('score',pymongo.ASCENDING)])



for id, items in groupby(data, lambda s: s['student_id']):
    lowest_score = min(i['score'] for i in items)

    print lowest_score
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1 回答 1

3

groupby做你想做的事：

from itertools import groupby

data = getResultFromDatabase()

for id, items in groupby(data, lambda s: s['student_id']):
    lowest_score_entry = min(items, key=lambda i: i['score'])
    print lowest_score_entry['score'], lowest_score_entry['_id']
于 2012-11-09T21:49:43.823 回答