mongoengine + django如何计算相同内容的项目数 [英] mongoengine +django how to count the number of items of same content

问题描述

我想计算一个字段中所有不同值的数量.mongoengine中没有要使用的"annotate()"，那么我该如何计算数量并按数字排序

I want to count all the numbers of different values in a field.there is no "annotate()" to use in mongoengine ,then how could i count the number and order them by numbers

我想解决的唯一方法是使用"distinct()"找出不同的值，然后使用"count()"对每个值进行计数 这是实现我想要的结果的愚蠢方式

The only way i thought out to solve this is use"distinct()"to find out the different values and then use "count()"to count each of the values it's a stupid way to realize the result i want

您还有其他方法吗?

推荐答案

MongoEngine提供了一些应满足您需求的地图简化助手. Queryset方法item_frequencies [1]将满足您的需求.新的聚合框架没有任何特殊支持，但将来可能会添加支持.

MongoEngine has some map reduce helpers that should meet your needs. The Queryset method item_frequencies[1] will meet your needs. There isnt any special support for the new aggregation framework but support could be added in the future.

示例用法:

BlogPost.objects.item_frequencies('tags')

[1] http://docs.mongoengine.org/en/latest/apireference.html?highlight=item_frequencies#mongoengine.queryset.QuerySet.item_frequencies

这篇关于mongoengine + django如何计算相同内容的项目数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！