优化Django REST Framework演员统计接口：解决数据重复与统计需求

一、需求明确 #

需要实现以下三个核心功能：

• 统计指定演员参演电影的各类型数量（格式如 {'Drama':2, 'Horror':3}）
• 解决接口返回数据随参演电影数量重复的问题
• 获取该演员的热门类型（含名称，支持并列）、最常合作搭档的完整信息（ID、姓名、合作次数）

二、基础模型假设（基于常见电影数据集设计） #

from django.db import models

class Actor(models.Model):
    id = models.IntegerField(primary_key=True)
    name = models.CharField(max_length=100)
    # 其他字段（如出生日期、简介等）

class Genre(models.Model):
    id = models.IntegerField(primary_key=True)
    name = models.CharField(max_length=50, unique=True)

class Movie(models.Model):
    id = models.IntegerField(primary_key=True)
    title = models.CharField(max_length=200)
    # 其他字段（如上映日期、评分等）

class MovieActor(models.Model):
    movie = models.ForeignKey(Movie, on_delete=models.CASCADE, related_name='actors')
    actor = models.ForeignKey(Actor, on_delete=models.CASCADE, related_name='movies')
    role = models.CharField(max_length=100, blank=True)

class MovieGenre(models.Model):
    movie = models.ForeignKey(Movie, on_delete=models.CASCADE, related_name='genres')
    genre = models.ForeignKey(Genre, on_delete=models.CASCADE, related_name='movies')

三、核心优化实现 #

1. 替换View解决数据重复问题 #

原ListAPIView会返回每条电影的重复演员数据，改用APIView返回聚合统计结果，彻底避免重复：

from rest_framework.views import APIView
from rest_framework.response import Response
from rest_framework import status
from django.db.models import Count
from .models import Actor, MovieActor, MovieGenre
from .serializers import ActorStatsSerializer

class ActorStatsView(APIView):
    def get(self, request, actor_id):
        try:
            actor = Actor.objects.get(id=actor_id)
        except Actor.DoesNotExist:
            return Response({"error": "演员不存在"}, status=status.HTTP_404_NOT_FOUND)
        
        # 后续统计逻辑见下文
        return Response(serializer.data, status=status.HTTP_200_OK)

2. 统计各类型电影数量 #

通过Django ORM的annotate和values直接关联类型名称，生成键值对格式的统计结果：

# 在ActorStatsView的get方法中添加
genre_stats = MovieGenre.objects.filter(
    movie__actors__actor=actor
).values('genre__name').annotate(
    count=Count('genre__id')
).order_by('-count')

# 转换为需求格式的字典
genre_counts = {item['genre__name']: item['count'] for item in genre_stats}

3. 获取支持并列的热门类型 #

先提取最大类型数量，再筛选所有符合该数量的类型：

# 在ActorStatsView的get方法中添加
top_genres = []
if genre_stats:
    max_count = genre_stats[0]['count']
    top_genres = [item['genre__name'] for item in genre_stats if item['count'] == max_count]

4. 获取最常合作搭档（含完整信息） #

通过关联表统计共同参演的演员，排除自身后按合作次数排序，支持并列：

# 在ActorStatsView的get方法中添加
co_actor_stats = MovieActor.objects.filter(
    movie__actors__actor=actor
).exclude(actor=actor).values(
    'actor__id', 'actor__name'
).annotate(
    合作次数=Count('movie__id', distinct=True)
).order_by('-合作次数')

top_co_actors = []
if co_actor_stats:
    max_co_count = co_actor_stats[0]['合作次数']
    top_co_actors = [
        {
            'id': item['actor__id'],
            'name': item['actor__name'],
            '合作次数': item['合作次数']
        } for item in co_actor_stats if item['合作次数'] == max_co_count
    ]

5. 专用序列化器设计 #

创建针对统计结果的序列化器，避免与基础演员序列化器混淆：

from rest_framework import serializers

class CoActorSerializer(serializers.Serializer):
    id = serializers.IntegerField()
    name = serializers.CharField()
    合作次数 = serializers.IntegerField()

class ActorStatsSerializer(serializers.Serializer):
    actor_id = serializers.IntegerField(source='id')
    actor_name = serializers.CharField(source='name')
    genre_counts = serializers.DictField(child=serializers.IntegerField())
    top_genres = serializers.ListField(child=serializers.CharField())
    top_co_actors = serializers.ListField(child=CoActorSerializer())

6. URL配置更新 #

替换原getActors路由，指向新的统计接口：

from django.urls import path
from .views import ActorStatsView

urlpatterns = [
    path('actors/<int:actor_id>/stats/', ActorStatsView.as_view(), name='actor-stats'),
]

四、完整View实现 #

整合所有逻辑后的最终View代码：

from rest_framework.views import APIView
from rest_framework.response import Response
from rest_framework import status
from django.db.models import Count
from .models import Actor, MovieActor, MovieGenre
from .serializers import ActorStatsSerializer

class ActorStatsView(APIView):
    def get(self, request, actor_id):
        try:
            actor = Actor.objects.get(id=actor_id)
        except Actor.DoesNotExist:
            return Response({"error": "演员不存在"}, status=status.HTTP_404_NOT_FOUND)
        
        # 统计各类型数量
        genre_stats = MovieGenre.objects.filter(
            movie__actors__actor=actor
        ).values('genre__name').annotate(
            count=Count('genre__id')
        ).order_by('-count')
        genre_counts = {item['genre__name']: item['count'] for item in genre_stats}
        
        # 获取热门类型（支持并列）
        top_genres = []
        if genre_stats:
            max_count = genre_stats[0]['count']
            top_genres = [item['genre__name'] for item in genre_stats if item['count'] == max_count]
        
        # 获取最常合作搭档
        co_actor_stats = MovieActor.objects.filter(
            movie__actors__actor=actor
        ).exclude(actor=actor).values(
            'actor__id', 'actor__name'
        ).annotate(
            合作次数=Count('movie__id', distinct=True)
        ).order_by('-合作次数')
        top_co_actors = []
        if co_actor_stats:
            max_co_count = co_actor_stats[0]['合作次数']
            top_co_actors = [
                {
                    'id': item['actor__id'],
                    'name': item['actor__name'],
                    '合作次数': item['合作次数']
                } for item in co_actor_stats if item['合作次数'] == max_co_count
            ]
        
        # 序列化返回
        serializer_data = {
            'actor_id': actor.id,
            'actor_name': actor.name,
            'genre_counts': genre_counts,
            'top_genres': top_genres,
            'top_co_actors': top_co_actors
        }
        serializer = ActorStatsSerializer(data=serializer_data)
        serializer.is_valid(raise_exception=True)
        
        return Response(serializer.data, status=status.HTTP_200_OK)

五、关键优化说明 #

• 数据重复解决：放弃ListAPIView的逐条电影返回，改用聚合统计结果，彻底消除重复
• 类型名称关联：通过values('genre__name')直接关联类型名称，避免仅返回ID的问题
• 并列场景处理：先获取最大统计值，再筛选所有符合条件的条目，支持多个热门类型/搭档
• 性能优化：使用distinct=True避免重复统计同一部电影的合作关系，降低数据库查询压力

内容的提问来源于stack exchange，提问作者Vagner