How do you return 404 when resource is not found in Django REST Framework

Simply way to do it, you can use raise Http404, here is your views.py

from django.http import Http404

from rest_framework import status
from rest_framework.response import Response
from rest_framework.views import APIView

from yourapp.models import Snippet
from yourapp.serializer import SnippetSerializer


class SnippetDetailView(APIView):

    def get_object(self, pk):
        try:
            return Snippet.objects.get(pk=pk)
        except Snippet.DoesNotExist:
            raise Http404

    def get(self, request, pk, format=None):
        snippet = self.get_object(pk)
        serializer = SnippetSerializer(snippet)
        return Response(serializer.data, status=status.HTTP_200_OK)

You also can handle it with Response(status=status.HTTP_404_NOT_FOUND), this answer is how to do with it: https://stackoverflow.com/a/24420524/6396981

But previously, inside your serializer.py

from rest_framework import serializers

from yourapp.models import Snippet


class SnippetSerializer(serializers.ModelSerializer):
    user = serializers.CharField(
        source='user.pk',
        read_only=True
    )
    photo = serializers.ImageField(
        max_length=None,
        use_url=True
    )
    ....

    class Meta:
        model = Snippet
        fields = ('user', 'title', 'photo', 'description')

    def create(self, validated_data):
        return Snippet.objects.create(**validated_data)

To test it, an example using curl command;

$ curl -X GET http://localhost:8000/snippets/<pk>/

# example;

$ curl -X GET http://localhost:8000/snippets/99999/

Hope it can help..

---

Update

If you want to handle for all error 404 urls with DRF, DRF also provide about it with APIException, this answer may help you; https://stackoverflow.com/a/30628065/6396981

I'll give an example how do with it;

1. views.py

from rest_framework.exceptions import NotFound

def error404(request):
    raise NotFound(detail="Error 404, page not found", code=404)

2. urls.py

from django.conf.urls import (
  handler400, handler403, handler404, handler500)

from yourapp.views import error404

handler404 = error404

Makesure your DEBUG = False

from rest_framework import status    
from rest_framework.response import Response

# return 404 status code    
return Response({'status': 'details'}, status=status.HTTP_404_NOT_FOUND)

An easier way is to use get_object_or_404 method in django:
as described in this link:

get_object_or_404(klass, *args, kwargs)
- Calls get() on a given model manager, but it raises Http404 instead of the model’s DoesNotExist exception.
- klass: A Model class, a Manager, or a QuerySet instance from which to get the object.

As an example, pay attention to

obj = get_object_or_404(Snippet, pk=pk)
return obj

in the following code:

from django.shortcuts import get_object_or_404
from snippets.models import Snippet
from snippets.serializers import SnippetSerializer
from rest_framework.views import APIView
from rest_framework.response import Response

class SnippetDetail(APIView):
    """
    Retrieve, update or delete a snippet instance.
    """
    def get_object(self, pk):
        obj = get_object_or_404(Snippet, pk=pk)
        return obj

    def get(self, request, pk, format=None):
        snippet = self.get_object(pk)
        serializer = SnippetSerializer(snippet)
        return Response(serializer.data)
    ...