Currently I have a lot of python objects in my code similar to the following:

class MyClass():
  def __init__(self, name, friends):
      self.myName = name
      self.myFriends = [str(x) for x in friends]

Now I want to turn this into a Django model, where self.myName is a string field, and self.myFriends is a list of strings.

from django.db import models

class myDjangoModelClass():
    myName = models.CharField(max_length=64)
    myFriends = ??? # what goes here?

Since the list is such a common data structure in python, I sort of expected there to be a Django model field for it. I know I can use a ManyToMany or OneToMany relationship, but I was hoping to avoid that extra indirection in the code.

Edit:

I added this related question, which people may find useful.

Would this relationship not be better expressed as a one-to-many foreign key relationship to a Friends table? I understand that myFriends are just strings but I would think that a better design would be to create a Friend model and have MyClass contain a foreign key realtionship to the resulting table.

“Premature optimization is the root of all evil.”

With that firmly in mind, let’s do this! Once your apps hit a certain point, denormalizing data is very common. Done correctly, it can save numerous expensive database lookups at the cost of a little more housekeeping.

To return a list of friend names we’ll need to create a custom Django Field class that will return a list when accessed.

David Cramer posted a guide to creating a SeperatedValueField on his blog. Here is the code:

from django.db import models

class SeparatedValuesField(models.TextField):
    __metaclass__ = models.SubfieldBase

    def __init__(self, *args, **kwargs):
        self.token = kwargs.pop('token', ',')
        super(SeparatedValuesField, self).__init__(*args, **kwargs)

    def to_python(self, value):
        if not value: return
        if isinstance(value, list):
            return value
        return value.split(self.token)

    def get_db_prep_value(self, value):
        if not value: return
        assert(isinstance(value, list) or isinstance(value, tuple))
        return self.token.join([unicode(s) for s in value])

    def value_to_string(self, obj):
        value = self._get_val_from_obj(obj)
        return self.get_db_prep_value(value)

The logic of this code deals with serializing and deserializing values from the database to Python and vice versa. Now you can easily import and use our custom field in the model class:

from django.db import models
from custom.fields import SeparatedValuesField 

class Person(models.Model):
    name = models.CharField(max_length=64)
    friends = SeparatedValuesField()

A simple way to store a list in Django is to just convert it into a JSON string, and then save that as Text in the model. You can then retrieve the list by converting the (JSON) string back into a python list. Here’s how:

The “list” would be stored in your Django model like so:

class MyModel(models.Model):
    myList = models.TextField(null=True) # JSON-serialized (text) version of your list

In your view/controller code:

Storing the list in the database:

import simplejson as json # this would be just 'import json' in Python 2.7 and later
...
...

myModel = MyModel()
listIWantToStore = [1,2,3,4,5,'hello']
myModel.myList = json.dumps(listIWantToStore)
myModel.save()

Retrieving the list from the database:

jsonDec = json.decoder.JSONDecoder()
myPythonList = jsonDec.decode(myModel.myList)

Conceptually, here’s what’s going on:

>>> myList = [1,2,3,4,5,'hello']
>>> import simplejson as json
>>> myJsonList = json.dumps(myList)
>>> myJsonList
'[1, 2, 3, 4, 5, "hello"]'
>>> myJsonList.__class__
<type 'str'>
>>> jsonDec = json.decoder.JSONDecoder()
>>> myPythonList = jsonDec.decode(myJsonList)
>>> myPythonList
[1, 2, 3, 4, 5, u'hello']
>>> myPythonList.__class__
<type 'list'>

If you are using Django >= 1.9 with Postgres you can make use of ArrayField advantages

A field for storing lists of data. Most field types can be used, you simply pass another field instance as the base_field. You may also specify a size. ArrayField can be nested to store multi-dimensional arrays.

It is also possible to nest array fields:

from django.contrib.postgres.fields import ArrayField
from django.db import models

class ChessBoard(models.Model):
    board = ArrayField(
        ArrayField(
            models.CharField(max_length=10, blank=True),
            size=8,
        ),
        size=8,
    )

As @thane-brimhall mentioned it is also possible to query elements directly. Documentation reference

As this is an old question, and Django techniques must have changed significantly since, this answer reflects Django version 1.4, and is most likely applicable for v 1.5.

Django by default uses relational databases; you should make use of ’em. Map friendships to database relations (foreign key constraints) with the use of ManyToManyField. Doing so allows you to use RelatedManagers for friendlists, which use smart querysets. You can use all available methods such as filter or values_list.

Using ManyToManyField relations and properties:

class MyDjangoClass(models.Model):
    name = models.CharField(...)
    friends = models.ManyToManyField("self")

    @property
    def friendlist(self):
        # Watch for large querysets: it loads everything in memory
        return list(self.friends.all())

You can access a user’s friend list this way:

joseph = MyDjangoClass.objects.get(name="Joseph")
friends_of_joseph = joseph.friendlist

Note however that these relations are symmetrical: if Joseph is a friend of Bob, then Bob is a friend of Joseph.

class Course(models.Model):
   name = models.CharField(max_length=256)
   students = models.ManyToManyField(Student)

class Student(models.Model):
   first_name = models.CharField(max_length=256)
   student_number = models.CharField(max_length=128)
   # other fields, etc...

   friends = models.ManyToManyField('self')

Remember that this eventually has to end up in a relational database. So using relations really is the common way to solve this problem. If you absolutely insist on storing a list in the object itself, you could make it for example comma-separated, and store it in a string, and then provide accessor functions that split the string into a list. With that, you will be limited to a maximum number of strings, and you will lose efficient queries.

In case you’re using postgres, you can use something like this:

class ChessBoard(models.Model):

    board = ArrayField(
        ArrayField(
            models.CharField(max_length=10, blank=True),
            size=8,
        ),
        size=8,
    )

if you need more details you can read in the link below: https://docs.djangoproject.com/pt-br/1.9/ref/contrib/postgres/fields/

You can store virtually any object using a Django Pickle Field, ala this snippet:

http://www.djangosnippets.org/snippets/513/

Storing a list of strings in Django model:

class Bar(models.Model):
    foo = models.TextField(blank=True)

    def set_list(self, element):
        if self.foo:
            self.foo = self.foo + "," + element
        else:
            self.foo = element

    def get_list(self):
        if self.foo:
            return self.foo.split(",")
        else:
            None

and you can call it like this:

bars = Bar()
bars.set_list("str1")
bars.set_list("str2")
list = bars.get_list()
if list is not None:
    for bar in list:
        print bar
else:
    print "List is empty."      

My solution, may be it helps someone:

import json
from django.db import models


class ExampleModel(models.Model):
    _list = models.TextField(default='[]')

    @property
    def list(self):
        return json.loads(self._list)

    @list.setter
    def list(self, value):
        self._list = json.dumps(self.list + value)

Using one-to-many relation (FK from Friend to parent class) will make your app more scalable (as you can trivially extend the Friend object with additional attributes beyond the simple name). And thus this is the best way

Ok, I’ve tried about near everything and I cannot get this to work.

• I have a Django model with an ImageField on it
• I have code that downloads an image via HTTP (tested and works)
• The image is saved directly into the ‘upload_to’ folder (the upload_to being the one that is set on the ImageField)
• All I need to do is associate the already existing image file path with the ImageField

I’ve written this code about 6 different ways.

The problem I’m running into is all of the code that I’m writing results in the following behavior: (1) Django will make a 2nd file, (2) rename the new file, adding an _ to the end of the file name, then (3) not transfer any of the data over leaving it basically an empty re-named file. What’s left in the ‘upload_to’ path is 2 files, one that is the actual image, and one that is the name of the image,but is empty, and of course the ImageField path is set to the empty file that Django try to create.

In case that was unclear, I’ll try to illustrate:

## Image generation code runs.... 
/Upload
     generated_image.jpg     4kb

## Attempt to set the ImageField path...
/Upload
     generated_image.jpg     4kb
     generated_image_.jpg    0kb

ImageField.Path = /Upload/generated_image_.jpg

How can I do this without having Django try to re-store the file? What I’d really like is something to this effect…

model.ImageField.path = generated_image_path

…but of course that doesn’t work.

And yes I’ve gone through the other questions here like this one as well as the django doc on File

UPDATE After further testing, it only does this behavior when running under Apache on Windows Server. While running under the ‘runserver’ on XP it does not execute this behavior.

I am stumped.

Here is the code which runs successfully on XP…

f = open(thumb_path, 'r')
model.thumbnail = File(f)
model.save()

I have some code that fetches an image off the web and stores it in a model. The important bits are:

from django.core.files import File  # you need this somewhere
import urllib


# The following actually resides in a method of my model

result = urllib.urlretrieve(image_url) # image_url is a URL to an image

# self.photo is the ImageField
self.photo.save(
    os.path.basename(self.url),
    File(open(result[0], 'rb'))
    )

self.save()

That’s a bit confusing because it’s pulled out of my model and a bit out of context, but the important parts are:

• The image pulled from the web is not stored in the upload_to folder, it is instead stored as a tempfile by urllib.urlretrieve() and later discarded.
• The ImageField.save() method takes a filename (the os.path.basename bit) and a django.core.files.File object.

Let me know if you have questions or need clarification.

Edit: for the sake of clarity, here is the model (minus any required import statements):

class CachedImage(models.Model):
    url = models.CharField(max_length=255, unique=True)
    photo = models.ImageField(upload_to=photo_path, blank=True)

    def cache(self):
        """Store image locally if we have a URL"""

        if self.url and not self.photo:
            result = urllib.urlretrieve(self.url)
            self.photo.save(
                    os.path.basename(self.url),
                    File(open(result[0], 'rb'))
                    )
            self.save()

Super easy if model hasn’t been created yet:

First, copy your image file to the upload path (assumed = ‘path/’ in following snippet).

Second, use something like:

class Layout(models.Model):
    image = models.ImageField('img', upload_to='path/')

layout = Layout()
layout.image = "path/image.png"
layout.save()

tested and working in django 1.4, it might work also for an existing model.

Just a little remark. tvon answer works but, if you’re working on windows, you probably want to open() the file with 'rb'. Like this:

class CachedImage(models.Model):
    url = models.CharField(max_length=255, unique=True)
    photo = models.ImageField(upload_to=photo_path, blank=True)

    def cache(self):
        """Store image locally if we have a URL"""

        if self.url and not self.photo:
            result = urllib.urlretrieve(self.url)
            self.photo.save(
                    os.path.basename(self.url),
                    File(open(result[0], 'rb'))
                    )
            self.save()

or you’ll get your file truncated at the first 0x1A byte.

Here is a method that works well and allows you to convert the file to a certain format as well (to avoid “cannot write mode P as JPEG” error):

import urllib2
from django.core.files.base import ContentFile
from PIL import Image
from StringIO import StringIO

def download_image(name, image, url):
    input_file = StringIO(urllib2.urlopen(url).read())
    output_file = StringIO()
    img = Image.open(input_file)
    if img.mode != "RGB":
        img = img.convert("RGB")
    img.save(output_file, "JPEG")
    image.save(name+".jpg", ContentFile(output_file.getvalue()), save=False)

where image is the django ImageField or your_model_instance.image here is a usage example:

p = ProfilePhoto(user=user)
download_image(str(user.id), p.image, image_url)
p.save()

Hope this helps

Ok, If all you need to do is associate the already existing image file path with the ImageField, then this solution may be helpfull:

from django.core.files.base import ContentFile

with open('/path/to/already/existing/file') as f:
  data = f.read()

# obj.image is the ImageField
obj.image.save('imgfilename.jpg', ContentFile(data))

Well, if be earnest, the already existing image file will not be associated with the ImageField, but the copy of this file will be created in upload_to dir as ‘imgfilename.jpg’ and will be associated with the ImageField.

What I did was to create my own storage that will just not save the file to the disk:

from django.core.files.storage import FileSystemStorage

class CustomStorage(FileSystemStorage):

    def _open(self, name, mode='rb'):
        return File(open(self.path(name), mode))

    def _save(self, name, content):
        # here, you should implement how the file is to be saved
        # like on other machines or something, and return the name of the file.
        # In our case, we just return the name, and disable any kind of save
        return name

    def get_available_name(self, name):
        return name

Then, in my models, for my ImageField, I’ve used the new custom storage:

from custom_storage import CustomStorage

custom_store = CustomStorage()

class Image(models.Model):
    thumb = models.ImageField(storage=custom_store, upload_to='/some/path')

If you want to just “set” the actual filename, without incurring the overhead of loading and re-saving the file (!!), or resorting to using a charfield (!!!), you might want to try something like this —

model_instance.myfile = model_instance.myfile.field.attr_class(model_instance, model_instance.myfile.field, 'my-filename.jpg')

This will light up your model_instance.myfile.url and all the rest of them just as if you’d actually uploaded the file.

Like @t-stone says, what we really want, is to be able to set instance.myfile.path = ‘my-filename.jpg’, but Django doesn’t currently support that.

THe simplest solution in my opinion:

from django.core.files import File

with open('path_to_file', 'r') as f:   # use 'rb' mode for python3
    data = File(f)
    model.image.save('filename', data, True)

A lot of these answers were outdated, and I spent many hours in frustration (I’m fairly new to Django & web dev in general). However, I found this excellent gist by @iambibhas: https://gist.github.com/iambibhas/5051911

import requests

from django.core.files import File
from django.core.files.temp import NamedTemporaryFile


def save_image_from_url(model, url):
    r = requests.get(url)

    img_temp = NamedTemporaryFile(delete=True)
    img_temp.write(r.content)
    img_temp.flush()

    model.image.save("image.jpg", File(img_temp), save=True)

This is might not be the answer you are looking for. but you can use charfield to store the path of the file instead of ImageFile. In that way you can programmatically associate uploaded image to field without recreating the file.

You can try:

model.ImageField.path = os.path.join('/Upload', generated_image_path)

class tweet_photos(models.Model):
upload_path='absolute path'
image=models.ImageField(upload_to=upload_path)
image_url = models.URLField(null=True, blank=True)
def save(self, *args, **kwargs):
    if self.image_url:
        import urllib, os
        from urlparse import urlparse
        file_save_dir = self.upload_path
        filename = urlparse(self.image_url).path.split('/')[-1]
        urllib.urlretrieve(self.image_url, os.path.join(file_save_dir, filename))
        self.image = os.path.join(file_save_dir, filename)
        self.image_url = ''
    super(tweet_photos, self).save()

class Pin(models.Model):
    """Pin Class"""
    image_link = models.CharField(max_length=255, null=True, blank=True)
    image = models.ImageField(upload_to='images/', blank=True)
    title = models.CharField(max_length=255, null=True, blank=True)
    source_name = models.CharField(max_length=255, null=True, blank=True)
    source_link = models.CharField(max_length=255, null=True, blank=True)
    description = models.TextField(null=True, blank=True)
    tags = models.ForeignKey(Tag, blank=True, null=True)

    def __unicode__(self):
        """Unicode class."""
        return unicode(self.image_link)

    def save(self, *args, **kwargs):
        """Store image locally if we have a URL"""
        if self.image_link and not self.image:
            result = urllib.urlretrieve(self.image_link)
            self.image.save(os.path.basename(self.image_link), File(open(result[0], 'r')))
            self.save()
            super(Pin, self).save()

Working! You can save image by using FileSystemStorage. check the example below

def upload_pic(request):
if request.method == 'POST' and request.FILES['photo']:
    photo = request.FILES['photo']
    name = request.FILES['photo'].name
    fs = FileSystemStorage()
##### you can update file saving location too by adding line below #####
    fs.base_location = fs.base_location+'/company_coverphotos'
##################
    filename = fs.save(name, photo)
    uploaded_file_url = fs.url(filename)+'/company_coverphotos'
    Profile.objects.filter(user=request.user).update(photo=photo)

Your can use Django REST framework and python Requests library to Programmatically saving image to Django ImageField

Here is a Example:

import requests


def upload_image():
    # PATH TO DJANGO REST API
    url = "http://127.0.0.1:8080/api/gallery/"

    # MODEL FIELDS DATA
    data = {'first_name': "Rajiv", 'last_name': "Sharma"}

    #  UPLOAD FILES THROUGH REST API
    photo = open('/path/to/photo'), 'rb')
    resume = open('/path/to/resume'), 'rb')
    files = {'photo': photo, 'resume': resume}

    request = requests.post(url, data=data, files=files)
    print(request.status_code, request.reason) 

With Django 3, with a model such as this one:

class Item(models.Model):
   name = models.CharField(max_length=255, unique=True)
   photo= models.ImageField(upload_to='image_folder/', blank=True)

if the image has already been uploaded, we can directly do :

Item.objects.filter(...).update(photo='image_folder/sample_photo.png')

or

my_item = Item.objects.get(id=5)
my_item.photo='image_folder/sample_photo.png'
my_item.save()

I have a model that represents paintings I present on my site. On the main webpage I’d like to show some of them: newest, one that was not visited for most time, most popular one and a random one.

I’m using Django 1.0.2.

While first 3 of them are easy to pull using django models, last one (random) causes me some trouble. I can ofc code it in my view, to something like this:

number_of_records = models.Painting.objects.count()
random_index = int(random.random()*number_of_records)+1
random_paint = models.Painting.get(pk = random_index)

It doesn’t look like something I’d like to have in my view tho – this is entirely part of database abstraction and should be in the model. Also, here I need to take care of removed records (then number of all records won’t cover me all the possible key values) and probably lots of other things.

Any other options how I can do it, preferably somehow inside the model abstraction?

Using order_by('?') will kill the db server on the second day in production. A better way is something like what is described in Getting a random row from a relational database.

from django.db.models.aggregates import Count
from random import randint

class PaintingManager(models.Manager):
    def random(self):
        count = self.aggregate(count=Count('id'))['count']
        random_index = randint(0, count - 1)
        return self.all()[random_index]

Simply use:

MyModel.objects.order_by('?').first()

It is documented in QuerySet API.

The solutions with order_by(‘?’)[:N] are extremely slow even for medium-sized tables if you use MySQL (don’t know about other databases).

order_by('?')[:N] will be translated to SELECT ... FROM ... WHERE ... ORDER BY RAND() LIMIT N query.

It means that for every row in table the RAND() function will be executed, then the whole table will be sorted according to value of this function and then first N records will be returned. If your tables are small, this is fine. But in most cases this is a very slow query.

I wrote simple function that works even if id’s have holes (some rows where deleted):

def get_random_item(model, max_id=None):
    if max_id is None:
        max_id = model.objects.aggregate(Max('id')).values()[0]
    min_id = math.ceil(max_id*random.random())
    return model.objects.filter(id__gte=min_id)[0]

It is faster than order_by(‘?’) in almost all cases.

Here’s a simple solution:

from random import randint

count = Model.objects.count()
random_object = Model.objects.all()[randint(0, count - 1)] #single random object

You could create a manager on your model to do this sort of thing. To first understand what a manager is, the Painting.objects method is a manager that contains all(), filter(), get(), etc. Creating your own manager allows you to pre-filter results and have all these same methods, as well as your own custom methods, work on the results.

EDIT: I modified my code to reflect the order_by['?'] method. Note that the manager returns an unlimited number of random models. Because of this I’ve included a bit of usage code to show how to get just a single model.

from django.db import models

class RandomManager(models.Manager):
    def get_query_set(self):
        return super(RandomManager, self).get_query_set().order_by('?')

class Painting(models.Model):
    title = models.CharField(max_length=100)
    author = models.CharField(max_length=50)

    objects = models.Manager() # The default manager.
    randoms = RandomManager() # The random-specific manager.

Usage

random_painting = Painting.randoms.all()[0]

Lastly, you can have many managers on your models, so feel free to create a LeastViewsManager() or MostPopularManager().

The other answers are either potentially slow (using order_by('?')) or use more than one SQL query. Here’s a sample solution with no ordering and just one query (assuming Postgres):

random_instance_or_none = Model.objects.raw('''
    select * from {0} limit 1
    offset floor(random() * (select count(*) from {0}))
'''.format(Model._meta.db_table)).first()

Be aware that this will raise an index error if the table is empty. Write yourself a model-agnostic helper function to check for that.

Just a simple idea how I do it:

def _get_random_service(self, professional):
    services = Service.objects.filter(professional=professional)
    i = randint(0, services.count()-1)
    return services[i]

Hi I needed to select a random record from a queryset who’s length I also needed to report (ie web page produced described item and said records left)

q = Entity.objects.filter(attribute_value='this or that')
item_count = q.count()
random_item = q[random.randomint(1,item_count+1)]

took half as long(0.7s vs 1.7s) as:

item_count = q.count()
random_item = random.choice(q)

I’m guessing it avoids pulling down the whole query before selecting the random entry and made my system responsive enough for a page that is accessed repeatedly for a repetitive task where users want to see the item_count count down.

Randomization in DB feels nasty and better in python. But at the same time, it’s not a good idea to bring all the data from DB to python memory just to ignore most of the results (especially in the production environment). we might need some sort of filtering also.

1. So Basically we have data at DB,
2. we wanna use the rand function of python
3. and afterwords bring up the whole required data from DB.

Basically using 2 queries will be much less expensive than picking random in DB CPU (computing in DB) or loading whole data (heavy Network Utilization). Solutions explained must need a scalable nature trying to plan here won’t work for a production environment espicially with filters, soft/hard deletes, or even with an is_public flag. because probably random id we generated might be deleted from the database or will be cut down in filters. Its a bad practice to assume max_id(records) == count(records).

(Ofcouce, If you do’not delete a percentage of data which is comparable to query uses, or if you dont wanna use any kond of filters, and if you are confident, random id which you can proceed with a random )

if you want only one items. Refer ( @Valter Silva )

import random

mgr = models.Painting.objects
qs = mgr.filter(...)
random_id = random.choice(1, qs.count())-1        # <--- [ First Query Hit ]

random_paint = qs[random_id] ## <-- [ Second Query Hit ]

if you want ‘n’ items.

import random

req_no_of_random_items = 8        ## i need 8 random items.
qs = models.Painting.objects.filter(...)

## if u prefer to use random values often, you can keep this in cache. 
possible_ids = list(qs.values_list('id', flat=True))        # <--- [ First Query Hit ]

possible_ids = random.choices(possible_ids, k=8)
random_paint = qs.filter(pk__in=possible_ids) ## in a generic case to get 'n' items.

or if you want to have a more optimized code for production, use a cachefunction to get ids of products:

from django.core.cache import cache

def id_set_cache(qs):
    key = "some_random_key_for_cache"
    id_set =  cache.get(key)
    if id_set is None:
        id_set = list(qs.values_list('id', flat=True)
        cache.set(key, id_set)
    retrun id_set

Just to note a (fairly common) special case, if there is a indexed auto-increment column in the table with no deletes, the optimum way to do a random select is a query like:

SELECT * FROM table WHERE id = RAND() LIMIT 1

that assumes such a column named id for table. In django you can do this by:

Painting.objects.raw('SELECT * FROM appname_painting WHERE id = RAND() LIMIT 1')

in which you must replace appname with your application name.

In General, with an id column, the order_by(‘?’) can be done much faster with:

Paiting.objects.raw(
        'SELECT * FROM auth_user WHERE id>=RAND() * (SELECT MAX(id) FROM auth_user) LIMIT %d' 
    % needed_count)

This is Highly recomended Getting a random row from a relational database

Because using django orm to do such a thing like that, will makes your db server angry specially if you have big data table :|

And the solution is provide a Model Manager and write the SQL query by hand ;)

Update:

Another solution which works on any database backend even non-rel ones without writing custom ModelManager. Getting Random objects from a Queryset in Django

You may want to use the same approach that you’d use to sample any iterator, especially if you plan to sample multiple items to create a sample set. @MatijnPieters and @DzinX put a lot of thought into this:

def random_sampling(qs, N=1):
    """Sample any iterable (like a Django QuerySet) to retrieve N random elements

    Arguments:
      qs (iterable): Any iterable (like a Django QuerySet)
      N (int): Number of samples to retrieve at random from the iterable

    References:
      @DZinX:  https://stackoverflow.com/a/12583436/623735
      @MartinPieters: https://stackoverflow.com/a/12581484/623735
    """
    samples = []
    iterator = iter(qs)
    # Get the first `N` elements and put them in your results list to preallocate memory
    try:
        for _ in xrange(N):
            samples.append(iterator.next())
    except StopIteration:
        raise ValueError("N, the number of reuested samples, is larger than the length of the iterable.")
    random.shuffle(samples)  # Randomize your list of N objects
    # Now replace each element by a truly random sample
    for i, v in enumerate(qs, N):
        r = random.randint(0, i)
        if r < N:
            samples[r] = v  # at a decreasing rate, replace random items
    return samples

One much easier approach to this involves simply filtering down to the recordset of interest and using random.sample to select as many as you want:

from myapp.models import MyModel
import random

my_queryset = MyModel.objects.filter(criteria=True)  # Returns a QuerySet
my_object = random.sample(my_queryset, 1)  # get a single random element from my_queryset
my_objects = random.sample(my_queryset, 5)  # get five random elements from my_queryset

Note that you should have some code in place to verify that my_queryset is not empty; random.sample returns ValueError: sample larger than population if the first argument contains too few elements.

Method for auto-incrementing primary key with no deletes

If you have a table where the primary key is a sequential integer with no gaps, then the following method should work:

import random
max_id = MyModel.objects.last().id
random_id = random.randint(0, max_id)
random_obj = MyModel.objects.get(pk=random_id)

This method is much more efficient than other methods here that iterate through all rows of the table. While it does require two database queries, both are trivial. Furthermore, it’s simple and doesn’t require defining any extra classes. However, it’s applicability is limited to tables with an auto-incrementing primary key where rows have never deleted, such that there are no gaps in the sequence of ids.

In the case where rows have been deleted such that are gaps, this method could still work if it is retried until an existing primary key is randomly selected.

References

• https://stackoverflow.com/a/10836811/4651668
• https://stackoverflow.com/a/2118712/4651668
• https://stackoverflow.com/a/39751708/4651668
• https://github.com/greenelab/hetmech-backend/pull/48

I got very simple solution, make custom manager:

class RandomManager(models.Manager):
    def random(self):
        return random.choice(self.all())

and then add in model:

class Example(models.Model):
    name = models.CharField(max_length=128)
    objects = RandomManager()

Now, you can use it:

Example.objects.random()