How do you convert a PIL `Image` to a Django `File`?

The way to do this without having to write back to the filesystem, and then bring the file back into memory via an open call, is to make use of StringIO and Django InMemoryUploadedFile. Here is a quick sample on how you might do this. This assumes that you already have a thumbnailed image named 'thumb':

import StringIO

from django.core.files.uploadedfile import InMemoryUploadedFile

# Create a file-like object to write thumb data (thumb data previously created
# using PIL, and stored in variable 'thumb')
thumb_io = StringIO.StringIO()
thumb.save(thumb_io, format='JPEG')

# Create a new Django file-like object to be used in models as ImageField using
# InMemoryUploadedFile.  If you look at the source in Django, a
# SimpleUploadedFile is essentially instantiated similarly to what is shown here
thumb_file = InMemoryUploadedFile(thumb_io, None, 'foo.jpg', 'image/jpeg',
                                  thumb_io.len, None)

# Once you have a Django file-like object, you may assign it to your ImageField
# and save.
...

Let me know if you need more clarification. I have this working in my project right now, uploading to S3 using django-storages. This took me the better part of a day to properly find the solution here.

This is actual working example for python 3.5 and django 1.10

in views.py:

from io import BytesIO
from django.core.files.base import ContentFile
from django.core.files.uploadedfile import InMemoryUploadedFile

def pill(image_io):
    im = Image.open(image_io)
    ltrb_border = (0, 0, 0, 10)
    im_with_border = ImageOps.expand(im, border=ltrb_border, fill='white')

    buffer = BytesIO()
    im_with_border.save(fp=buffer, format='JPEG')
    buff_val = buffer.getvalue()
    return ContentFile(buff_val)

def save_img(request)
    if request.POST:
       new_record = AddNewRecordForm(request.POST, request.FILES)
       pillow_image = pill(request.FILES['image'])
       image_file = InMemoryUploadedFile(pillow_image, None, 'foo.jpg', 'image/jpeg', pillow_image.tell, None)
       request.FILES['image'] = image_file  # really need rewrite img in POST for success form validation
       new_record.image = request.FILES['image']
       new_record.save()
       return redirect(...)

To complete for those who, like me, want to couple it with Django's FileSystemStorage: (What I do here is upload an image, resize it to 2 dimensions and save both files.

utils.py

def resize_and_save(file):
    size = 1024, 1024
    thumbnail_size = 300, 300
    uploaded_file_url = getURLforFile(file, size, MEDIA_ROOT)
    uploaded_thumbnail_url = getURLforFile(file, thumbnail_size, THUMBNAIL_ROOT)
    return [uploaded_file_url, uploaded_thumbnail_url]

def getURLforFile(file, size, location):
    img = Image.open(file)
    img.thumbnail(size, Image.ANTIALIAS)
    thumb_io = BytesIO()
    img.save(thumb_io, format='JPEG')
    thumb_file = InMemoryUploadedFile(thumb_io, None, file.name, 'image/jpeg', thumb_io.tell, None)
    fs = FileSystemStorage(location=location)
    filename = fs.save(file.name, thumb_file)
    return fs.url(filename)  

In views.py

if request.FILES:
        fl, thumbnail = resize_and_save(request.FILES['avatar'])
        #delete old profile picture before saving new one
        try:
            os.remove(BASE_DIR + user.userprofile.avatarURL)
        except Exception as e:
            pass         
        user.userprofile.avatarURL = fl
        user.userprofile.thumbnailURL = thumbnail
        user.userprofile.save()