Context
I have made a simple web app for uploading content to a blog. The front sends AJAX requests (using FormData) to the backend which is Bottle running on Python 3.7. Text content is saved to a MySQL database and images are saved to a folder on the server. Everything works fine.
Image processing and PIL/Pillow
Now, I want to enable processing of uploaded images to standardise them (I need them all resized and/or cropped to 700x400px).
I was hoping to use Pillow for this. My problem is creating a PIL Image object from the file object in Bottle. I cannot initialise a valid Image object.
Code
# AJAX sends request to this route
#post('/update')
def update():
# Form data
title = request.forms.get("title")
body = request.forms.get("body")
image = request.forms.get("image")
author = request.forms.get("author")
# Image upload
file = request.files.get("file")
if file:
extension = file.filename.split(".")[-1]
if extension not in ('png', 'jpg', 'jpeg'):
return {"result" : 0, "message": "File Format Error"}
save_path = "my/save/path"
file.save(save_path)
The problem
This all works as expected, but I cannot create a valid Image object with pillow for processing. I even tried reloading the saved image using the save path but this did not work either.
Other attempts
The code below did not work. It caused an internal server error, though I am having trouble setting up more detailed Python debugging.
path = save_path + "/" + file.filename
image_data = open(path, "rb")
image = Image.open(image_data)
When logged manually, the path is a valid relative URL ("../domain-folder/images") and I have checked that I am definitely importing PIL (Pillow) correctly using PIL.PILLOW_VERSION.
I tried adapting this answer:
image = Image.frombytes('RGBA', (128,128), image_data, 'raw')
However, I won’t know the size until I have created the Image object. I also tried using io:
image = Image.open(io.BytesIO(image_data))
This did not work either. In each case, it is only the line trying to initialise the Image object that causes problems.
Summary
The Bottle documentation says the uploaded file is a file-like object, but I am not having much success in creating an Image object that I can process.
How should I go about this? I do not have a preference about processing before or after saving. I am comfortable with the processing, it is initialising the Image object that is causing the problem.
Edit - Solution
I got this to work by adapting the answer from eatmeimadanish. I had to use a io.BytesIO object to save the file from Bottle, then load it with Pillow from there. After processing, it could be saved in the usual way.
obj = io.BytesIO()
file.save(obj) # This saves the file retrieved by Bottle to the BytesIO object
path = save_path + "/" + file.filename
# Image processing
im = Image.open(obj) # Reopen the object with PIL
im = im.resize((700,400))
im.save(path, optimize=True)
I found this from the Pillow documentation about a different function that may also be of use.
PIL.Image.frombuffer(mode, size, data, decoder_name='raw', *args)
Note that this function decodes pixel data only, not entire images.
If you have an entire image file in a string, wrap it in a BytesIO object, and use open() to load it.
Use StringIO instead.
From PIL import Image
try:
import cStringIO as StringIO
except ImportError:
import StringIO
s = StringIO.StringIO()
#save your in memory file to this instead of a regular file
file = request.files.get("file")
if file:
extension = file.filename.split(".")[-1]
if extension not in ('png', 'jpg', 'jpeg'):
return {"result" : 0, "message": "File Format Error"}
file.save(s)
im = Image.open(s)
im.resize((700,400))
im.save(s, 'png', optimize=True)
s64 = base64.b64encode(s.getvalue())
From what I understand, you're trying to resize the image after it has been saved locally (note that you could try to do the resize before it is saved). If this is what you want to achieve here, you can open the image directly using Pillow, it does the job for you (you do not have to open(path, "rb"):
image = Image.open(path)
image.resize((700,400)).save(path)
Related
I can do this OK both in js and php but not in python. I'm trying to pull a thumbnail image from google books api into a python variable.
The text objects are fine eg
newTitle = (parsed_json['items'][0]['volumeInfo']['title'])
isbn10 = (parsed_json['items'][0]['volumeInfo']['industryIdentifiers'][1]['identifier'])
isbn13 = (parsed_json['items'][0]['volumeInfo']['industryIdentifiers'][0]['identifier'])
The image is supplied in the api as follows. (if you put the http// url into a browser you see the image):
"imageLinks": {
"smallThumbnail": "http://books.google.com/books/content?id=XUnNDwAAQBAJ&printsec=frontcover&img=1&zoom=5&edge=curl&source=gbs_api",
"thumbnail": "http://books.google.com/books/content?id=XUnNDwAAQBAJ&printsec=frontcover&img=1&zoom=1&edge=curl&source=gbs_api"
I have tried the simple:
myImage = (parsed_json['items'][0]['volumeInfo']['imageLinks'][thumbnail])
which doesn't work.
I have installed pillow to provide image management:
from PIL import Image
img = Image.open("parsed_json['items'][0]['volumeInfo']['imageLinks'][thumbnail]") or
img = Image.open(parsed_json['items'][0]['volumeInfo']['imageLinks'][thumbnail])
which doesn't work. I have tried more complex arrangements:
from PIL import Image
import requests
from io import BytesIO
response = requests.get(parsed_json['items'][0]['volumeInfo']['imageLinks'][thumbnail])
img = Image.open(BytesIO(response.content))
but nothing seems to work. I have tried many other variations on these attempts. I have also, unsuccessfully tried to load the text that points to the thumbnail to try another route. I am confident that the "['items'][0]['volumeInfo']['imageLinks'][thumbnail]" is correct though my only way of testing whether the variable is properly loaded is to save it or if the line of code isn't working.
I didn't have problems downloading and opening the image.
I have use the following code
import requests
from PIL import Image
image_url = "https://books.google.com/books/content?id=XUnNDwAAQBAJ&printsec=frontcover&img=1&zoom=5&edge=curl&source=gbs_api"
r = requests.get(image_url)
with open("demo_image",'wb') as f:
f.write(r.content)
img = Image.open("demo_image")
I am using the Drive API to download an image. Following their file downloading documentation in Python, I end up with a variable fh that is a populated io.BytesIO instance. I try to save it as an image:
file_id = "0BwyLGoHzn5uIOHVycFZpSEwycnViUjFYQXR5Nnp6QjBrLXJR"
request = service.files().get_media(fileId=file_id)
fh = io.BytesIO()
downloader = MediaIoBaseDownload(fh, request)
done = False
while done is False:
status, done = downloader.next_chunk()
print('Download {} {}%.'.format(file['name'],
int(status.progress() * 100)))
fh.seek(0)
image = Image.open(fh) # error
The error is: cannot identify image file <_io.BytesIO object at 0x106cba890>. Actually, the error does not occur with another image but is thrown with most images, including the one I linked at the beginning of this post.
After reading this answer I change that last line to:
byteImg = fh.read()
dataBytesIO = io.BytesIO(byteImg)
image = Image.open(dataBytesIO) # still the same error
I've also tried this answer, where I change the last line of my first code block to
byteImg = fh.read()
image = Image.open(StringIO(byteImg))
But I still get a cannot identify image file <StringIO.StringIO instance at 0x106471e60> error.
I've tried using alternates (requests, urllib) with no fruition. I can Image.open the the image if I download it manually.
This error was not present a month ago, and has recently popped up into the application this code is in. I've spent days debugging this error with no success and have finally brought the issue to Stack Overflow. I am using from PIL import Image.
Ditch the Drive service's MediaIOBaseDownload. Instead, use the webContentLink property of a media file (a link for downloading the content of the file in a browser, only available for files with binary content). Read more here.
With that content link, we can use an alternate form of streaming—the requests and shutil libraries and the —to get the image.
import requests
import shutil
r = requests.get(file['webContentLink'], stream=True)
with open('output_file', 'wb') as f:
shutil.copyfileobj(r.raw, f)
I am using the Pillow fork of PIL and keep receiving the error
OSError: cannot identify image file <_io.BytesIO object at 0x103a47468>
when trying to open an image. I am using virtualenv with python 3.4 and no installation of PIL.
I have tried to find a solution to this based on others encountering the same problem, however, those solutions did not work for me. Here is my code:
from PIL import Image
import io
# This portion is part of my test code
byteImg = Image.open("some/location/to/a/file/in/my/directories.png").tobytes()
# Non test code
dataBytesIO = io.BytesIO(byteImg)
Image.open(dataBytesIO) # <- Error here
The image exists in the initial opening of the file and it gets converted to bytes. This appears to work for almost everyone else but I can't figure out why it fails for me.
EDIT:
dataBytesIO.seek(0)
does not work as a solution (tried it) since I'm not saving the image via a stream, I'm just instantiating the BytesIO with data, therefore (if I'm thinking of this correctly) seek should already be at 0.
(This solution is from the author himself. I have just moved it here.)
SOLUTION:
# This portion is part of my test code
byteImgIO = io.BytesIO()
byteImg = Image.open("some/location/to/a/file/in/my/directories.png")
byteImg.save(byteImgIO, "PNG")
byteImgIO.seek(0)
byteImg = byteImgIO.read()
# Non test code
dataBytesIO = io.BytesIO(byteImg)
Image.open(dataBytesIO)
The problem was with the way that Image.tobytes()was returning the byte object. It appeared to be invalid data and the 'encoding' couldn't be anything other than raw which still appeared to output wrong data since almost every byte appeared in the format \xff\. However, saving the bytes via BytesIO and using the .read() function to read the entire image gave the correct bytes that when needed later could actually be used.
image = Image.open(io.BytesIO(decoded))
# File "C:\Users\14088\anaconda3\envs\tensorflow\lib\site-packages\PIL\Image.py", line 2968, in open
# "cannot identify image file %r" % (filename if filename else fp)
# PIL.UnidentifiedImageError: cannot identify image file <_io.BytesIO object at 0x000002B733BB11C8>
===
I fixed as worked:
message = request.get_json(force=True)
encoded = message['image']
# https://stackoverflow.com/questions/26070547/decoding-base64-from-post-to-use-in-pil
#image_data = re.sub('^data:image/.+;base64,', '', message['image'])
image_data = re.sub('^data:image/.+;base64,', '', encoded)
# Remove extra "data:image/...'base64" is Very important
# If "data:image/...'base64" is not remove, the following line generate an error message:
# File "C:\Work\SVU\950_SVU_DL_TF\sec07_TF_Flask06_09\32_KerasFlask06_VisualD3\32_predict_app.py", line 69, in predict
# image = Image.open(io.BytesIO(decoded))
# File "C:\Users\14088\anaconda3\envs\tensorflow\lib\site-packages\PIL\Image.py", line 2968, in open
# "cannot identify image file %r" % (filename if filename else fp)
# PIL.UnidentifiedImageError: cannot identify image file <_io.BytesIO object at 0x000002B733BB11C8>
# image = Image.open(BytesIO(base64.b64decode(image_data)))
decoded = base64.b64decode(image_data)
image = Image.open(io.BytesIO(decoded))
# return json.dumps({'result': 'success'}), 200, {'ContentType': 'application/json'}
#print('#app.route => image:')
#print()
processed_image = preprocess_image(image, target_size=(224, 224))
prediction = model.predict(processed_image).tolist()
#print('prediction:', prediction)
response = {
'prediction': {
'dog': prediction[0][0],
'cat': prediction[0][1]
}
}
print('response:', response)
return jsonify(response)
On some cases the same error happens when you are dealing with a Raw Image file such CR2. Example: http://www.rawsamples.ch/raws/canon/g10/RAW_CANON_G10.CR2
when you try to run:
byteImg = Image.open("RAW_CANON_G10.CR2")
You will get this error:
OSError: cannot identify image file 'RAW_CANON_G10.CR2'
So you need to convert the image using rawkit first, here is an example how to do it:
from io import BytesIO
from PIL import Image, ImageFile
import numpy
from rawkit import raw
def convert_cr2_to_jpg(raw_image):
raw_image_process = raw.Raw(raw_image)
buffered_image = numpy.array(raw_image_process.to_buffer())
if raw_image_process.metadata.orientation == 0:
jpg_image_height = raw_image_process.metadata.height
jpg_image_width = raw_image_process.metadata.width
else:
jpg_image_height = raw_image_process.metadata.width
jpg_image_width = raw_image_process.metadata.height
jpg_image = Image.frombytes('RGB', (jpg_image_width, jpg_image_height), buffered_image)
return jpg_image
byteImg = convert_cr2_to_jpg("RAW_CANON_G10.CR2")
Code credit if for mateusz-michalik on GitHub (https://github.com/mateusz-michalik/cr2-to-jpg/blob/master/cr2-to-jpg.py)
While reading Dicom files the problem might be caused due to Dicom compression.
Make sure both gdcm and pydicom are installed.
GDCM is usually the one that's more difficult to install. The latest way to easily install the same is
conda install -U conda-forge gdcm
When dealing with url, this error can arise from a wrong extension of the downloaded
file or just a corrupted file.
So to avoid that use a try/except bloc so you app doesn't crash and will continue its job.
In the except part, you can retrieve the file in question for analysis:
A snippet here:
for url in urls:
with closing(urllib.request.urlopen(url)) as f:
try:
img = Image(f, 30*mm, 30*mm)
d_img.append(img)
except Exception as e:
print(url) #here you get the file causing the exception
print(e)
Here a related answer
The image file itself might be corrupted. So if you were to process a considerable amount of image files, then simply enclose the line that processes each image file with a try catch statement.
I have a stored picture on my computer. I open it using the Python Image module. Then I crop this image into several pieces using this module. To conclude, I would like to upload the image via a POST request on a website.
Because that small images are PIL object, I converted each of them into StringIO to be able to send the form without having to save them on my PC.
Unfortunately, I encounter an error, whereas if the images are physically stored on my PC, there is no problem. I do not understand why.
You can visit the website here: http://www.noelshack.com/api.php
This is a very basic API that returns the link to the uploaded picture.
In my case, the problem is that returns nothing, at the end of the second image (no problem for the first).
Here is the programming code to crop the image into 100 pieces.
import requests
import Image
import StringIO
import os
image = Image.open("test.jpg")
width, height = image.size
images = []
for i in range(10):
for j in range(10):
crop = image.crop((i * 10, j * 10, (i + 1) * 10, (j + 1) * 10))
images.append(crop)
The function to upload an image:
def upload(my_file):
api_url = 'http://www.noelshack.com/api.php'
r = requests.post(api_url, files={'fichier': my_file})
if not 'www.noelshack.com' in r.text:
raise Exception(r.text)
return r.text
Now we have two possibilities. The first is to save each of the 100 images on disk and upload them.
if not os.path.exists("directory"):
os.makedirs("directory")
i = 0
for img in images:
img.save("directory/" + str(i) + ".jpg")
i += 1
for file in os.listdir("directory"):
with open("directory/" + file, "rb") as f:
print upload(f)
It works like a charm, but it is not very convenient. So, I thought to use StringIO.
for img in images:
my_file = StringIO.StringIO()
img.save(my_file, "JPEG")
print upload(my_file.getvalue())
# my_file.close() -> Does not change anything
The first link is printed, but the function raise the exception then.
I think the problem lies in the img.save(), because the same kind of for loop was not working to save to disk and then upload. In addition, if you add a time.sleep(1) between the uploads, it seems to work.
Any help would be welcome please, because I'm really stuck.
my_file.getvalue() returns a string. What you need is a file-like object, which my_file already is. And file like objects have a cursor, so to speak, which says where to read from or write to. So, if you do my_file.seek(0) before the upload, it should get fixed.
modify the code to:
for img in images:
my_file = StringIO.StringIO()
img.save(my_file, "JPEG")
my_file.seek(0)
print upload(my_file)
I've been attempting to work at this for hours but decided to turn to the experts here on stackoverflow.
I'm trying to download an image from a url:
import urllib
originalphoto = urllib.urlretrieve(bundle.obj.url)
#originalphoto is being saved to the tmp directory in Ubuntu
This works and it saves the image in the tmp directory, but I need to modify this image by resizing it to a 250px by 250px image and then save it to a folder on my Desktop: /home/ubuntu/Desktop/resizedshots
The name of the original image is in bundle.obj.url, for example if bundle.obj.url is:
http://photographs.500px.com/kyle/09-09-201315-47-571378756077.jpg the name of the image is "09-09-201315-47-571378756077.jpg"
After the image is resized, I need to save is to this folder as 09-09-201315-47-571378756077small.jpg
As you can see, I'm adding in the word "small" to the end the file name. Once all of this is done, I would like to delete the temporary image file that was downloaded so that it doesn't take up the disk.
Any ideas on how this can be done?
Thanks
This is the definition:
def urlretrieve(url, filename=None, reporthook=None, data=None):
You can set the second argument to something you know and then do
import os
os.remove(something_you_know)
If you do not set the second argument you do this:
import urllib, os
url = 'http://photographs.500px.com/kyle/09-09-201315-47-571378756077.jpg'
file, headers = urllib.urlretrieve(url)
# do something
os.remove(file)
if os.remove does not work you still have the file open.