I'm trying to use OpenCV2 to read images withing a zipfile with cv2.imread and then resize them and save them but I can't figure out how to read those images within the zipfile. This is my code so far:
import cv2
from zipfile import ZipFile
with ZipFile("sample_images.zip", "r") as zipFile:
imgs = zipFile.namelist()
img = zipFile.open(imgs[0])
processedImg = cv2.imread(img)
This is the error I get:
Traceback (most recent call last):
File "C:/Users/Yash/Desktop/Programming/opencv test/test.py", line 8, in <module>
processedImg = cv2.imread(img)
SystemError: <built-in function imread> returned NULL without setting an error
Try this
from io import BytesIO
from zipfile import ZipFile
from PIL import Image
with ZipFile("sample_images.zip", "r") as zipFile:
imgs = zipFile.namelist()
print(imgs[0])
img = zipFile.read(imgs[0])
processedImg = Image.open(BytesIO(img))
processedImg.show()
Related
I have an image and when I process it, I can´t get the colors.
Here is my code:
import cv2
from PIL import Image
im = Image.open ('imag.jpg')
ret,thresh1 = cv2.threshold(im,127,255,cv2.THRESH_BINARY)
from collections import defaultdict
by_color = defaultdict(int)
for pixel in thresh1.getdata():
by_color[pixel] += 1
print (by_color)
If I run just the original image (im), works well. But when I run processed image (thresh1), the following error appears:
Traceback (most recent call last):
File "file.py", line 62, in <module>
ret,thresh1 = cv2.threshold(im,127,255,cv2.THRESH_BINARY)
TypeError: Expected cv::UMat for argument 'src'
For this function:
ret,thresh1 = cv2.threshold(im,127,255,cv2.THRESH_BINARY)
to work you should read image using cv2.imread (returns numpy.array) rather than PIL.Image.open (returns Image object).
I am trying to display an image from a Url however i am not sure how to do so.
Below is my attempt:
imageFile = "http://photo.elsoar.com/wp-content/images/Personal-computer.jpg"
image1 =PhotoImage(open(imageFile))
image1.grid(row=1, column=5)
This just produces an error
Traceback (most recent call last):
File "C:\Users\Derren\Desktop\Online Shopper.py", line 131, in <module>
image1 =PhotoImage(open(imageFile))
IOError: [Errno 22] invalid mode ('r') or filename: 'http://photo.elsoar.com/wp-content/images/Personal-computer.jpg'
This is the image i want to have
http://photo.elsoar.com/wp-content/images/Personal-computer.jpg
However it is essential that it is resourced from the internet and not local files
The function open() only works on locals files. You can see doc here part 7.2. If you want to work with an online image, you can use this :
from PIL import Image
import requests
from io import BytesIO
response = requests.get("url")
img = Image.open(BytesIO(response.content))
img.save("my_img.png") #with the good filename extension
If you particularly want a PhotoImage based solution that this should work:
import urllib
f = urllib.urlopen("http://photo.elsoar.com/wp-content/images/Personal-computer.jpg")
data=f.read()
root = Tkinter.Tk()
image1 =PhotoImage(data)
Having trouble with this error code regarding the following code for Pytesseract. (Python 3.6.1, Mac OSX)
import pytesseract
import requests
from PIL import Image
from PIL import ImageFilter
from io import StringIO, BytesIO
def process_image(url):
image = _get_image(url)
image.filter(ImageFilter.SHARPEN)
return pytesseract.image_to_string(image)
def _get_image(url):
r = requests.get(url)
s = BytesIO(r.content)
img = Image.open(s)
return img
process_image("https://www.prepressure.com/images/fonts_sample_ocra_medium.png")
Error:
/usr/local/Cellar/python3/3.6.0_1/Frameworks/Python.framework/Versions/3.6/bin/python3.6 /Users/g/pyfo/reddit/ocr.py
Traceback (most recent call last):
File "/Users/g/pyfo/reddit/ocr.py", line 20, in <module>
process_image("https://www.prepressure.com/images/fonts_sample_ocra_medium.png")
File "/Users/g/pyfo/reddit/ocr.py", line 10, in process_image
image.filter(ImageFilter.SHARPEN)
File "/usr/local/lib/python3.6/site-packages/PIL/Image.py", line 1094, in filter
return self._new(filter.filter(self.im))
File "/usr/local/lib/python3.6/site-packages/PIL/ImageFilter.py", line 53, in filter
raise ValueError("cannot filter palette images")
ValueError: cannot filter palette images
Process finished with exit code 1
Seems simple enough, but is not working. Any help would be greatly appreciated.
The image you have is a pallet-based image. You need to convert it to a full RGB image in order to use the PIL filters.
import pytesseract
import requests
from PIL import Image, ImageFilter
from io import StringIO, BytesIO
def process_image(url):
image = _get_image(url)
image = image.convert('RGB')
image = image.filter(ImageFilter.SHARPEN)
return pytesseract.image_to_string(image)
def _get_image(url):
r = requests.get(url)
s = BytesIO(r.content)
img = Image.open(s)
return img
process_image("https://www.prepressure.com/images/fonts_sample_ocra_medium.png")
You should also note that the the .convert() and .filter() methods return a copy of the image, they don't change the existing image object. You need to assign the return value to a variable as shown in the code above.
NOTE: I don't have pytesseract, so I can't check the last line of process_image().
from PIL import Image
image = Image.open("image.jpg")
file_path = io.BytesIO();
image.save(file_path,'JPEG');
image2 = Image.open(file_path.getvalue());
I get this error TypeError: embedded NUL character on the last statement Image.open on running the program
What is the correct way to open a file from streams?
http://effbot.org/imagingbook/introduction.htm#more-on-reading-images
from PIL import Image
import StringIO
buffer = StringIO.StringIO()
buffer.write(open('image.jpeg', 'rb').read())
buffer.seek(0)
image = Image.open(buffer)
print image
# <PIL.JpegImagePlugin.JpegImageFile image mode=RGB size=800x600 at 0x7FE2EEE2B098>
# if we try open again
image = Image.open(buffer)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/lib/python2.7/dist-packages/PIL/Image.py", line 2028, in open
raise IOError("cannot identify image file")
IOError: cannot identify image file
Make sure you call buff.seek(0) before reading any StringIO objects. Otherwise you'll be reading from the end of the buffer, which will look like an empty file and is likely causing the error you're seeing.
Using BytesIO is much more simple, it took me a while to figure out. This allows you to read and write to zip files for example.
from PIL import Image
from io import BytesIO
# bytes of a simple 2x2 gif file
gif_bytes = b'\x47\x49\x46\x38\x39\x61\x02\x00\x02\x00\x80\x00\x00\x00\xFF\xFF\xFF\x21\xF9\x04\x00\x00\x00\x00\x00\x2C\x00\x00\x00\x00\x02\x00\x02\x00\x00\x02\x03\x44\x02\x05\x00\x3B'
gif_bytes_io = BytesIO() # or io.BytesIO()
# store the gif bytes to the IO and open as image
gif_bytes_io.write(gif_bytes)
image = Image.open(gif_bytes_io)
# optional proof of concept:
# image.show()
# save as png through a stream
png_bytes_io = BytesIO() # or io.BytesIO()
image.save(png_bytes_io, format='PNG')
print(png_bytes_io.getvalue()) # outputs the byte stream of the png
How to read image from StringIO into PIL in python? I will have a stringIO object. How to I read from it with a image in it? I cant event have ot read a image from a file. Wow!
from StringIO import StringIO
from PIL import Image
image_file = StringIO(open("test.gif",'rb').readlines())
im = Image.open(image_file)
print im.format, "%dx%d" % im.size, im.mode
Traceback (most recent call last):
File "/home/ubuntu/workspace/receipt/imap_poller.py", line 22, in <module>
im = Image.open(image_file)
File "/usr/local/lib/python2.7/dist-packages/Pillow-2.3.1-py2.7-linux-x86_64.egg/PIL/Image.py", line 2028, in open
raise IOError("cannot identify image file")
IOError: cannot identify image file
Don't use readlines(), it returns a list of strings which is not what you want. To retrieve the bytes from the file, use read() function instead.
Your example worked out of the box with read() and a JPG file on my PC:
# Python 2.x
>>>from StringIO import StringIO
# Python 3.x
>>>from io import StringIO
>>>from PIL import Image
>>>image_file = StringIO(open("test.jpg",'rb').read())
>>>im = Image.open(image_file)
>>>print im.size, im.mode
(2121, 3508) RGB