I am trying to read an img from a url. My code works when the img is coded jpeg, but, when It's .svg for example, the function imdecode from openCV just don't work. Here it is the code:
url_response = urllib.urlopen("https://upload.wikimedia.org/wikipedia/commons/a/ac/BanderaEstepa.svg")
img_array = array(bytearray(url_response.read()), dtype=uint8)
img_url = cv2.imdecode(img_array, 1)
Need help please. Thanks in advance.
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 trying to resolve captcha for the following image
!https://ibb.co/35X723J
I have tried using tessaract
data = br.open(captchaurl).read()
b = bytearray(data)
save = open(filename, 'wb')
save.write(data)
save.close()
ctext= pytesseract.image_to_string(Image.open(filename))
Here is a workaround. You need to clear a bit the image but you wont get a perfect result. Try the following:
try:
from PIL import Image
except ImportError:
import Image
import pytesseract
import cv2
file = 'sample.jpg'
img = cv2.imread(file, cv2.IMREAD_GRAYSCALE)
img = cv2.resize(img, None, fx=10, fy=10, interpolation=cv2.INTER_LINEAR)
img = cv2.medianBlur(img, 9)
th, img = cv2.threshold(img, 185, 255, cv2.THRESH_BINARY)
kernel = cv2.getStructuringElement(cv2.MORPH_RECT, (4,8))
img = cv2.morphologyEx(img, cv2.MORPH_CLOSE, kernel)
cv2.imwrite("sample2.jpg", img)
file = 'sample2.jpg'
text = pytesseract.image_to_string(file)
print(''.join(x for x in text if x.isdigit()))
Option 1:
I think using Pytesseract should solve the issue. I tried out your code and it gave me the following result when i gave in the exact cropped captcha image as input into pytesseract:
Input Image:
Output:
print(ctext)
'436359 oS'
I suggest you don't give the full page url as input into pytesseract. Instead give the exact image url as "https://i.ibb.co/RGn9fF5/Jpeg-Image-CS2.jpg" which will take in only the image.
And regarding the extra 'oS' characters in the output, you can do a string manipulation to chop off the characters other than numbers in the output.
re.sub("[^0-9]", "", ctext)
Option 2:
You can also use google's OCR to accomplish this which gives you the exact result without errors. Though I have shown you the web interface of it, google has nice python libraries through which you can accomplish this using python itself. Looks like this:
I want to do a practice that consists of capturing webs in jpg, but it did not just work (I am newbie), this is the code I use.
import numpy as np
import urllib
import cv2
def url_to_image("http://www.hereiputweb.com"):
resp = urllib.urlopen("http://www.hereiputweb.com")
image = np.asarray(bytearray(resp.read()), dtype="uint8")
image = cv2.imdecode(image, cv2.IMREAD_COLOR)
return image
The code I got it from a manual but it gives me fault in the line:
def url_to_image("http://www.hereiputweb.com"):
I think I indicated the web incorrectly, very far I should not be .. tried several forms but nothing .. what do I do wrong?
regards
There is a really brief tutorial (https://docs.python.org/3/tutorial/).
The relevant part would be https://docs.python.org/3/tutorial/controlflow.html#defining-functions
So, you should define your function as follows:
def url_to_image(url):
resp = urllib.urlopen(url)
image = np.asarray(bytearray(resp.read()), dtype="uint8")
image = cv2.imdecode(image, cv2.IMREAD_COLOR)
return image
I have not checked the implementation works ;)
Then you can use your function:
url = "http://www.hereiputweb.com"
my_image = url_to_image(url)
The problem is not with your implementation, it's with your URL!
This method require a functioning URL that returns an image. The URL you're using is not an image.
Try using an URL of an image (e.g: some URLs that end with .jpg) and it shall work!
Remember that the URL must be on-line!
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 have no problem getting the opencv face detection using haar feature based cascades working on saved images:
from PIL import Image
face_cascade = cv2.CascadeClassifier('haarcascade_frontalface_default.xml')
eye_cascade = cv2.CascadeClassifier('haarcascade_eye.xml')
img = cv2.imread('pic.png')
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
faces = face_cascade.detectMultiScale(gray, 1.3, 5)
but I can't figure out how to open a url image and pass it into face_cascade. I've been playing around with cStringIO, but I don't know what to do with it...
import cv2.cv as cv
import urllib, cStringIO
img = 'http://scontent-b.cdninstagram.com/hphotos-prn/t51.2885-15/10424498_582114441904402_1105042543_n.png'
file = cStringIO.StringIO(urllib.urlopen(img).read())
source = Image.open(file).convert("RGB")
bitmap = cv.CreateImageHeader(source.size, cv.IPL_DEPTH_8U, 3)
cv.SetData(bitmap, source.tostring())
cv.CvtColor(bitmap, bitmap, cv.CV_RGB2BGR)
is it possible to work with a numpy array instead?
source2 = Image.open(file)
imarr=numpy.array(source2,dtype=numpy.uint8)
I'm a beginner, so I apologize for the poor explanation.
thanks a lot in advance!!
In your first example you are using OpenCV2.imread to read your image in the second you are presumably using PIL.Image then trying to convert.
Why not simply save the file to a temp directory and then use OpenCV2.imread again?
Or in another way you can use VideoCapture() class to open url image.
See the C++ code below,
VideoCapture cap;
if(!cap.open("http://docs.opencv.org/trunk/_downloads/opencv-logo.png")){
cout<<"Cannot open image"<<endl;
return -1;
}
Mat src;
cap>>src;
imshow("src",src);
waitKey();