I'm trying to make this welcome card thingy in discord.py and I'm running my bot on replit.com as of now.
font = ImageFont.truetype("arial.ttf", 28)
I got some examples and it worked great as long as I'm running it on my PC but wen I get to replit.com it gives the error saying
Command raised an exception: OSError: cannot open resource
How should I go about correcting this?
I don't know discord or replit but presume the issue is that you can't upload your binary font file.
If so, you have a couple of options:
find your desired font online somewhere and use requests.get(URL) to grab it in your code on replit, or
make a base64 variable in your code and decode it
The first option is covered here.
Let's look at the second. Say your font is called /Fonts/funky.ttf on your PC. Now you want that in base64 which you can do with a commandline tool on your local PC:
base64 < /Fonts/funky.txt
That will make a long string of characters. Copy it, and in your Python code add a string called font64 and set it equal to the pasted string, i.e.
font64 = 'PASTED STRING'
Now in your code you can convert the string back from base64 to binary, then wrap it in a BytesIO to make it look like a file and load it:
import base64
import io
from PIL import ImageFont
font64 = 'PASTED STRING'
# decode from base64 to binary
binary = base64.b64decode(font64)
# wrap in BytesIO to make file-like object
FileLike = io.BytesIO(binary)
# load font
font = ImageFont.truetype(FileLike, 28)
Related
Bonjour
Here is my problem:
I made a tkinter App to transform entries (letters or numbers) in barcode or Qrcode... very simple thing...
The program function like a charm when I run it in pycharm but when I get the final App for Windows, only the Qrcode part works
I use PytoExe
Here is the part that create the barcode
import qrcode
def creer_barrecode():
number = entryMonUrl.get().strip()
desktop_path = str(Path.home() / "Desktop")
fila = desktop_path + '/bar_code.png'
if number:
# my_code = EAN13(number, writer=ImageWriter())
# my_code.save(fila)
with open(fila, 'wb') as f:
Code128(number, writer=ImageWriter()).write(f)
When I check the folder, only an empty image is there (0 octet) with the correct name
Thank you for your help
I tried EAN and Code128 but same things
What do I do wrong?
I just used this code to convert a picture to binary:
import io
tme = input("Name: ")
with io.open(tme, "rb") as se:
print(se.read())
se.close()
Now it looks like this:
5MEMMMMMMMMMMMMM777777777777777777777777777777777\x95\x95\x95\x95\x95\x95\x95\x95\x95\x95\x95\x95\x95MEEMMMMEEMM\x96\x97\x97\x97\x97\x97\x97\x97\x97\x97\x97\x97\x97\x97\x97\x97\x97
And now I want to be able to interpret what this binary code is telling me exactly... I know it ruffly but not enough to be able to do anything on purpose. I searched the web but I didn't find anything that could help me on that point. Can you tell me how is it working or send me a link where I can read how it's done?
You can't just change random bytes in an image. There's a header at the start with the height and width, probably the date and a palette and information about the number of channels and bits per pixel. Then there is the image data which is often padded, and/or compressed.
So you need an imaging library like PIL/Pillow and code something like this:
from PIL import Image
im = Image.open('image.bmp').convert('RGB')
px = im.load()
# Look at pixel[4,4]
print (px[4,4])
# Make it red
px[4,4] = (255,0,0)
# Save to disk
im.save('result.bmp')
Documentation and examples available here.
The output should be printed in hexadecimal format. The first bytes in bitmap file are 'B' and 'M'.
You are attempting to print the content in ASCII. Moreover, it doesn't show the first bytes because the content has scrolled further down. Add print("start\n") to make sure you see the start of the output.
import io
import binascii
tme = 'path.bmp'
print("start") # make sure this line appears in console output
with io.open(tme, "rb") as se:
content = se.read()
print(binascii.hexlify(content))
Now you should see something like
start
b'424d26040100000000003...
42 is hex value for B
4d is hex value for M ...
The first 14 bytes in the file are BITMAPFILEHEADER
The next 40 bytes are BITMAPINFOHEADER
The bytes after that are color table (if any) and finally the actual pixels.
See BITMAPFILEHEADER and BITMAPINFOHEADER
I have a function in my code that takes a string representing the url of an image and creates a File object from that string, to be attached to a Tweet. This seems to work about 90% of the time, but occasionally fails.
require 'open-uri'
attachment_url = "https://s3.amazonaws.com/FirmPlay/photos/images/000/002/443/medium/applying_too_many_jobs_-_daniel.jpg?1448392757"
image = File.new(open(attachment_url))
If I run the above code it returns TypeError: no implicit conversion of StringIO into String. If I change open(attachment_url) to open(attachment_url).read I get ArgumentError: string contains null byte. I also tried stripping out the null bytes from the file like so, but that also made no difference.
image = File.new(open(attachment_url).read.gsub("\u0000", ''))
Now if I try the original code with a different image, such as the one below, it works fine. It returns a File object as expected:
attachment_url = "https://s3.amazonaws.com/FirmPlay/photos/images/000/002/157/medium/mike_4.jpg"
I thought maybe it had something to do with the params in the original url, so I stripped those out, but it made no difference. If I open the images in Chrome they appear to be fine.
I'm not sure what I'm missing here. How can I resolve this issue?
Thanks!
Update
Here is the working code I have in my app:
filename = self.attachment_url.split(/[\/]/)[-1].split('?')[0]
stream = open(self.attachment_url)
image = File.open(filename, 'w+b') do |file|
stream.respond_to?(:read) ? IO.copy_stream(stream, file) : file.write(stream)
open(file)
end
Jordan's answer works except that calling File.new returns an empty File object, whereas File.open returns a File object containing the image data from stream.
The reason you're getting TypeError: no implicit conversion of StringIO into String is that open sometimes returns a String object and sometimes returns a StringIO object, which is unfortunate and confusing. Which it does depends on the size of the file. See this answer for more information: open-uri returning ASCII-8BIT from webpage encoded in iso-8859 (Although I don't recommend using the ensure-encoding gem mentioned therein, since it hasn't been updated since 2010 and Ruby has had significant encoding-related changes since then.)
The reason you're getting ArgumentError: string contains null byte is that you're trying to pass the image data as the first argument to File.new:
image = File.new(open(attachment_url))
The first argument of File.new should be a filename, and null bytes aren't allowed in filenames on most systems. Try this instead:
image_data = open(attachment_url)
filename = 'some-filename.jpg'
File.new(filename, 'wb') do |file|
if image_data.respond_to?(:read)
IO.copy_stream(image_data, file)
else
file.write(image_data)
end
end
The above opens the file (creating it if it doesn't exist; the b in 'wb' tells Ruby that you're going to write binary data), then writes the data from image_data to it using IO.copy_stream if it's a StreamIO object or File#write otherwise, then closes the file again.
If you use Paperclip, they have a method to copy to disk.
def raw_image_data
attachment.copy_to_local_file.read
end
change attachment to what ever variable you used of course.
Using python 3.4 in linux and windows, I'm trying to create qr code images from a list of string objects. I don't want to just store the image as a file because the list of strings may change frequently. I want to then tile all the objects and display the resulting image on screen for the user to scan with a barcode scanner. For the user to know which code to scan I need to add some text to the qr code image.
I can create the list of image objects correctly and they are in a list and calling .show on these objects displays them properly but I don't know how to treat these objects as a file object to open them. The object that is given to the open function, (img_list[0] in my case), in my add_text_to_img needs to support read, seek and tell methods. When I try this as is I get an attribute error. I've tried BytesIO and StringIO but I get an error message that Image.open does not support buffer interface. Maybe I am not doing that part correctly.
I'm sure there are several ways to do this, but what is the best way to open in memory objects as a file object?
from io import BytesIO
import qrcode
from PIL import ImageFont, ImageDraw, Image
def make_qr_image_list(code_list):
"""
:param code_list: a list of string objects to encode into QR code image
:return: a list of image or some type of other data objects
"""
img_list = []
for item in code_list:
qr = qrcode.QRCode(
version=None,
error_correction=qrcode.ERROR_CORRECT_L,
box_size=4,
border=10
)
qr.add_data(item)
qr_image = qr.make_image(fit=True)
img_list.append(qr_image)
return img_list
def add_text_to_img(text_list, img_list):
"""
While I was working on this, I am only saving the first image. Once
it's working, I'll save the rest of the images to a list.
:param text_list: a list of strings to add to the corresponding image.
:param img_list: the list containing the images already created from
the text_list
:return:
"""
base = Image.open(img_list[0])
# img = Image.frombytes(mode='P', size=(164,164), data=img_list[0])
text_img = Image.new('RGBA', base.size, (255,255,255,0))
font = ImageFont.truetype('sans-serif.ttf', 10)
draw = ImageDraw.Draw(text_img)
draw.text((0,-20),text_list[0], (0,0,255,128), font=font)
# include some method to save the images after the text
# has been added here. Shouldn't actually save to a file.
# Should be saved to memory/img_list
output = Image.alpha_composite(base,text_img)
output.show()
if __name__ == '__main__':
test_list = ['AlGaN','n-AlGaN','p-AlGaN','MQW','LED AlN-AlGaN']
image_list = make_qr_image_list(test_list)
add_text_to_img(test_list, image_list)
im = image_list[0]
im.save('/my_save_path/test_image.png')
im.show()
Edit: I've been using python for about a year and I feel like this is a pretty common thing to do but I'm not even sure that I'm looking up/searching for the right terms. What topics would you search for to answer this? If anyone can post a link or two to what I need to read up on regarding this, that would be very appreciated.
You already have PIL image objects; qr.make_image() returns the (a wrapper around) the right type of object and you do not need to open them again.
As such, all you need to do is:
base = img_list[0]
and go from there.
You do need to match image modes when compositing; QR codes are black-and-white images (mode 1), so either convert that or use the same mode in your text_img image object. The Image.alpha_composite() operation does require that both images have an alpha channel. Converting the base is easy:
base = img_list[0].convert('RGBA')
I was hoping to be able to generate montages using PythonMagick. The documentation seems very sparse, but I've been trying to hunt it down using the code completion part of Eclipse at least, as well as a few other questions' suggestions here on Stack Overflow. It seems that the MagickWand API has the function I am looking for, according to this:
http://www.imagemagick.org/api/MagickWand/montage_8c.html
However, I cannot seem to find it in PythonMagick. Is this simply unavailable? If so I might just ditch the rest of my PythonMagick code and rely on subprocess.call on a portable ImageMagick distribution or something like that (this program will have to be portable, and run on Windows with an easy port to Mac OS... so far I have a few other PythonMagick commands working so I'd like to keep this route going if possible).
Thanks!
Using the python imagemagick/graphicsmagick bindings helps a lot, but unfortunately not all of the functionality is there yet. I actually had the same problem with #FizxMike. I needed to use montage and then do some further operations, but saving the file on hard disk and then reloading it in a proper pgmagick object in order to do the rest of the operations and saving it again was slow.
Eventually I used the subprocess solution, but instead of saving in a file, I redirect the output in stdout. Then, I use the stdout to load the image from a pgmagick.Blob in a pgmagick.Image object and do the rest of the processing in python code.
The procedure looks like this in code:
import os
import pgmagick
import subprocess
my_files = []
# Dir with the images that you want to operate on
dir_with_images = "."
for file in os.listdir(dir_with_images):
if file.endswith(".png"):
my_files.append(os.path.join(dir_with_images, file))
montage_cmd = ['gm', 'montage']
montage_cmd.extend(my_files)
# The trick is in the next line of code. Instead of saving in a file, e.g. myimage.png
# the montaged file will just be "printed" in the stdout with 'png:-'
montage_cmd.extend(['-tile', '2x2', '-background', 'none', '-geometry', '+0+0', 'png:-'])
# Use the command line 'gm montage' since there are not python bindings for it :(
p = subprocess.Popen(montage_cmd, stdout=subprocess.PIPE, stderr=subprocess.PIPE)
# Get the stdout in a variable
stdout, stderr = p.communicate()
# Load the stdout in a python pgmagick Image object using the pgmagick.Blob
# and do the rest of the editing on python code
img = pgmagick.Image(pgmagick.Blob(stdout))
# Display the image
img.display()
geometry = pgmagick.Geometry(300, 200)
geometry.aspect(True)
# Resize the montaged image to 300x200, but keep the aspect ratio
img.scale(geometry)
# Display it again
img.display()
# And finally save it <- Only once disk access at this point.
img.write('myimage.png')
I have the same problem, even pgmagick lacks the montageImage() function needed (Magick++ montage example)
This is what I do (in a Django View):
#ImageMagick CLI is better documented anyway (-background none preserves transparency)
subprocess.call("montage -border 0 -geometry "+str(cols)+"x -tile 1x"+str(len(pages))+" "+target_path[0:len(target_path)-4]+"[0-9]*.png -background none "+target_path,shell=True)`
Not fun because I have to juggle around a bunch of files first... writing to hard disk is not the fastest thing to do, then delete the temp files.
I would much rather do it all in ram.
I am still in search of a better answer myself.