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Is there a way of attaching an image on a python code in such a way that it becomes part of the soure code?

I'm a beginner in python and I'm trying to send someone my small python program together with a picture that'll display when the code is run.
I tried to first convert the image to a binary file thinking that I'd be able to paste it in the source code but I'm not sure if that's even possible as I failed to successfully do it.

You can base64-encode your JPEG/PNG image which will make it into a regular (non-binary string) like this:
base64 -w0 IMAGE.JPG
Then you want to get the result into a Python variable, so repeat the command but copy the output to your clipboard:
base64 -w0 IMAGE.JPG | xclip -selection clipboard # Linux
base64 -w0 IMAGE.JPG | pbcopy # macOS
Now start Python and make a variable called img and paste the clipboard into it:
img = 'PASTE'
It will look like this:
img = '/9j/4AAQSk...' # if your image was JPEG
img = 'iVBORw0KGg...' # if your image was PNG
Now do some imports:
from PIL import Image
import base64
import io
# Make PIL Image from base64 string
pilImage = Image.open(io.BytesIO(base64.b64decode(img)))
Now you can do what you like with your image:
# Print its description and size
print(pilImage)
<PIL.JpegImagePlugin.JpegImageFile image mode=RGB size=200x100>
# Save it to local disk
pilImage.save('result.jpg')

You can save a picture in byte format inside a variable in your program. You can then convert the bytes back into a file-like object using the BytesIO function of the io module and plot that object using the Image module from the Pillow library.
import io
import PIL.Image
with open("filename.png", "rb") as file:
img_binary = file.read()
img = PIL.Image.open(io.BytesIO(img_binary))
img.show()
To save the binary data inside your program without having to read from the source file you need to encode it with something like base64, use print() and then simply copy the output into a new variable and remove the file reading operation from your code.
That would look like this:
img_encoded = base64.encodebytes(img_binary)
print(img_binary)
img_encoded = " " # paste the output from the console into the variable
the output will be very long, especially if you are using a big image. I only used a very small png for testing.
This is how the program should look like at the end:
import io
import base64
import PIL.Image
# with open("filename.png", "rb") as file:
# img_binary = file.read()
# img_encoded = base64.encodebytes(img_binary)
img_encoded = b'iVBORw0KGgoAAAANSUhEUgAAADAAAAAwCAYAAABX[...]'
img = PIL.Image.open(io.BytesIO(base64.decodebytes(img_encoded)))
img.show()

You could perhaps have your Python program download the image from a site where you upload files such as Google Drive, Mega, or Imgur. That way, you can always access and view the image easily without the need of running the program or for example converting the binary back into the image in the method you mentioned.
Otherwise, you could always store the image as bytes in a variable and have your program read this variable. I'm assuming that you really wish to do it this way as it would be easier to distribute as there is only one file that needs to be downloaded and run.
Or you could take a look at pyinstaller which is made for python programs to be easily distributed across machines without the need to install Python by packaging it as an executable (.exe) file! That way you can include the image file together by embedding it into the program. There are plenty of tutorials for pyinstaller you could google up. Note: Include the '--onefile' in your parameters when running pyinstaller as this will package the executable into a single file that the person you're sending it to can easily open whoever it may be-- granted the executable file can run on the user's operating system. :)

Converting .pgm to .png in Python

Problem:
I want to import all ".pgm" files from several folders and store them as ".png".
For this I use the following lines of code:
from glob import glob
files=glob('yaleB*/*.pgm')
print 'number of files is',len(files)
count=0
for f in files:
new_f=f[:-3]+'png'
!convert $f $new_f
count += 1
if count % 100==0:
print count,f,new_f
Unfortunately, I always get this error: "command not found: convert"
Is there a simple fix for this?
My assumption is, that this is an "old" command (Jupyter Notebook where I took this code from is from 2014..)

you can try using os and pillow
import os
from PIL import Image
for file in os.listdir():
filename, extension = os.path.splitext(file)
if extension == ".pgm":
new_file = "{}.png".format(filename)
with Image.open(file) as im:
im.save(new_file)

The exclamation mark (!) means it is an external command, and convert is part of the ImageMagick package.
So you would need to install ImageMagick for it to work. Note that if you install v7 or newer, the command would become:
!magick INPUT.PGM OUTPUT.PNG
However, if you want to convert all PGM files in the current directory to PNG files, you don't even need a loop, you can do them all in one go with:
magick mogrify -format png *.pgm
Yet more caveats... there is no real need from a Python point of view to convert PGM files to PNG, since OpenCV, PIL/Pillow, wand and scikit-image can all read PGM files anyway. So can GIMP, feh, eog, Photoshop.

Issue with removing files in python

I'm currently having some issues removing a file in python. I am creating a temporary file for pdf to image conversion. It is housed in a folder that holds a .ppm file and converts it to a .jpg file. It then deletes the temporary .ppm file. Here is the code:
import pdf2image
from PIL import Image
import os
images = pdf2image.convert_from_path('Path to pdf.pdf', output_folder='./folder name')
file = ''
for files in os.listdir('./folder name'):
if files.endswith(".ppm"):
file = files
path = os.path.join('folder name',file)
im = Image.open(path)
im.save("Path to image.jpg")
im.close()
os.remove(path)
The issue is at the end in the os.remove(path). I get the following error:
PermissionError: [WinError 32] The process cannot access the file because it is being used by another process: 'path to ppm file'
I would appreciate any help and thanks in advance.

Not really the answer to your question, but you can just output in the correct format at the start, and avoid the issue in the first place:
pdf2image.convert_from_path('Path to pdf.pdf', output_folder='./folder name', fmt='jpg')
To actually answer your question, I'm not sure why you're having the issue, because really the close() should prevent this problem. Perhaps check out this answer and try using a with statement? Or maybe the permissions release is just delayed, I'm curious what throwing that remove in a loop for as long as it throws an exception would do.
Edit: To set the name, you'll want to do something like:
images = pdf2image.convert_from_path('Path to pdf.pdf', output_folder='./folder name', fmt='jpg')
for image in images:
# Save the image
The pdf2image documentation looks like it recommends using a temporary folder, like in this example, and then you can just .save(...) the PIL image:
import tempfile
with tempfile.TemporaryDirectory() as path:
images_from_path = convert_from_path('/home/kankroc/example.pdf', output_folder=path)
# Do something here
Edit: I realized that the reason it was in use is probably because you need to close() all the images in images. You should read up on the pdf2image documentation and about the PIL images that it spits out for more details.

python reading a video

I am using Python 2.7.11 and OpenCV 2.4.9. I cannot read a video by using cv2.imread() or cv2.VideoCapture().
import cv2
cap = cv2.VideoCapture('cam.avi')
print ("open = ",cap.isOpened())
OR
import cv2
cap = cv2.imread('cam.avi')
print ("open = ",cap.isOpened())
It will return false.
I don't know why. I am sure that the cam.avi is here.

imread() does not support reading from video files directly.
See also the documentation of OpenCV.
If you want to read a video with imread you will first have to convert it to single images, either via a serperate program (ffmpeg comes to mind) or using OpenCV and store the images in memory.

Try providing full path to video, like:
import cv2
cap = cv2.VideoCapture(r'C:\Users\e01069\Downloads\drop.avi')
print ("open = ",cap.isOpened())
If you run following in your same file, you would know that python is looking for your file on some different location.
import os
print os.path.abspath(__file__) #this is your current working directory
Note: .imread wouldn't work this way.

Using MemoryFS file-like object with PIL Image

I'm trying to open Image file from PyFileSystem MemoryFS using PIL, I tried to do that like this example and i got the error below:
from PIL import Image
from fs.memoryfs import MemoryFS
fs=MemoryFS()
fs.makedir("test")
out=fs.open("test/file.jpg",'wb')
out.write(someimagefile.read())
out.close()
in=fs.open("test/file.jpg",'rb')
im=Image.open(in) #error: cannot identify image file <IO wrapper for <MemoryFile in <MemoryFS> test/file.jpg>>
however if I don't use a directory (ex. out=fs.open("file.jpg",'wb')) It does work as expected.
What am I doing wrong?
Thanks for your help.

I already get an error using the following line:
from fs.memoryfs import MemoryFS
Probably I don't have that library installed. Do you need this library? You can just open the image without opening it with MemoryFS:
im = Image.open("test/file.jpg")
Side note: I wouldn't use in as variable name, since it is also a Python keyword.