How can I convert canvas content to an image?

from Tkinter import *
root = Tk()
cv = Canvas(root)
cv.create_rectangle(10,10,50,50)
cv.pack()
root.mainloop()

I want to convert canvas content to a bitmap or other image, and then do other operations, such as rotating or scaling the image, or changing its coordinates.

Bitmaps can improve efficiency to show if I am no longer drawing.

What should I do?

Answers:

Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.

Method 1

I have found a great way of doing this which is really helpful. For it, you need the PIL module. Here is the code:

from PIL import ImageGrab

def getter(widget):
    x=root.winfo_rootx()+widget.winfo_x()
    y=root.winfo_rooty()+widget.winfo_y()
    x1=x+widget.winfo_width()
    y1=y+widget.winfo_height()
    ImageGrab.grab().crop((x,y,x1,y1)).save("file path here")

What this does is you pass a widget name into the function. The command root.winfo_rootx() and the root.winfo_rooty() get the pixel position of the top left of the overall root window.

Then, the widget.winfo_x() and widget.winfo_y() are added to, basically just get the pixel coordinate of the top left hand pixel of the widget which you want to capture (at pixels (x,y) of your screen).

I then find the (x1,y1) which is the bottom left pixel of the widget. The ImageGrab.grab() makes a printscreen, and I then crop it to only get the bit containing the widget. Although not perfect, and won’t make the best possible image, this is a great tool for just getting a image of any widget and saving it.

If you have any questions, post a comment! Hope this helped!

Method 2

You can either generate a postscript document (to feed into some other tool: ImageMagick, Ghostscript, etc):

from Tkinter import *
root = Tk()
cv = Canvas(root)
cv.create_rectangle(10,10,50,50)
cv.pack()
root.mainloop()

cv.update()
cv.postscript(file="file_name.ps", colormode='color')

root.mainloop()

or draw the same image in parallel on PIL and on Tkinter’s canvas (see: Saving a Tkinter Canvas Drawing (Python)). For example (inspired by the same article):

from Tkinter import *
import Image, ImageDraw

width = 400
height = 300
center = height//2
white = (255, 255, 255)
green = (0,128,0)

root = Tk()

# Tkinter create a canvas to draw on
cv = Canvas(root, width=width, height=height, bg='white')
cv.pack()

# PIL create an empty image and draw object to draw on
# memory only, not visible
image1 = Image.new("RGB", (width, height), white)
draw = ImageDraw.Draw(image1)

# do the Tkinter canvas drawings (visible)
cv.create_line([0, center, width, center], fill='green')

# do the PIL image/draw (in memory) drawings
draw.line([0, center, width, center], green)

# PIL image can be saved as .png .jpg .gif or .bmp file (among others)
filename = "my_drawing.jpg"
image1.save(filename)

root.mainloop()

Method 3

Use Pillow to convert from Postscript to PNG

from PIL import Image

def save_as_png(canvas,fileName):
    # save postscipt image 
    canvas.postscript(file = fileName + '.eps') 
    # use PIL to convert to PNG 
    img = Image.open(fileName + '.eps') 
    img.save(fileName + '.png', 'png')

Method 4

Maybe you can try to use widget_winfo_id to get the HWND of the canvas.

import win32gui

from PIL import ImageGrab

HWND = canvas.winfo_id()  # get the handle of the canvas

rect = win32gui.GetWindowRect(HWND)  # get the coordinate of the canvas

im = ImageGrab.grab(rect)  # get image of the current location

Method 5

A better way for @B.Jenkins’s answer that doesn’t need a reference to the root object:

from PIL import ImageGrab


def save_widget_as_image(widget, file_name):
    ImageGrab.grab(bbox=(
        widget.winfo_rootx(),
        widget.winfo_rooty(),
        widget.winfo_rootx() + widget.winfo_width(),
        widget.winfo_rooty() + widget.winfo_height()
    )).save(file_name)


All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0

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