Tkinter canvas create image multiple times
I use this function in my code to create an image on a canvas:
def _create_image(self, coord):
(x,y) = coord
self.one = ImageTk.PhotoImage(Image.open("test.jpg"))
root.one = self.one
self.canvas.create_image(x-25, y-25, image=self.one, anchor='nw', tags="image")
My problem is that everytime i call this function the old image is deleted and a new one is created.
How do i prevent the old image from being deleted? I want to create the image multiple times on my canvas.
python canvas tkinter tkinter-canvas
edited Nov 16 '18 at 20:48
petezurich
3,50581734
asked Nov 16 '18 at 20:45
K-Doe
8911
add a comment |
I use this function in my code to create an image on a canvas:
def _create_image(self, coord):
(x,y) = coord
self.one = ImageTk.PhotoImage(Image.open("test.jpg"))
root.one = self.one
self.canvas.create_image(x-25, y-25, image=self.one, anchor='nw', tags="image")
My problem is that everytime i call this function the old image is deleted and a new one is created.
How do i prevent the old image from being deleted? I want to create the image multiple times on my canvas.
python canvas tkinter tkinter-canvas
edited Nov 16 '18 at 20:48
petezurich
3,50581734
asked Nov 16 '18 at 20:45
K-Doe
8911
Possible duplicate of Using an tkinter image canvas in a loop
– stovfl
Nov 16 '18 at 21:27
add a comment |
I use this function in my code to create an image on a canvas:
def _create_image(self, coord):
(x,y) = coord
self.one = ImageTk.PhotoImage(Image.open("test.jpg"))
root.one = self.one
self.canvas.create_image(x-25, y-25, image=self.one, anchor='nw', tags="image")
My problem is that everytime i call this function the old image is deleted and a new one is created.
How do i prevent the old image from being deleted? I want to create the image multiple times on my canvas.
python canvas tkinter tkinter-canvas
edited Nov 16 '18 at 20:48
petezurich
3,50581734
asked Nov 16 '18 at 20:45
K-Doe
8911
I use this function in my code to create an image on a canvas:
def _create_image(self, coord):
(x,y) = coord
self.one = ImageTk.PhotoImage(Image.open("test.jpg"))
root.one = self.one
self.canvas.create_image(x-25, y-25, image=self.one, anchor='nw', tags="image")
My problem is that everytime i call this function the old image is deleted and a new one is created.
How do i prevent the old image from being deleted? I want to create the image multiple times on my canvas.
python canvas tkinter tkinter-canvas
python canvas tkinter tkinter-canvas
edited Nov 16 '18 at 20:48
petezurich
3,50581734
asked Nov 16 '18 at 20:45
K-Doe
8911
edited Nov 16 '18 at 20:48
petezurich
3,50581734
asked Nov 16 '18 at 20:45
K-Doe
8911
edited Nov 16 '18 at 20:48
petezurich
3,50581734
edited Nov 16 '18 at 20:48
petezurich
3,50581734
edited Nov 16 '18 at 20:48
petezurich
3,50581734
3,50581734
asked Nov 16 '18 at 20:45
K-Doe
8911
asked Nov 16 '18 at 20:45
K-Doe
8911
asked Nov 16 '18 at 20:45
K-Doe
8911
8911
Possible duplicate of Using an tkinter image canvas in a loop
– stovfl
Nov 16 '18 at 21:27
add a comment |
Possible duplicate of Using an tkinter image canvas in a loop
– stovfl
Nov 16 '18 at 21:27
Possible duplicate of Using an tkinter image canvas in a loop
– stovfl
Nov 16 '18 at 21:27
Possible duplicate of Using an tkinter image canvas in a loop
– stovfl
Nov 16 '18 at 21:27
add a comment |
3 Answers
3
active
oldest
votes
First create a list in for example __init__().
self.img_ref =
Then append each new image to this list as you create them:
def _create_image(self, coord):
(x,y) = coord
self.one = ImageTk.PhotoImage(Image.open("test.jpg"))
root.one = self.one
self.canvas.create_image(x-25, y-25, image=self.one,
anchor='nw', tags="image")
self.img_ref.append(self.one) # Keep reference to image
Even if every image is the same image you have to keep a reference to each one.
answered Nov 16 '18 at 21:22
figbeam
2,765137
Perfect! Thank you very much.
– K-Doe
Nov 16 '18 at 21:30
add a comment |
You don't need to modify the __init__() method nor store a list of references since it's always the same image file. Here something that avoids using unnecessary resource, so would therefore use less memory (and likely be faster, too).
It accomplishes this by testing to see whether the one attribute already exists or not, and if not, only then reads the image data and creates the PImageTk.PhotoImagehotoImage that first time.
This approach allows you to create multiple Canvas widget image objects from the same ImageTk.PhotoImage, instead of loading multiple copies of it into memory.
def _create_image(self, coord):
(x,y) = coord
if not getattr(self, 'one', None): # First call?
pil_img = Image.open("test.jpg")
self.one = ImageTk.PhotoImage(pil_img)
self.canvas.create_image(x-25, y-25, image=self.one,
anchor='nw', tags="image")
You could also do the creation of the ImageTk.PhotoImage all in one line:
# pil_img = Image.open("test.jpg") # Leave out.
self.one = ImageTk.PhotoImage(file="test.jpg")
answered Nov 16 '18 at 22:14
martineau
65.9k989177
add a comment |
Image.open() will rewrite the image each time.
edited Nov 17 '18 at 7:48
Benyamin Jafari
2,77732036
answered Nov 16 '18 at 21:14
Amar Nabhani
1
CallingImage.open()multiple times wouldn't rewrite anything, it will return a new PIL image object each time.
– martineau
Nov 16 '18 at 22:19
@martineau ah yes i apologize i must have confused this with something else
– Amar Nabhani
Nov 16 '18 at 22:21
1
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
– Sebastian 506563
Nov 16 '18 at 22:59
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
First create a list in for example __init__().
self.img_ref =
Then append each new image to this list as you create them:
def _create_image(self, coord):
(x,y) = coord
self.one = ImageTk.PhotoImage(Image.open("test.jpg"))
root.one = self.one
self.canvas.create_image(x-25, y-25, image=self.one,
anchor='nw', tags="image")
self.img_ref.append(self.one) # Keep reference to image
Even if every image is the same image you have to keep a reference to each one.
answered Nov 16 '18 at 21:22
figbeam
2,765137
Perfect! Thank you very much.
– K-Doe
Nov 16 '18 at 21:30
add a comment |
First create a list in for example __init__().
self.img_ref =
Then append each new image to this list as you create them:
def _create_image(self, coord):
(x,y) = coord
self.one = ImageTk.PhotoImage(Image.open("test.jpg"))
root.one = self.one
self.canvas.create_image(x-25, y-25, image=self.one,
anchor='nw', tags="image")
self.img_ref.append(self.one) # Keep reference to image
Even if every image is the same image you have to keep a reference to each one.
answered Nov 16 '18 at 21:22
figbeam
2,765137
Perfect! Thank you very much.
– K-Doe
Nov 16 '18 at 21:30
add a comment |
First create a list in for example __init__().
self.img_ref =
Then append each new image to this list as you create them:
def _create_image(self, coord):
(x,y) = coord
self.one = ImageTk.PhotoImage(Image.open("test.jpg"))
root.one = self.one
self.canvas.create_image(x-25, y-25, image=self.one,
anchor='nw', tags="image")
self.img_ref.append(self.one) # Keep reference to image
Even if every image is the same image you have to keep a reference to each one.
answered Nov 16 '18 at 21:22
figbeam
2,765137
First create a list in for example __init__().
self.img_ref =
Then append each new image to this list as you create them:
def _create_image(self, coord):
(x,y) = coord
self.one = ImageTk.PhotoImage(Image.open("test.jpg"))
root.one = self.one
self.canvas.create_image(x-25, y-25, image=self.one,
anchor='nw', tags="image")
self.img_ref.append(self.one) # Keep reference to image
Even if every image is the same image you have to keep a reference to each one.
answered Nov 16 '18 at 21:22
figbeam
2,765137
answered Nov 16 '18 at 21:22
figbeam
2,765137
answered Nov 16 '18 at 21:22
figbeam
2,765137
answered Nov 16 '18 at 21:22
figbeam
2,765137
2,765137
Perfect! Thank you very much.
– K-Doe
Nov 16 '18 at 21:30
add a comment |
Perfect! Thank you very much.
– K-Doe
Nov 16 '18 at 21:30
Perfect! Thank you very much.
– K-Doe
Nov 16 '18 at 21:30
Perfect! Thank you very much.
– K-Doe
Nov 16 '18 at 21:30
add a comment |
You don't need to modify the __init__() method nor store a list of references since it's always the same image file. Here something that avoids using unnecessary resource, so would therefore use less memory (and likely be faster, too).
It accomplishes this by testing to see whether the one attribute already exists or not, and if not, only then reads the image data and creates the PImageTk.PhotoImagehotoImage that first time.
This approach allows you to create multiple Canvas widget image objects from the same ImageTk.PhotoImage, instead of loading multiple copies of it into memory.
def _create_image(self, coord):
(x,y) = coord
if not getattr(self, 'one', None): # First call?
pil_img = Image.open("test.jpg")
self.one = ImageTk.PhotoImage(pil_img)
self.canvas.create_image(x-25, y-25, image=self.one,
anchor='nw', tags="image")
You could also do the creation of the ImageTk.PhotoImage all in one line:
# pil_img = Image.open("test.jpg") # Leave out.
self.one = ImageTk.PhotoImage(file="test.jpg")
answered Nov 16 '18 at 22:14
martineau
65.9k989177
add a comment |
You don't need to modify the __init__() method nor store a list of references since it's always the same image file. Here something that avoids using unnecessary resource, so would therefore use less memory (and likely be faster, too).
It accomplishes this by testing to see whether the one attribute already exists or not, and if not, only then reads the image data and creates the PImageTk.PhotoImagehotoImage that first time.
This approach allows you to create multiple Canvas widget image objects from the same ImageTk.PhotoImage, instead of loading multiple copies of it into memory.
def _create_image(self, coord):
(x,y) = coord
if not getattr(self, 'one', None): # First call?
pil_img = Image.open("test.jpg")
self.one = ImageTk.PhotoImage(pil_img)
self.canvas.create_image(x-25, y-25, image=self.one,
anchor='nw', tags="image")
You could also do the creation of the ImageTk.PhotoImage all in one line:
# pil_img = Image.open("test.jpg") # Leave out.
self.one = ImageTk.PhotoImage(file="test.jpg")
answered Nov 16 '18 at 22:14
martineau
65.9k989177
add a comment |
You don't need to modify the __init__() method nor store a list of references since it's always the same image file. Here something that avoids using unnecessary resource, so would therefore use less memory (and likely be faster, too).
It accomplishes this by testing to see whether the one attribute already exists or not, and if not, only then reads the image data and creates the PImageTk.PhotoImagehotoImage that first time.
This approach allows you to create multiple Canvas widget image objects from the same ImageTk.PhotoImage, instead of loading multiple copies of it into memory.
def _create_image(self, coord):
(x,y) = coord
if not getattr(self, 'one', None): # First call?
pil_img = Image.open("test.jpg")
self.one = ImageTk.PhotoImage(pil_img)
self.canvas.create_image(x-25, y-25, image=self.one,
anchor='nw', tags="image")
You could also do the creation of the ImageTk.PhotoImage all in one line:
# pil_img = Image.open("test.jpg") # Leave out.
self.one = ImageTk.PhotoImage(file="test.jpg")
answered Nov 16 '18 at 22:14
martineau
65.9k989177
You don't need to modify the __init__() method nor store a list of references since it's always the same image file. Here something that avoids using unnecessary resource, so would therefore use less memory (and likely be faster, too).
It accomplishes this by testing to see whether the one attribute already exists or not, and if not, only then reads the image data and creates the PImageTk.PhotoImagehotoImage that first time.
This approach allows you to create multiple Canvas widget image objects from the same ImageTk.PhotoImage, instead of loading multiple copies of it into memory.
def _create_image(self, coord):
(x,y) = coord
if not getattr(self, 'one', None): # First call?
pil_img = Image.open("test.jpg")
self.one = ImageTk.PhotoImage(pil_img)
self.canvas.create_image(x-25, y-25, image=self.one,
anchor='nw', tags="image")
You could also do the creation of the ImageTk.PhotoImage all in one line:
# pil_img = Image.open("test.jpg") # Leave out.
self.one = ImageTk.PhotoImage(file="test.jpg")
answered Nov 16 '18 at 22:14
martineau
65.9k989177
edited Nov 16 '18 at 23:03
answered Nov 16 '18 at 22:14
martineau
65.9k989177
answered Nov 16 '18 at 22:14
martineau
65.9k989177
answered Nov 16 '18 at 22:14
martineau
65.9k989177
65.9k989177
add a comment |
add a comment |
Image.open() will rewrite the image each time.
edited Nov 17 '18 at 7:48
Benyamin Jafari
2,77732036
answered Nov 16 '18 at 21:14
Amar Nabhani
1
CallingImage.open()multiple times wouldn't rewrite anything, it will return a new PIL image object each time.
– martineau
Nov 16 '18 at 22:19
@martineau ah yes i apologize i must have confused this with something else
– Amar Nabhani
Nov 16 '18 at 22:21
1
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
– Sebastian 506563
Nov 16 '18 at 22:59
add a comment |
Image.open() will rewrite the image each time.
edited Nov 17 '18 at 7:48
Benyamin Jafari
2,77732036
answered Nov 16 '18 at 21:14
Amar Nabhani
1
CallingImage.open()multiple times wouldn't rewrite anything, it will return a new PIL image object each time.
– martineau
Nov 16 '18 at 22:19
@martineau ah yes i apologize i must have confused this with something else
– Amar Nabhani
Nov 16 '18 at 22:21
1
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
– Sebastian 506563
Nov 16 '18 at 22:59
add a comment |
Image.open() will rewrite the image each time.
edited Nov 17 '18 at 7:48
Benyamin Jafari
2,77732036
answered Nov 16 '18 at 21:14
Amar Nabhani
1
Image.open() will rewrite the image each time.
edited Nov 17 '18 at 7:48
Benyamin Jafari
2,77732036
answered Nov 16 '18 at 21:14
Amar Nabhani
1
edited Nov 17 '18 at 7:48
Benyamin Jafari
2,77732036
edited Nov 17 '18 at 7:48
Benyamin Jafari
2,77732036
edited Nov 17 '18 at 7:48
Benyamin Jafari
2,77732036
2,77732036
answered Nov 16 '18 at 21:14
Amar Nabhani
1
answered Nov 16 '18 at 21:14
Amar Nabhani
1
answered Nov 16 '18 at 21:14
Amar Nabhani
1
1
CallingImage.open()multiple times wouldn't rewrite anything, it will return a new PIL image object each time.
– martineau
Nov 16 '18 at 22:19
@martineau ah yes i apologize i must have confused this with something else
– Amar Nabhani
Nov 16 '18 at 22:21
1
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
– Sebastian 506563
Nov 16 '18 at 22:59
add a comment |
CallingImage.open()multiple times wouldn't rewrite anything, it will return a new PIL image object each time.
– martineau
Nov 16 '18 at 22:19
@martineau ah yes i apologize i must have confused this with something else
– Amar Nabhani
Nov 16 '18 at 22:21
1
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
– Sebastian 506563
Nov 16 '18 at 22:59
Calling
Image.open() multiple times wouldn't rewrite anything, it will return a new PIL image object each time.– martineau
Nov 16 '18 at 22:19
Calling
Image.open() multiple times wouldn't rewrite anything, it will return a new PIL image object each time.– martineau
Nov 16 '18 at 22:19
@martineau ah yes i apologize i must have confused this with something else
– Amar Nabhani
Nov 16 '18 at 22:21
@martineau ah yes i apologize i must have confused this with something else
– Amar Nabhani
Nov 16 '18 at 22:21
1
1
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
– Sebastian 506563
Nov 16 '18 at 22:59
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
– Sebastian 506563
Nov 16 '18 at 22:59
add a comment |
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Possible duplicate of Using an tkinter image canvas in a loop
– stovfl
Nov 16 '18 at 21:27