Open PIL image from byte file
Question:
I have this image with size 128 x 128 pixels and RGBA stored as byte values in my memory. But
from PIL import Image
image_data = ... # byte values of the image
image = Image.frombytes('RGBA', (128,128), image_data)
image.show()
throws the exception
ValueError: not enough image data
Why? What am I doing wrong?
Answers:
The documentation for Image.open says that it can accept a file-like object, so you should be able to pass in a io.BytesIO object created from the bytes object containing the encoded image:
from PIL import Image
import io
image_data = ... # byte values of the image
image = Image.open(io.BytesIO(image_data))
image.show()
You can try this:
image = Image.frombytes('RGBA', (128,128), image_data, 'raw')
Source Code:
def frombytes(mode, size, data, decoder_name="raw", *args):
param mode: The image mode.
param size: The image size.
param data: A byte buffer containing raw data for the given mode.
param decoder_name: What decoder to use.
I have this image with size 128 x 128 pixels and RGBA stored as byte values in my memory. But
from PIL import Image
image_data = ... # byte values of the image
image = Image.frombytes('RGBA', (128,128), image_data)
image.show()
throws the exception
ValueError: not enough image data
Why? What am I doing wrong?
The documentation for Image.open says that it can accept a file-like object, so you should be able to pass in a io.BytesIO object created from the bytes object containing the encoded image:
from PIL import Image
import io
image_data = ... # byte values of the image
image = Image.open(io.BytesIO(image_data))
image.show()
You can try this:
image = Image.frombytes('RGBA', (128,128), image_data, 'raw')
Source Code:
def frombytes(mode, size, data, decoder_name="raw", *args): param mode: The image mode. param size: The image size. param data: A byte buffer containing raw data for the given mode. param decoder_name: What decoder to use.