英文:
How would I read a raw RGBA4444 image using Pillow?
问题
我尝试读取一个'.waltex'图像，这是一种'walaber图像'。它基本上是以原始图像格式存在的。问题是它使用'RGBA8888'、'RGBA4444'、'RGB565'和'RGB5551'（这些都可以从文件头中确定），但我找不到在PIL中使用这些颜色规格的方法。
我尝试过以下方法：
from PIL import Imagewith open('Carl.waltex', 'rb') as file:rawdata = file.read()image = Image.frombytes('RGBA', (1024,1024), rawdata, 'raw', 'RGBA;4B')image.show()
我尝试了所有16位原始模式，但都没有找到适用的。要明确一下，这个文件具体是'RGBA4444'，采用小端序，每像素8个字节。
如果你需要这个文件，我可以提供链接。
英文:
I'm trying to read a '.waltex' image, which is a 'walaber image'. It's basically just in raw image format. The problem is, it uses 'RGBA8888', 'RGBA4444', 'RGB565' and 'RGB5551' (all of which can be determined from the header), and I could not find a way to use these color specs in PIL.
I've tried doing this
from PIL import Imagewith open('Carl.waltex', 'rb') as file:rawdata = file.read()image = Image.frombytes('RGBA', (1024,1024), rawdata, 'raw', 'RGBA;4B')image.show()
I've tried all the 16-bit raw modes in the last input, and I couldn't find a single one that worked. To be clear, this file is specifically 'RGBA4444' with little endian, 8 bytes per pixel.
If you need the file, then I can link it.
答案1
得分: 3
我已经对原始代码进行了一些更改，以便：
- 您可以传递要读取的文件名作为参数
- 它解析标头并检查魔术字符串，自动推导格式（RGBA8888或RGBA4444）、高度和宽度
- 它现在处理RGBA8888，就像您最新分享的示例图像一样
所以，代码如下所示：
#!/usr/bin/env python3import structimport sysimport numpy as npfrom PIL import Imagedef loadWaltex(filename):# 以二进制模式打开输入文件with open(filename, 'rb') as fd:# 读取16字节的标头并提取元数据# https://zenhax.com/viewtopic.php?t=14164header = fd.read(16)magic, vers, fmt, w, h, _ = struct.unpack('4sBBHH6s', header)if magic != b'WALT':sys.exit(f'ERROR: {filename} does not start with "WALT" magic string')# 检查格式是否为0（RGBA8888）或3（RGBA4444）if fmt == 0:fmtdesc = "RGBA8888"# 读取文件余下的部分（标头后面的部分）data = np.fromfile(fd, dtype=np.uint8)R = data[0::4].reshape((h, w))G = data[1::4].reshape((h, w))B = data[2::4].reshape((h, w))A = data[3::4].reshape((h, w))# 堆叠通道以生成RGBA图像RGBA = np.dstack((R, G, B, A))else:fmtdesc = "RGBA4444"# 读取文件余下的部分（标头后面的部分）data = np.fromfile(fd, dtype=np.uint16).reshape((h, w))# 从uint16中提取RGBA4444R = (data >> 12) & 0xfG = (data >> 8) & 0xfB = (data >> 4) & 0xfA = data & 0xf# 堆叠通道以生成RGBA图像RGBA = np.dstack((R, G, B, A)).astype(np.uint8) << 4# 用户的调试信息print(f'Filename: {filename}, version: {vers}, format: {fmtdesc} ({fmt}), w: {w}, h: {h}')# 转化为PIL图像im = Image.fromarray(RGBA)return imif __name__ == "__main__":# 加载由第一个参数指定的图像im = loadWaltex(sys.argv[1])im.save('result.png')
当您运行它时：
./decodeRGBA.py objects.waltex
您将得到：
对于您的两个示例图像，调试输出如下：
Filename: Carl.waltex, version: 1, format: RGBA4444 (3), w: 1024, h: 1024Filename: objects.waltex, version: 1, format: RGBA8888 (0), w: 256, h: 1024
原始答案
我发现使用Numpy是处理这种情况的最简单方法，而且性能非常好：
#!/usr/bin/env python3import numpy as npfrom PIL import Image# 定义图像的已知参数并读入Numpy数组h, w, offset = 1024, 1024, 16data = np.fromfile('Carl.waltex', dtype=np.uint16, offset=offset).reshape((h,w))# 从uint16中提取RGBA4444R = (data >> 12) & 0xfG = (data >> 8) & 0xfB = (data >> 4) & 0xfA = data & 0xf# 堆叠4个单独的通道以生成RGBA图像RGBA = np.dstack((R, G, B, A)).astype(np.uint8) << 4# 转化为PIL图像im = Image.fromarray(RGBA)im.save('result.png')
注意: 您的图像在开头有16字节的填充。有时候这个数量是可变的。在这种情况下的一个有用技巧是读取整个文件，计算出有多少个有效的像素数据样本（在您的情况下是1024 * 1024），然后切片数据以取最后N个样本，从而忽略开头的可变填充。这将看起来像这样：
# 定义图像的已知参数并读入Numpy数组h, w = 1024, 1024data = np.fromfile('Carl.waltex', dtype=np.uint16)[-h*w:].reshape((h,w))
如果您不喜欢Numpy，更喜欢使用列表和结构来处理，可以使用以下代码获得完全相同的结果：
#!/usr/bin/env python3import structfrom PIL import Image# 定义图像的已知参数h, w, offset = 1024, 1024, 16data = open('Carl.waltex', 'rb').read()[offset:]# 解包成一堆h*w个无符号短整数uint16s = struct.unpack("H" * h * w, data)# 创建一个RGBA元组的列表pixels = []for RGBA4444 in uint16s:R = (RGBA4444 >> 8) & 0xf0G = (RGBA4444 >> 4) & 0xf0B = RGBA4444 & 0xf0A = (RGBA4444 & 0xf) << 4pixels.append((R, G, B, A))# 将RGBA元组的列表推入空图像中RGBA = Image.new('RGBA', (w,h))RGBA.putdata(pixels)RGBA.save('result.png')
请注意，与基于Numpy的方法相比，基于列表的方法速度要
英文:
Updated Answer
I have made some changes to my original code so that:
- you can pass a filename to read as parameter
- it parses the header and checks the magic string and derives the format (RGBA8888 or RGBA4444) and height and width automatically
- it now handles RGBA8888 like your newly-shared sample image
So, it looks like this:
#!/usr/bin/env python3import structimport sysimport numpy as npfrom PIL import Imagedef loadWaltex(filename):# Open input file in binary modewith open(filename, 'rb') as fd:# Read 16 byte header and extract metadata# https://zenhax.com/viewtopic.php?t=14164header = fd.read(16)magic, vers, fmt, w, h, _ = struct.unpack('4sBBHH6s', header)if magic != b'WALT':sys.exit(f'ERROR: {filename} does not start with "WALT" magic string')# Check if fmt=0 (RGBA8888) or fmt=3 (RGBA4444)if fmt == 0:fmtdesc = "RGBA8888"# Read remainder of file (part following header)data = np.fromfile(fd, dtype=np.uint8)R = data[0::4].reshape((h,w))G = data[1::4].reshape((h,w))B = data[2::4].reshape((h,w))A = data[3::4].reshape((h,w))# Stack the channels to make RGBA imageRGBA = np.dstack((R,G,B,A))else:fmtdesc = "RGBA4444"# Read remainder of file (part following header)data = np.fromfile(fd, dtype=np.uint16).reshape((h,w))# Split the RGBA444 out from the uint16R = (data>>12) & 0xfG = (data>>8) & 0xfB = (data>>4) & 0xfA = data & 0xf# Stack the channels to make RGBA imageRGBA = np.dstack((R,G,B,A)).astype(np.uint8) << 4# Debug info for userprint(f'Filename: {filename}, version: {vers}, format: {fmtdesc} ({fmt}), w: {w}, h: {h}')# Make into PIL Imageim = Image.fromarray(RGBA)return imif __name__ == "__main__":# Load image specified by first parameterim = loadWaltex(sys.argv[1])im.save('result.png')
And when you run it with:
./decodeRGBA.py objects.waltex
You get:
The debug output for your two sample images is:
Filename: Carl.waltex, version: 1, format: RGBA4444 (3), w: 1024, h: 1024Filename: objects.waltex, version: 1, format: RGBA8888 (0), w: 256, h: 1024
Original Answer
I find using Numpy is the easiest approach for this type of thing, and it is also highly performant:
#!/usr/bin/env python3import numpy as npfrom PIL import Image# Define the known parameters of the image and read into Numpy arrayh, w, offset = 1024, 1024, 16data = np.fromfile('Carl.waltex', dtype=np.uint16, offset=offset).reshape((h,w))# Split the RGBA4444 out from the uint16R = (data >> 12) & 0xfG = (data >> 8) & 0xfB = (data >> 4) & 0xfA = data & 0xf# Stack the 4 individual channels to make an RGBA imageRGBA = np.dstack((R,G,B,A)).astype(np.uint8) << 4# Make into PIL Imageim = Image.fromarray(RGBA)im.save('result.png')
Note: Your image has 16 bytes of padding at the start. Sometimes that amount is variable. A useful technique in that case is to read the entire file, work out how many useful samples of pixel data there are (in your case 1024*1024), and then slice the data to take the last N samples - thereby ignoring any variable padding at the start. That would look like this:
# Define the known parameters of the image and read into Numpy arrayh, w = 1024, 1024data = np.fromfile('Carl.waltex', dtype=np.uint16)[-h*w:].reshape((h,w))
If you don't like Numpy and prefer messing about with lists and structs, you can get exactly the same result like this:
#!/usr/bin/env python3import structfrom PIL import Image# Define the known parameters of the imageh, w, offset = 1024, 1024, 16data = open('Carl.waltex', 'rb').read()[offset:]# Unpack into bunch of h*w unsigned shortsuint16s = struct.unpack("H" * h *w, data)# Build a list of RGBA tuplespixels = []for RGBA4444 in uint16s:R = (RGBA4444 >> 8) & 0xf0G = (RGBA4444 >> 4) & 0xf0B = RGBA4444 & 0xf0A = ( RGBA4444 & 0xf) << 4pixels.append((R,G,B,A))# Push the list of RGBA tuples into an empty imageRGBA = Image.new('RGBA', (w,h))RGBA.putdata(pixels)RGBA.save('result.png')
Note that the Numpy approach is 60x faster than the list-based approach:
Numpy: 3.6 ms ± 73.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)listy: 213 ms ± 712 µs per loop (mean ± std. dev. of 7 runs, 1 loop each)
Note: These images and the waltex file format seem to be from the games "Where's My Water?" and "Where's My Perry?". I got some hints as to the header format from ZENHAX.
答案2
得分: 1
这是您要翻译的代码部分：
Now that I have a better idea of how your Waltex files work, I attempted to write a custom PIL Plugin for them - a new experience for me. I've put it as a different answer because the approach is very different.You use it very simply like this:from PIL import Imageimport WaltexImagePluginim = Image.open('objects.waltex')im.show()
You need to save the following as `WaltexImagePlugin.py` in the directory beside your main Python program:from PIL import Image, ImageFileimport structdef _accept(prefix):return prefix[:4] == b"WALT"class WaltexImageFile(ImageFile.ImageFile):format = "Waltex"format_description = "Waltex texture image"def _open(self):header = self.fp.read(HEADER_LENGTH)magic, vers, fmt, w, h, _ = struct.unpack('4sBBHH6s', header)# size in pixels (width, height)self._size = w, h# mode settingself.mode = 'RGBA'# Decoderif fmt == 0:# RGBA8888# Just use built-in raw decoderself.tile = [("raw", (0, 0) + self.size, HEADER_LENGTH, (self.mode,0, 1))]elif fmt == 3:# RGBA4444# Use raw decoder with custom RGBA;4B unpackerself.tile = [("raw", (0, 0) + self.size, HEADER_LENGTH, ('RGBA;4B',0, 1))]Image.register_open(WaltexImageFile.format, WaltexImageFile, _accept)Image.register_extensions(WaltexImageFile.format,[".waltex"],)HEADER_LENGTH = 16
It works perfectly for your RGBA888 images, but cannot quite handle the byte ordering of your RGBA444 file, so you need to reverse it for those images. I used this:......im = Image.open(...)# Split channels and recombine in correct ordera, b, c, d = im.split()im = Image.merge((c,d,a,b))
请注意，上述内容包括一些代码和注释，我已经按您的要求提供了代码部分的翻译。如果您有任何其他翻译需求或问题，请随时提出。
英文:
Now that I have a better idea of how your Waltex files work, I attempted to write a custom PIL Plugin for them - a new experience for me. I've put it as a different answer because the approach is very different.
You use it very simply like this:
from PIL import Imageimport WaltexImagePluginim = Image.open('objects.waltex')im.show()
You need to save the following as WaltexImagePlugin.py in the directory beside your main Python program:
from PIL import Image, ImageFileimport structdef _accept(prefix):return prefix[:4] == b"WALT"class WaltexImageFile(ImageFile.ImageFile):format = "Waltex"format_description = "Waltex texture image"def _open(self):header = self.fp.read(HEADER_LENGTH)magic, vers, fmt, w, h, _ = struct.unpack('4sBBHH6s', header)# size in pixels (width, height)self._size = w, h# mode settingself.mode = 'RGBA'# Decoderif fmt == 0:# RGBA8888# Just use built-in raw decoderself.tile = [("raw", (0, 0) + self.size, HEADER_LENGTH, (self.mode,0, 1))]elif fmt == 3:# RGBA4444# Use raw decoder with custom RGBA;4B unpackerself.tile = [("raw", (0, 0) + self.size, HEADER_LENGTH, ('RGBA;4B',0, 1))]Image.register_open(WaltexImageFile.format, WaltexImageFile, _accept)Image.register_extensions(WaltexImageFile.format,[".waltex"],)HEADER_LENGTH = 16
It works perfectly for your RGBA888 images, but cannot quite handle the byte ordering of your RGBA444 file, so you need to reverse it for those images. I used this:
......im = Image.open(...)# Split channels and recombine in correct ordera, b, c, d = im.split()im = Image.merge((c,d,a,b))
If anyone knows how to use something in the Unpack.c file to do this correctly, please ping me. Thank you.
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