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Python 和 16 位 PGM

发布于 2024-12-04 05:03:21 字数 386 浏览 1 评论 0 原文

我有 16 位 PGM 图像，我正在尝试用 Python 读取它们。好像（？）PIL不支持这种格式？

import Image
im = Image.open('test.pgm')
im.show()

大致显示图像，但不正确。整个过程中都有暗带，据报道 img 具有 mode=L。我认为这与我早期关于 16 位 TIFF 文件 的问题有关。难道 16 位如此罕见以至于 PIL 不支持它吗？关于如何使用 PIL 或其他标准库或自制代码在 Python 中读取 16 位 PGM 文件，有什么建议吗？

原文

I have 16-bit PGM images that I am trying to read in Python. It seems (?) like PIL does not support this format?

import Image
im = Image.open('test.pgm')
im.show()

Shows roughly the image, but it isn't right. There are dark bands throughout and img is reported to have mode=L. I think this is related to an early question I had about 16-bit TIFF files. Is 16-bit that rare that PIL just does not support it? Any advice how I can read 16-bit PGM files in Python, using PIL or another standard library, or home-grown code?

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评论（3）

愛放△進行李 2024-12-11 05:03:22

您需要"L;16"模式；然而，在加载 PGM 时，PIL 似乎有一种 "L" 模式硬编码到 File.c 中。如果你想能够，你必须编写自己的解码器读取 16 位 PGM。

然而，16 位图像支持似乎仍然不稳定：

>>> im = Image.fromstring('I;16', (16, 16), '\xCA\xFE' * 256, 'raw', 'I;16') 
>>> im.getcolors()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/lib/python2.6/dist-packages/PIL/Image.py", line 866, in getcolors
    return self.im.getcolors(maxcolors)
ValueError: image has wrong mode

我认为 PIL 能够读取 16 位图像，但实际上存储和操作它们仍然处于实验阶段。

>>> im = Image.fromstring('L', (16, 16), '\xCA\xFE' * 256, 'raw', 'L;16') 
>>> im
<Image.Image image mode=L size=16x16 at 0x27B4440>
>>> im.getcolors()
[(256, 254)]

看，它只是将 0xCAFE 值解释为 0xFE，这并不完全正确。

You need a mode of "L;16"; however it looks like PIL has a mode of "L" hardcoded into File.c when loading a PGM. You’d have to write your own decoder if you want to be able to read a 16-bit PGM.

However, 16-bit image support still seems flaky:

>>> im = Image.fromstring('I;16', (16, 16), '\xCA\xFE' * 256, 'raw', 'I;16') 
>>> im.getcolors()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/lib/python2.6/dist-packages/PIL/Image.py", line 866, in getcolors
    return self.im.getcolors(maxcolors)
ValueError: image has wrong mode

I think PIL is capable of reading images with 16 bits, but actually storing and manipulating them is still experimental.

>>> im = Image.fromstring('L', (16, 16), '\xCA\xFE' * 256, 'raw', 'L;16') 
>>> im
<Image.Image image mode=L size=16x16 at 0x27B4440>
>>> im.getcolors()
[(256, 254)]

See, it just interpreted the 0xCAFE value as 0xFE, which isn’t exactly correct.

回复 原文
我最亲爱的 2024-12-11 05:03:22

以下仅依赖numpy来加载图像，可以是8位或16位位原始 PGM/PPM。我还展示了几种不同的查看图像的方式。使用PIL（导入图像）的要求首先将数据转换为8位。

#!/usr/bin/python2 -u

from __future__ import print_function
import sys, numpy

def read_pnm_from_stream( fd ):
   pnm = type('pnm',(object,),{}) ## create an empty container
   pnm.header = fd.readline()
   pnm.magic = pnm.header.split()[0]
   pnm.maxsample = 1 if ( pnm.magic == 'P4' ) else 0
   while ( len(pnm.header.split()) < 3+(1,0)[pnm.maxsample] ): s = fd.readline() ; pnm.header += s if ( len(s) and s[0] != '#' ) else ''
   pnm.width, pnm.height = [int(item) for item in pnm.header.split()[1:3]]
   pnm.samples = 3 if ( pnm.magic == 'P6' ) else 1
   if ( pnm.maxsample == 0 ): pnm.maxsample = int(pnm.header.split()[3])
   pnm.pixels = numpy.fromfile( fd, count=pnm.width*pnm.height*pnm.samples, dtype='u1' if pnm.maxsample < 256 else '>u2' )
   pnm.pixels = pnm.pixels.reshape(pnm.height,pnm.width) if pnm.samples==1 else pnm.pixels.reshape(pnm.height,pnm.width,pnm.samples)
   return pnm

if __name__ == '__main__':

## read image
 # src = read_pnm_from_stream( open(filename) )
   src = read_pnm_from_stream( sys.stdin )
 # print("src.header="+src.header.strip(), file=sys.stderr )
 # print("src.pixels="+repr(src.pixels), file=sys.stderr )

## write image
   dst=src
   dst.pixels = numpy.array([ dst.maxsample-i for i in src.pixels ],dtype=dst.pixels.dtype) ## example image processing
 # print("dst shape: "+str(dst.pixels.shape), file=sys.stderr )
   sys.stdout.write(("P5" if dst.samples==1 else "P6")+"\n"+str(dst.width)+" "+str(dst.height)+"\n"+str(dst.maxsample)+"\n");
   dst.pixels.tofile( sys.stdout ) ## seems to work, I'm not sure how it decides about endianness

## view using Image
   import Image
   viewable = dst.pixels if dst.pixels.dtype == numpy.dtype('u1') else numpy.array([ x>>8 for x in dst.pixels],dtype='u1')
   Image.fromarray(viewable).show()

## view using scipy
   import scipy.misc
   scipy.misc.toimage(dst.pixels).show()

使用说明

  • 我最终弄清楚了“它如何决定字节序”——它实际上是将图像作为大字节序（而不是本机）存储在内存中。这种方案可能会减慢任何重要的图像处理速度——尽管 Python 的其他性能问题可能会将此问题归为琐事（见下文）。

  • 我在此处提出了一个与字节顺序相关的问题。我还遇到了一些与字节序相关的有趣的困惑，因为我正在通过使用 pnmdepth 65535 预处理图像来进行测试，这对于测试字节序来说（本身）并不好，因为低字节和高字节可能最终会出现是相同的（我没有立即注意到，因为 print(array) 输出十进制）。我还应该应用 pnmgamma 来避免一些混乱。

  • 由于 Python 速度太慢，numpy 试图偷偷摸摸聪明地了解如何应用某些操作（请参阅 广播）。使用 numpy 提高效率的第一条经验法则是让 numpy 为您处理迭代（或者换句话说，不要编写自己的 for 循环）。上面代码中有趣的是，在执行“示例图像处理”时，它仅部分遵循此规则，因此该行的性能极大地依赖于为 reshape 提供的参数.

  • 下一个大numpy字节序之谜：为什么newbyteorder()似乎返回一个数组，当记录返回dtype.如果您想使用 dst.pixels=dst.pixels.byteswap(True).newbyteorder() 转换为本机字节序，这是相关的。

  • 移植到 Python 3 的提示：带有 ASCII 文本标头的二进制输入，从标准输入读取

The following depends only on numpy to load the image, which can be 8-bit or 16-bit raw PGM/PPM. I also show a couple different ways to view the image. The one that uses PIL (import Image) requires that the data first be converted to 8-bit.

#!/usr/bin/python2 -u

from __future__ import print_function
import sys, numpy

def read_pnm_from_stream( fd ):
   pnm = type('pnm',(object,),{}) ## create an empty container
   pnm.header = fd.readline()
   pnm.magic = pnm.header.split()[0]
   pnm.maxsample = 1 if ( pnm.magic == 'P4' ) else 0
   while ( len(pnm.header.split()) < 3+(1,0)[pnm.maxsample] ): s = fd.readline() ; pnm.header += s if ( len(s) and s[0] != '#' ) else ''
   pnm.width, pnm.height = [int(item) for item in pnm.header.split()[1:3]]
   pnm.samples = 3 if ( pnm.magic == 'P6' ) else 1
   if ( pnm.maxsample == 0 ): pnm.maxsample = int(pnm.header.split()[3])
   pnm.pixels = numpy.fromfile( fd, count=pnm.width*pnm.height*pnm.samples, dtype='u1' if pnm.maxsample < 256 else '>u2' )
   pnm.pixels = pnm.pixels.reshape(pnm.height,pnm.width) if pnm.samples==1 else pnm.pixels.reshape(pnm.height,pnm.width,pnm.samples)
   return pnm

if __name__ == '__main__':

## read image
 # src = read_pnm_from_stream( open(filename) )
   src = read_pnm_from_stream( sys.stdin )
 # print("src.header="+src.header.strip(), file=sys.stderr )
 # print("src.pixels="+repr(src.pixels), file=sys.stderr )

## write image
   dst=src
   dst.pixels = numpy.array([ dst.maxsample-i for i in src.pixels ],dtype=dst.pixels.dtype) ## example image processing
 # print("dst shape: "+str(dst.pixels.shape), file=sys.stderr )
   sys.stdout.write(("P5" if dst.samples==1 else "P6")+"\n"+str(dst.width)+" "+str(dst.height)+"\n"+str(dst.maxsample)+"\n");
   dst.pixels.tofile( sys.stdout ) ## seems to work, I'm not sure how it decides about endianness

## view using Image
   import Image
   viewable = dst.pixels if dst.pixels.dtype == numpy.dtype('u1') else numpy.array([ x>>8 for x in dst.pixels],dtype='u1')
   Image.fromarray(viewable).show()

## view using scipy
   import scipy.misc
   scipy.misc.toimage(dst.pixels).show()

Usage notes

  • I eventually figured out "how it decides about endianness" -- it is actually storing the image in memory as big-endian (rather than native). This scheme might slow down any non-trivial image processing -- although other performance issues with Python may relegate this concern to a triviality (see below).

  • I asked a question related to the endianness concern here. I also ran into some interesting confusion related to endianness with this because I was testing by preprocessing the image with pnmdepth 65535 which is not good (by itself) for testing endianness since the low and high bytes might end up being the same (I didn't notice right away because print(array) outputs decimal). I should have also applied pnmgamma to save myself some confusion.

  • Because Python is so slow, numpy tries to be sneakyclever about how it applies certain operations (see broadcasting). The first rule of thumb for efficiency with numpy is let numpy handle iteration for you (or put another way don't write your own for loops). The funny thing in the code above is that it only partially follows this rule when doing the "example image processing", and therefore the performance of that line has an extreme dependency on the parameters that were given to reshape.

  • The next big numpy endianness mystery: Why does newbyteorder() seem to return an array, when it's documented to return a dtype. This is relevant if you want to convert to native endian with dst.pixels=dst.pixels.byteswap(True).newbyteorder().

  • Hints on porting to Python 3: binary input with an ASCII text header, read from stdin

回复 原文
今天小雨转甜 2024-12-11 05:03:22

这是一个通用的 PNM/PAM 基于NumPy 和 PyPNG< 中未记录的函数/a>.

def read_pnm( filename, endian='>' ):
   fd = open(filename,'rb')
   format, width, height, samples, maxval = png.read_pnm_header( fd )
   pixels = numpy.fromfile( fd, dtype='u1' if maxval < 256 else endian+'u2' )
   return pixels.reshape(height,width,samples)

当然编写这种图像格式通常不需要库的帮助......

Here's a generic PNM/PAM reader based on NumPy and an undocumented function in PyPNG.

def read_pnm( filename, endian='>' ):
   fd = open(filename,'rb')
   format, width, height, samples, maxval = png.read_pnm_header( fd )
   pixels = numpy.fromfile( fd, dtype='u1' if maxval < 256 else endian+'u2' )
   return pixels.reshape(height,width,samples)

Of course writing this image format generally doesn't require the assistance of a library...

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