V4L2 10bit像素存储机制疑问及RAW数据解析求助

背景与预期存储格式 #

根据V4L2的10bit传感器文档，我原本认为单个10bit像素的存储格式应该是：

xxxxxxxx 000000xx

其中前8位是低字节，后8位是高字节，且高字节的前6位为0。

数据采集方式 #

我在Jetson Xavier NX上使用以下命令获取传感器数据：

v4l2-ctl -d /dev/video0 --set-fmt-video=width=1280,height=720,pixelformat=RG10 --set-ctrl=bypass_mode=0 --stream-mmap --stream-count=1 --stream-to=file.raw

数据读取代码 #

以下是读取RAW文件并打印数据及二进制形式的Python代码：

import numpy as np
raw_file_path = 'file.raw'
width = 3280
height = 2464
dtype = np.uint16
with open(raw_file_path, 'rb') as f:
    data = np.fromfile(f, dtype=dtype)

if data.size != width * height:
    raise ValueError(f"File size doesn't match expected shape ({height}, {width}). Got {data.size} elements.")

image = data.reshape((height, width)) # shape (3280, 2482)

print(image)
print(image.shape)
print(image.dtype)


image_bin = np.vectorize(lambda x: format(x, '016b'))(image)

print(image_bin)
print(image_bin.shape)
print(image_bin.dtype)

输出结果 #

运行代码后得到的输出如下：

[[4548 4996 4484 ... 4676 4932 4612]
[5125 4420 4996 ... 0 0 0]
[4676 5125 4484 ... 4612 5060 4548]
...
[4676 4356 4676 ... 0 0 0]
[4164 4420 4292 ... 4292 4548 4292]
[4420 4292 4420 ... 0 0 0]]
(2464, 3280)
uint16
[['0001000111000100' '0001001110000100' '0001000110000100' ...
  '0001001001000100' '0001001101000100' '0001001000000100']
 ['0001010000000101' '0001000101000100' '0001001110000100' ...
  '0000000000000000' '0000000000000000' '0000000000000000']
 ['0001001001000100' '0001010000000101' '0001000110000100' ...
  '0001001000000100' '0001001111000100' '0001000111000100']
...
 ['0001001001000100' '0001000100000100' '0001001001000100' ...
  '0000000000000000' '0000000000000000' '0000000000000000']
 ['0001000001000100' '0001000101000100' '0001000011000100' ...
  '0001000011000100' '0001000111000100' '0001000011000100']
 ['0001000101000100' '0001000011000100' '0001000101000100' ...
  '0000000000000000' '0000000000000000' '0000000000000000']]
(2464, 3280)
<U16

疑问 #

从输出的二进制数据中看不到预期的连续6个零，请问10bit图像实际是如何存储的？

内容的提问来源于stack exchange，提问作者Talha Tahir