uint8小字节序到uint16大字节序 [英] uint8 little endian array to uint16 big endian

问题描述

在Python2.7中，通过USB批量传输，我从相机获得了图像帧:

In Python2.7, from an USB bulk transfer I get an image frame from a camera:

frame = dev.read(0x81, 0x2B6B0, 1000)

我知道一帧是342x260 = 88920像素的小尾数，因为我从批量传输中读取了2x88920 = 177840(0x2B6B0).

I know that one frame is 342x260 = 88920 pixels little endian, because that I read 2x88920 = 177840 (0x2B6B0) from the bulk transfer.

如何将typecode = B的帧数组的内容转换为uint16大字节序数组?

How can I convert the content of the frame array that is typecode=B into an uint16 big endian array?

推荐答案

这种方法应该可以解决问题:

Something like this should do the trick:

frame_short_swapped = array.array('H', ((j << 8) | i
                                        for (i,j)
                                        in zip(frame[::2], frame[1::2])))

它将来自frame的两个连续字节配对，并将其解压缩为i和j.将j左移一个字节，并用i将其移至or，有效地交换字节(2字节类型的aka字节顺序转换)并将其输入到H类型的数组中.我对此有点不安，因为它应该对应于C short类型(根据文档)，但是类型大小实际上只能保证最小长度.我想如果需要严格的话，您需要引入ctypes.c_uint16吗?

It pairs two consecutive bytes from frame and unpacks that pair into i and j. Shift j one byte left and or it with i, effectively swap bytes (aka endianness conversion for 2 byte type) and feed that into an H type array. I am a bit uneasy about that bit, since it should correspond to C short type (according to docs), but type sizes really only guarantee minimal length. I guess you would need to introduce ctypes.c_uint16 if strict about that?

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