问题
I would like to read in a JPEG-Header and analyze it.
According to Wikipedia, the header consists of a sequences of markers. Each Marker starts with FF xx, where xx is a specific Marker-ID.
So my idea, was to simply read in the image in binary format, and seek for the corresponding character-combinations in the binary stream. This should enable me to split the header in the corresponding marker-fields.
For instance, this is, what I receive, when I read in the first 20 bytes of an image:
binary_data = open('picture.jpg','rb').read(20)
print(binary_data)
b'\xff\xd8\xff\xe1-\xfcExif\x00\x00MM\x00*\x00\x00\x00\x08'
My questions are now:
1) Why does python not return me nice chunks of 2 bytes (in hex-format).
Somthing like this I would expect:
b'\xff \xd8 \xff \xe1 \x-' ... and so on. Some blocks delimited by '\x' are much longer than 2 bytes.
2) Why are there symbols like -, M, * in the returned string? Those are no characters of a hex representation I expect from a byte string (only: 0-9, a-f, I think).
Both observations hinder me in writing a simple parser. So ultimately my question summarizes to: How do I properly read-in and parse a JPEG Header in Python?
回答1:
You seem overly worried about how your binary data is represented on your console. Don't worry about that.
The default built-in string-based representation that print(..) applies to a bytes object is just "printable ASCII characters as such (except a few exceptions), all others as an escaped hex sequence". The exceptions are semi-special characters such as \, ", and ', which could mess up the string representation. But this alternative representation does not change the values in any way!
>>> a = bytes([1,2,4,92,34,39])
>>> a
b'\x01\x02\x04\\"\''
>>> a[0]
1
See how the entire object is printed 'as if' it's a string, but its individual elements are still perfectly normal bytes?
If you have a byte array and you don't like the appearance of this default, then you can write your own. But – for clarity – this still doesn't have anything to do with parsing a file.
>>> binary_data = open('iaijiedc.jpg','rb').read(20)
>>> binary_data
b'\xff\xd8\xff\xe0\x00\x10JFIF\x00\x01\x02\x01\x00H\x00H\x00\x00'
>>> ''.join(['%02x%02x ' % (binary_data[2*i],binary_data[2*i+1]) for i in range(len(binary_data)>>1)])
'ffd8 ffe0 0010 4a46 4946 0001 0201 0048 0048 0000 '
Why does python not return me nice chunks of 2 bytes (in hex-format)?
Because you don't ask it to. You are asking for a sequence of bytes, and that's what you get. If you want chunks of two-bytes, transform it after reading.
The code above only prints the data; to create a new list that contains 2-byte words, loop over it and convert each 2 bytes or use unpack (there are actually several ways):
>>> wd = [unpack('>H', binary_data[x:x+2])[0] for x in range(0,len(binary_data),2)]
>>> wd
[65496, 65504, 16, 19014, 18758, 1, 513, 72, 72, 0]
>>> [hex(x) for x in wd]
['0xffd8', '0xffe0', '0x10', '0x4a46', '0x4946', '0x1', '0x201', '0x48', '0x48', '0x0']
I'm using the little-endian specifier < and unsigned short H in unpack, because (I assume) these are the conventional ways to represent JPEG 2-byte codes. Check the documentation if you want to derive from this.
来源：https://stackoverflow.com/questions/49734233/reading-in-a-binary-jpeg-header-in-python