Python binary data reading

问题:

A urllib2 request receives binary response as below:

00 00 00 01 00 04 41 4D 54 44 00 00 00 00 02 41
97 33 33 41 99 5C 29 41 90 3D 71 41 91 D7 0A 47
0F C6 14 00 00 01 16 6A E0 68 80 41 93 B4 05 41
97 1E B8 41 90 7A E1 41 96 8F 57 46 E6 2E 80 00
00 01 16 7A 53 7C 80 FF FF

Its structure is:

DATA, TYPE, DESCRIPTION

00 00 00 01, 4 bytes, Symbol Count =1

00 04, 2 bytes, Symbol Length = 4

41 4D 54 44, 6 bytes, Symbol = AMTD

00, 1 byte, Error code = 0 (OK)

00 00 00 02, 4 bytes, Bar Count =  2

FIRST BAR

41 97 33 33, 4 bytes, Close = 18.90

41 99 5C 29, 4 bytes, High = 19.17

41 90 3D 71, 4 bytes, Low = 18.03

41 91 D7 0A, 4 bytes, Open = 18.23

47 0F C6 14, 4 bytes, Volume = 3,680,608

00 00 01 16 6A E0 68 80, 8 bytes, Timestamp = November 23,2007

SECOND BAR

41 93 B4 05, 4 bytes, Close = 18.4629

41 97 1E B8, 4 bytes, High = 18.89

41 90 7A E1, 4 bytes, Low = 18.06

41 96 8F 57, 4 bytes, Open = 18.82

46 E6 2E 80, 4 bytes, Volume = 2,946,325

00 00 01 16 7A 53 7C 80, 8 bytes, Timestamp = November 26,2007

TERMINATOR

FF FF, 2 bytes,

How to read binary data like this?

Thanks in advance.

Update:

I tried struct module on first 6 bytes with following code:

struct.unpack('ih', response.read(6))

(16777216, 1024)

But it should output (1, 4). I take a look at the manual but have no clue what was wrong.

回答1:

>>> data
'\x00\x00\x00\x01\x00\x04AMTD\x00\x00\x00\x00\x02A\x9733A\x99\\)A\x90=qA\x91\xd7\nG\x0f\xc6\x14\x00\x00\x01\x16j\xe0h\x80A\x93\xb4\x05A\x97\x1e\xb8A\x90z\xe1A\x96\x8fWF\xe6.\x80\x00\x00\x01\x16zS|\x80\xff\xff'
>>> from struct import unpack, calcsize
>>> scount, slength = unpack("!IH", data[:6])
>>> assert scount == 1
>>> symbol, error_code = unpack("!%dsb" % slength, data[6:6+slength+1])
>>> assert error_code == 0
>>> symbol
'AMTD'
>>> bar_count = unpack("!I", data[6+slength+1:6+slength+1+4])
>>> bar_count
(2,)
>>> bar_format = "!5fQ"                                                         
>>> from collections import namedtuple
>>> Bar = namedtuple("Bar", "Close High Low Open Volume Timestamp")             
>>> b = Bar(*unpack(bar_format, data[6+slength+1+4:6+slength+1+4+calcsize(bar_format)]))
>>> b
Bar(Close=18.899999618530273, High=19.170000076293945, Low=18.030000686645508, Open=18.229999542236328, Volume=36806.078125, Timestamp=1195794000000L)
>>> import time
>>> time.ctime(b.Timestamp//1000)
'Fri Nov 23 08:00:00 2007'
>>> int(b.Volume*100 + 0.5)
3680608

回答2:

So here's my best shot at interpreting the data you're giving...:

import datetime
import struct

class Printable(object):
  specials = ()
  def __str__(self):
    resultlines = []
    for pair in self.__dict__.items():
      if pair[0] in self.specials: continue
      resultlines.append('%10s %s' % pair)
    return '\n'.join(resultlines)

head_fmt = '>IH6sBH'
head_struct = struct.Struct(head_fmt)
class Header(Printable):
  specials = ('bars',)
  def __init__(self, symbol_count, symbol_length,
               symbol, error_code, bar_count):
    self.__dict__.update(locals())
    self.bars = []
    del self.self

bar_fmt = '>5fQ'
bar_struct = struct.Struct(bar_fmt)
class Bar(Printable):
  specials = ('header',)
  def __init__(self, header, close, high, low,
               open, volume, timestamp):
    self.__dict__.update(locals())
    self.header.bars.append(self)
    del self.self
    self.timestamp /= 1000.0
    self.timestamp = datetime.date.fromtimestamp(self.timestamp)

def showdata(data):
  terminator = '\xff' * 2
  assert data[-2:] == terminator
  head_data = head_struct.unpack(data[:head_struct.size])
  try:
    assert head_data[4] * bar_struct.size + head_struct.size == \
           len(data) - len(terminator)
  except AssertionError:
    print 'data length is %d' % len(data)
    print 'head struct size is %d' % head_struct.size
    print 'bar struct size is %d' % bar_struct.size
    print 'number of bars is %d' % head_data[4]
    print 'head data:', head_data
    print 'terminator:', terminator
    print 'so, something is wrong, since',
    print head_data[4] * bar_struct.size + head_struct.size, '!=',
    print len(data) - len(terminator)
    raise

  head = Header(*head_data)
  for i in range(head.bar_count):
    bar_substr = data[head_struct.size + i * bar_struct.size:
                      head_struct.size + (i+1) * bar_struct.size]
    bar_data = bar_struct.unpack(bar_substr)
    Bar(head, *bar_data)
  assert len(head.bars) == head.bar_count
  print head
  for i, x in enumerate(head.bars):
    print 'Bar #%s' % i
    print x

datas = '''
00 00 00 01 00 04 41 4D 54 44 00 00 00 00 02 41
97 33 33 41 99 5C 29 41 90 3D 71 41 91 D7 0A 47
0F C6 14 00 00 01 16 6A E0 68 80 41 93 B4 05 41
97 1E B8 41 90 7A E1 41 96 8F 57 46 E6 2E 80 00
00 01 16 7A 53 7C 80 FF FF
'''

data = ''.join(chr(int(x, 16)) for x in datas.split())
showdata(data)

this emits:

symbol_count 1
 bar_count 2
    symbol AMTD
error_code 0
symbol_length 4
Bar #0
    volume 36806.078125
 timestamp 2007-11-22
      high 19.1700000763
       low 18.0300006866
     close 18.8999996185
      open 18.2299995422
Bar #1
    volume 29463.25
 timestamp 2007-11-25
      high 18.8899993896
       low 18.0599994659
     close 18.4629001617
      open 18.8199901581

...which seems to be pretty close to what you want, net of some output formatting details. Hope this helps!-)

回答3:

>>> struct.unpack('ih', response.read(6))
(16777216, 1024)

You are unpacking big-endian data on a little-endian machine. Try this instead:

>>> struct.unpack('!IH', response.read(6))
(1L, 4)

This tells unpack to consider the data in network-order (big-endian). Also, the values of counts and lengths can not be negative, so you should should use the unsigned variants in your format string.

回答4:

Take a look at the struct.unpack in the struct module.

回答5:

Use pack/unpack functions from "struct" package. More info here http://docs.python.org/library/struct.html

Bye!

回答6:

As it was already mentioned, struct is the module you need to use.

Please read its documentation to learn about byte ordering, etc.

In your example you need to do the following (as your data is big-endian and unsigned):

>>> import struct
>>> x = '\x00\x00\x00\x01\x00\x04'
>>> struct.unpack('>IH', x)
(1, 4)