Python二进制数据读取

Question

urllib2请求接收如下的二进制响应：

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

其结构为：

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,

如何读取这样的二进制数据？

提前谢谢。

更新：

我在前6个字节上尝试了struct module，代码如下：

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

（167772161024）

但它应该输出（1，4）。我看了一下手册，但不知道出了什么问题。

Answer 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

Answer 2

所以这是我解释你提供的数据的最好机会…：

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)

它发出：

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

…这似乎很接近你想要的，除了一些输出格式的细节。希望这有帮助！-)

Answer 3

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

你正在一台小型的endian机器上解压big-endian数据。请改为：

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

这告诉unpack按网络顺序考虑数据（big endian）。此外，计数和长度的值不能为负，因此应在格式字符串中使用无符号变量。