3

我带着另一个冗长的问题回来了。在尝试了许多基于 Python 的 Damerau-Levenshtein 编辑距离实现后，我终于找到了下面列出的editdistance_reference(). 它似乎提供了正确的结果，并且似乎有一个有效的实施。

所以我开始着手将代码转换为 Cython。在我的测试数据中，参考方法能够提供 11,000 次比较的结果（对于 12 个字母长的词对），而 Cythonized 方法每秒进行超过 200,000 次比较。可悲的是，结果不正确：当您查看thisrow 我打印出来进行调试的变量时，无论我向其输入什么数据，我的版本都充满了变量，而参考输出显示了另一张图片。例如，测试'helo'产生'world' 以下输出（ED标记我的函数，EDR是正确工作的参考）：

来自editdistance()：

#ED  A [0, 0, 0, 0, 0, 1]
#ED  B [1, 0, 0, 0, 0, 1]
#ED  B [1, 1, 0, 0, 0, 1]
#ED  B [1, 1, 1, 0, 0, 1]
#ED  B [1, 1, 1, 1, 0, 1]
#ED  B [1, 1, 1, 1, 1, 1]

#ED  A [0, 0, 0, 0, 0, 2]
#ED  B [1, 0, 0, 0, 0, 2]
#ED  B [1, 1, 0, 0, 0, 2]
#ED  B [1, 1, 1, 0, 0, 2]
#ED  B [1, 1, 1, 1, 0, 2]
#ED  B [1, 1, 1, 1, 1, 2]

#ED  A [0, 0, 0, 0, 0, 3]
#ED  B [1, 0, 0, 0, 0, 3]
#ED  B [1, 1, 0, 0, 0, 3]
#ED  B [1, 1, 1, 0, 0, 3]
#ED  B [1, 1, 1, 1, 0, 3]
#ED  B [1, 1, 1, 1, 1, 3]

#ED  A [0, 0, 0, 0, 0, 4]
#ED  B [1, 0, 0, 0, 0, 4]
#ED  B [1, 1, 0, 0, 0, 4]
#ED  B [1, 1, 1, 0, 0, 4]
#ED  B [1, 1, 1, 1, 0, 4]
#ED  B [1, 1, 1, 1, 1, 4]

来自editdistance_reference()：

#EDR A [0, 0, 0, 0, 0, 1]
#EDR B [1, 0, 0, 0, 0, 1]
#EDR B [1, 2, 0, 0, 0, 1]
#EDR B [1, 2, 3, 0, 0, 1]
#EDR B [1, 2, 3, 4, 0, 1]
#EDR B [1, 2, 3, 4, 5, 1]

#EDR A [0, 0, 0, 0, 0, 2]
#EDR B [2, 0, 0, 0, 0, 2]
#EDR B [2, 2, 0, 0, 0, 2]
#EDR B [2, 2, 3, 0, 0, 2]
#EDR B [2, 2, 3, 4, 0, 2]
#EDR B [2, 2, 3, 4, 5, 2]

#EDR A [0, 0, 0, 0, 0, 3]
#EDR B [3, 0, 0, 0, 0, 3]
#EDR B [3, 3, 0, 0, 0, 3]
#EDR B [3, 3, 3, 0, 0, 3]
#EDR B [3, 3, 3, 3, 0, 3]
#EDR B [3, 3, 3, 3, 4, 3]

#EDR A [0, 0, 0, 0, 0, 4]
#EDR B [4, 0, 0, 0, 0, 4]
#EDR B [4, 4, 0, 0, 0, 4]
#EDR B [4, 4, 4, 0, 0, 4]
#EDR B [4, 4, 4, 4, 0, 4]
#EDR B [4, 4, 4, 4, 4, 4]

我一定很愚蠢，因为错误可能是那些非常明显的事情之一。但我似乎找不到它。

还有第二个问题：三个数组、、 和malloc的 i 空间，然后它们以循环方式交换。当我尝试等等时，我得到一条 glibc 抱怨的行。我用谷歌搜索过；会不会是指针交换业务让 glibc 有点头晕，从而无法正确释放内存？twoagooneagothisrowfree( twoago )double free or corruption

下面我首先列出setup.py进行编译所需的内容（/path/to/python3.1 ./setup.py build_ext --inplace），然后是适当的编辑距离代码，因此有兴趣的人会发现它更容易复制。

还有一件事：这是用 Python3.1 运行的；一件有趣的事情是，在*.pyx文件内部我们确实有裸 unicode 字符串，但print它仍然是一个语句，而不是一个函数。

是的，我知道这里有很多代码要粘贴，但问题是，当你把它剪得太多时，代码根本无法运行。我相信除了editdistance()正常工作之外的所有方法，但请随时指出您发现的任何问题。

setup.py：

from distutils.core import setup
from distutils.extension import Extension
from Cython.Distutils import build_ext

setup(
  name            = 'cython_dameraulevenshtein',
  ext_modules     = [
    Extension( 'cython_dameraulevenshtein', [ 'cython_dameraulevenshtein.pyx', ] ), ],
  cmdclass        = {
    'build_ext': build_ext }, )

cython_dameraulevenshtein.pyx（一直滚动到最后，看看有趣的东西）：

############################################################################################################
cdef extern from "stdlib.h":
  ctypedef  unsigned int size_t
  void      *malloc(size_t size)
  void      *realloc( void *ptr, size_t size )
  void      free(void *ptr)

#-----------------------------------------------------------------------------------------------------------
cdef inline unsigned int _minimum_of_two_uints( unsigned int a, unsigned int b ):
  if a < b: return a
  return b

#-----------------------------------------------------------------------------------------------------------
cdef inline unsigned int _minimum_of_three_uints( unsigned int a, unsigned int b, unsigned int c ):
  if a < b:
    if c < a:
      return c
    return a
  if c < b:
    return c
  return b

#-----------------------------------------------------------------------------------------------------------
cdef inline int _warp( unsigned int limit, int value ):
  return value if value >= 0 else limit + value

############################################################################################################
# ARRAYS THAT SAY SIZE ;-)
#-----------------------------------------------------------------------------------------------------------
cdef class Array_of_unsigned_int:
  cdef unsigned int *data
  cdef unsigned int length

  #---------------------------------------------------------------------------------------------------------
  def __cinit__( self, unsigned int length, fill_value = None ):
    self.length = length
    self.data   = <unsigned int *>malloc( length * sizeof( unsigned int ) )  ###OBS### must check malloc doesn't return NULL pointer
    if fill_value is not None:
      self.fill( fill_value )

  #---------------------------------------------------------------------------------------------------------
  cdef fill( self, unsigned int value ):
    cdef unsigned int idx
    cdef unsigned int *d    = self.data
    for idx from 0 <= idx < self.length:
      d[ idx ] = value

  #---------------------------------------------------------------------------------------------------------
  cdef resize( self, unsigned int length ):
    self.data   = <unsigned int *>realloc( self.data, length * sizeof( unsigned int ) )  ###OBS### must check realloc doesn't return NULL pointer
    self.length = length

  #---------------------------------------------------------------------------------------------------------
  def free( self ):
    """Always remember the milk: Free up memory."""
    free( self.data )  ###OBS### should free memory here

  #---------------------------------------------------------------------------------------------------------
  def as_list( self ):
    """Return the array as a Python list."""
    R                       = []
    cdef unsigned int idx
    cdef unsigned int *d    = self.data
    for idx from 0 <= idx < self.length:
      R.append( d[ idx ] )
    return R


############################################################################################################
# CONVERTING UNICODE TO CHARACTER IDs (CIDs)
#---------------------------------------------------------------------------------------------------------
cdef unsigned int _UMX_surrogate_lower_bound    = 0x10000
cdef unsigned int _UMX_surrogate_upper_bound    = 0x10ffff
cdef unsigned int _UMX_surrogate_hi_lower_bound = 0xd800
cdef unsigned int _UMX_surrogate_hi_upper_bound = 0xdbff
cdef unsigned int _UMX_surrogate_lo_lower_bound = 0xdc00
cdef unsigned int _UMX_surrogate_lo_upper_bound = 0xdfff
cdef unsigned int _UMX_surrogate_foobar_factor  = 0x400

#---------------------------------------------------------------------------------------------------------
cdef Array_of_unsigned_int _cids_from_text( text ):
  """Givn a ``text`` either as a Unicode string or as a ``bytes`` or ``bytearray``, return an instance of
  ``Array_of_unsigned_int`` that enumerates either the Unicode codepoints of each character or the value of
  each byte. Surrogate pairs will be condensed into single values, so on narrow Python builds the length of
  the array returned may be less than ``len( text )``."""
  #.........................................................................................................
  # Make sure ``text`` is either a Unicode string (``str``) or a ``bytes``-like thing:
  is_bytes = isinstance( text, ( bytes, bytearray, ) )
  assert is_bytes or isinstance( text, str ), '#121'
  #.........................................................................................................
  # Whether it is a ``str`` or a ``bytes``, we know the result can only have at most as many elements as
  # there are characters in ``text``, so we can already reserve that much space (in the case of a Unicode
  # text, there may be fewer CIDs if there happen to be surrogate characters):
  cdef unsigned int           length  = <unsigned int>len( text )
  cdef Array_of_unsigned_int  R       = Array_of_unsigned_int( length )
  #.........................................................................................................
  # If ``text`` is empty, we can return an empty array right away:
  if length == 0: return R
  #.........................................................................................................
  # Otherwise, prepare to copy data:
  cdef unsigned int idx               = 0
  #.........................................................................................................
  # If ``text`` is a ``bytes``-like thing, use simplified processing; we just have to copy over all byte
  # values and are done:
  if is_bytes:
    for idx from 0 <= idx < length:
      R.data[ idx ] = <unsigned int>text[ idx ]
    return R
  #.........................................................................................................
  cdef unsigned int cid               = 0
  cdef bool         is_surrogate      = False
  cdef unsigned int hi                = 0
  cdef unsigned int lo                = 0
  cdef unsigned int chr_count         = 0
  #.........................................................................................................
  # Iterate over all indexes in text:
  for idx from 0 <= idx < length:
    #.......................................................................................................
    # If we met with a surrogate CID in the last cycle, then that was a high surrogate CID, and the
    # corresponding low CID is on the current position. Having both, we can compute the intended CID
    # and reset the flag:
    if is_surrogate:
      lo = <unsigned int>ord( text[ idx ] )
      # IIRC, this formula was documented in Unicode 3:
      cid = ( ( hi - _UMX_surrogate_hi_lower_bound ) * _UMX_surrogate_foobar_factor
            + ( lo - _UMX_surrogate_lo_lower_bound ) + _UMX_surrogate_lower_bound )
      is_surrogate = False
    #.......................................................................................................
    else:
      # Otherwise, we retrieve the CID from the current position:
      cid = <unsigned int>ord( text[ idx ] )
      #.....................................................................................................
      if _UMX_surrogate_hi_lower_bound <= cid <= _UMX_surrogate_hi_upper_bound:
        # If this CID is a high surrogate CID, set ``hi`` to this value and set a flag so we'll come back
        # in the next cycle:
        hi                = cid
        is_surrogate      = True
        continue
    #.......................................................................................................
    R.data[ chr_count ] = cid
    chr_count     += 1
  #.........................................................................................................
  # Surrogate CIDs take up two characters but end up as a single resultant CID, so the return value may
  # have fewer elements than the naive string length indicated; in this case, we want to free some memory
  # and correct array length data:
  if chr_count != length:
    R.resize( chr_count )
  #.........................................................................................................
  return R

#---------------------------------------------------------------------------------------------------------
def cids_from_text( text ):
  cdef Array_of_unsigned_int c_R  =_cids_from_text( text )
  R                               = c_R.as_list()
  c_R.free() ###OBS### should free memory here
  return R


############################################################################################################
# SECOND-ORDER SIMILARITY
#-----------------------------------------------------------------------------------------------------------
cpdef float similarity( char *a, char *b ):
  """Given two byte strings ``a`` and ``b``, return their Damerau-Levenshtein similarity as a float between
  0.0 and 1.1. Similarity is computed as ``1 - relative_editdistance( a, b )``, so a result of ``1.0``
  indicates identity, while ``0.0`` indicates complete dissimilarity."""
  return 1.0 - relative_editdistance( a, b )

#-----------------------------------------------------------------------------------------------------------
cpdef float relative_editdistance( char *a, char *b ):
  """Given two byte strings ``a`` and ``b``, return their relative Damerau-Levenshtein distance. The return
  value is a float between 0.0 and 1.0; it is calculated as the absolute edit distance, divided by the
  length of the longer string. Therefore, ``0.0`` indicates identity, while ``1.0`` indicates complete
  dissimilarity."""
  cdef int length = max( len( a ), len( b ) )
  if length == 0: return 0.0
  return editdistance( a, b ) / <float>length

############################################################################################################
# EDIT DISTANCE
#-----------------------------------------------------------------------------------------------------------
cpdef unsigned int editdistance( text_a, text_b ):
  """Given texts as Unicode strings or ``bytes`` / ``bytearray`` objects, return their absolute
  Damerau-Levenshtein distance. Each deletion, insertion, substitution, and transposition is counted as one
  difference, so the edit distance between ``abc`` and ``ab``, ``abcx``, ``abx``, ``acb``, respectively, is
  ``1``."""
  #.........................................................................................................
  # This should be fast in Python, as it can (and probably is) implemented by doing an identity check in
  # the case of ``bytes`` and ``str`` objects:
  if text_a == text_b: return 0
  #.........................................................................................................
  # Convert Unicode text to C array of unsigned integers:
  cdef Array_of_unsigned_int a  = _cids_from_text( text_a )
  cdef Array_of_unsigned_int b  = _cids_from_text( text_b )
  R                             = c_editdistance( a, b )
  #.........................................................................................................
  # Always remember the milk:
  a.free()
  b.free()
  #.........................................................................................................
  return R

#-----------------------------------------------------------------------------------------------------------
cdef unsigned int c_editdistance( Array_of_unsigned_int cids_a, Array_of_unsigned_int cids_b ):
  # Conceptually, this is based on a len(a) + 1 * len(b) + 1 matrix.
  # However, only the current and two previous rows are needed at once,
  # so we only store those.
  #.........................................................................................................
  # This shortcut is pretty useless if comparison is not very fast; therefore, it is done in the function
  # that deals with the Python objects, q.v.
  # if cids_a.equals( cids_b ): return 0
  #.........................................................................................................
  cdef unsigned int a_length            = cids_a.length
  cdef unsigned int b_length            = cids_b.length
  #.........................................................................................................
  # Another shortcut: if one of the texts is empty, then the edit distance is trivially the length of the
  # other text. This also works for two empty texts, but those have already been taken care of by the
  # previous shortcut:
  #.........................................................................................................
  if a_length == 0: return b_length
  if b_length == 0: return a_length
  #.........................................................................................................
  cdef unsigned int row_length          = b_length   + 1
  cdef unsigned int row_length_1        = row_length - 1
  cdef unsigned int row_bytecount       = sizeof( unsigned int ) * row_length
  cdef unsigned int *oneago             = <unsigned int *>malloc( row_bytecount ) ###OBS### must check malloc doesn't return NULL pointer
  cdef unsigned int *twoago             = <unsigned int *>malloc( row_bytecount ) ###OBS### must check malloc doesn't return NULL pointer
  cdef unsigned int *thisrow            = <unsigned int *>malloc( row_bytecount ) ###OBS### must check malloc doesn't return NULL pointer
  cdef unsigned int idx                 = 0
  cdef unsigned int idx_a               = 0
  cdef unsigned int idx_b               = 0
  cdef          int idx_a_1_text        = 0
  cdef          int idx_b_1_row         = 0
  cdef          int idx_b_2_row         = 0
  cdef          int idx_b_1_text        = 0
  cdef unsigned int deletion_cost       = 0
  cdef unsigned int addition_cost       = 0
  cdef unsigned int substitution_cost   = 0
  #.........................................................................................................
  # Equivalent of ``thisrow = list( range( 1, b_length + 1 ) ) + [ 0 ]``:
  #print( '#305', cids_a.as_list(), cids_b.as_list(), a_length, b_length, row_length, row_length_1 )
  for idx from 1 <= idx < row_length:
    thisrow[ idx - 1 ] = idx
  thisrow[ row_length - 1 ] = 0
  #.........................................................................................................
  for idx_a from 0 <= idx_a < a_length:
    idx_a_1_text      = _warp(   a_length, idx_a - 1 )
    twoago, oneago = oneago, thisrow
    #.......................................................................................................
    # Equivalent of ``thisrow = [ 0 ] * b_length + [ idx_a + 1 ]``:
    for idx from 0 <= idx < row_length_1:
      thisrow[ idx ] = 0
    thisrow[ row_length - 1 ] = idx_a + 1
    #.......................................................................................................
    # some diagnostic output:
    x = []
    for idx from 0 <= idx < row_length: x.append( thisrow[ idx ] )
    print
    print '#ED  A', x
    #.......................................................................................................
    for idx_b from 0 <= idx_b < b_length:
      #.....................................................................................................
      idx_b_1_row       = _warp( row_length, idx_b - 1 )
      idx_b_1_text      = _warp(   b_length, idx_b - 1 )
      #.....................................................................................................
      assert 0 <= idx_b_1_row  < row_length, ( '#323', idx_b_1_row, )
      assert 0 <= idx_a_1_text <   a_length, ( '#324', idx_a_1_text, )
      assert 0 <= idx_b_1_text <   b_length, ( '#325', idx_b_1_text, )
      #.....................................................................................................
      deletion_cost     = oneago[  idx_b       ] + 1
      addition_cost     = thisrow[ idx_b_1_row ] + 1
      substitution_cost = oneago[  idx_b_1_row ] + ( 1 if    cids_a.data[ idx_a ]
                                                          != cids_b.data[ idx_b ] else 0 )
      thisrow[ idx_b ]  = _minimum_of_three_uints( deletion_cost, addition_cost, substitution_cost )
      #.....................................................................................................
      # Transpositions:
      if (  idx_a > 0
        and idx_b > 0
        and cids_a.data[ idx_a        ] == cids_b.data[ idx_b_1_text ]
        and cids_a.data[ idx_a_1_text ] == cids_b.data[ idx_b        ]
        and cids_a.data[ idx_a        ] != cids_b.data[ idx_b        ] ):
        #...................................................................................................
        idx_b_2_row       = _warp( row_length, idx_b - 2 )
        assert 0 <= idx_b_2_row  < row_length, ( '#340', idx_b_2_row, )
        thisrow[ idx_b ]  = _minimum_of_two_uints( thisrow[ idx_b ], twoago[ idx_b_2_row ] + 1 )
      #.....................................................................................................
      # some diagnostic output:
      x = []
      for idx from 0 <= idx < row_length: x.append( thisrow[ idx ] )
      print '#ED  B', x
  #.........................................................................................................
  # Here, ``b_length - 1`` can't become negative, since we already tested for ``b_length == 0`` in the
  # shortcut above:
  cdef unsigned int R = thisrow[ b_length - 1 ]
  #.........................................................................................................
  # Always remember the milk:
  # BUG: Activating below lines leads to glibc failing with ``double free or corruption``
  #free( twoago )
  #free( oneago )
  #free( thisrow )e
  #.........................................................................................................
  return R

#-----------------------------------------------------------------------------------------------------------
def editdistance_reference( text_a, text_b ):
  """This method is believed to compute a correct Damerau-Levenshtein edit distance, with deletions,
  insertions, substitutions, and transpositions. Do not touch it; it is here to validate results returned
  from the above method. Code adapted from
  http://mwh.geek.nz/2009/04/26/python-damerau-levenshtein-distance"""
  # Conceptually, the implementation is based on a ``( len( seq1 ) + 1 ) * ( len( seq2 ) + 1 )`` matrix.
  # However, only the current and two previous rows are needed at once, so we only store those. Python
  # lists wrap around for negative indices, so we put the leftmost column at the *end* of the list. This
  # matches with the zero-indexed strings and saves extra calculation.
  b_length  = len( text_b )
  oneago    = None
  thisrow   = list( range( 1, b_length + 1 ) ) + [ 0 ]
  for idx_a in range( len( text_a ) ):
    twoago, oneago, thisrow = oneago, thisrow, [ 0 ] * b_length + [ idx_a + 1 ]
    #.......................................................................................................
    # some diagnostic output:
    print
    print '#EDR A', thisrow
    #.......................................................................................................
    for idx_b in range( b_length ):
      deletion_cost     = oneago[  idx_b     ] + 1
      addition_cost     = thisrow[ idx_b - 1 ] + 1
      substitution_cost = oneago[  idx_b - 1 ] + ( text_a[ idx_a ] != text_b[ idx_b ] )
      thisrow[ idx_b ]  = min( deletion_cost, addition_cost, substitution_cost )
      if (  idx_a > 0
        and idx_b > 0
        and text_a[ idx_a     ] == text_b[ idx_b - 1 ]
        and text_a[ idx_a - 1 ] == text_b[ idx_b     ]
        and text_a[ idx_a     ] != text_b[ idx_b     ] ):
        thisrow[ idx_b ] = min( thisrow[ idx_b ], twoago[ idx_b - 2 ] + 1 )
      #.....................................................................................................
      # some diagnostic output:
      print '#EDR B', thisrow
      #.....................................................................................................
  return thisrow[ len( text_b ) - 1 ]

编辑我还将此文本发布到pastebin和 Cython 列表。

4

1 回答 1

2

做一些基本的调试。您知道在标记为 的第二个输出行中出错了#ED B。错误的值似乎表明它很早就找到了一个编辑，并且再也找不到了。这可能是因为其中一个min()args 以某种方式被限制在 1。 Print deletion_cost, substitution_cost, addition_cost... 这是错误的吗？为什么是错的？打印输入文本值。暂时禁用转置部分，看看是否可以解决问题。检查并重新检查_warp雀跃（如果我见过的话，这是一个狡猾的霍比特人噱头）及其用法。如果将“aaaaa”与“aaaaa”进行比较会发生什么？“qwerty”与“qwerty”？“xxxxx”与“yyyyy”？问题是否发生在所有bytes,bytearray和str输入上？

免费问题：我怀疑是腐败，而不是头晕。打印三个数组；他们的内容是否符合预期？尝试一次启用free()一个阵列 - 全部损坏？只有一个？哪一个？

关于内存管理的一些旁白：您可能想阅读这篇文章并考虑使用 Python 特定的例程而不是 malloc/free。如果有代理，缩小你的阵列似乎是最重要的。

更新：遵循我自己的建议。删除成本被塞满了。“oneago”与“thisrow”相同。导致错误答案和加倍（-！未损坏！-）免费的问题：指针的循环洗牌不是循环的。

# twoago, oneago = oneago, thisrow ### BUG ###
twoago, oneago, thisrow = oneago, thisrow, twoago ### FIXED ###

更新 2： [评论容量太小] 没有魔力，就像我建议的那样，只是普通的调试工作。“专注于我的修复”不是“超级易读”。参考代码确实为每次通过创建一个新列表，它可以这样做，因为thisrow指的是没有从上一次传递中结转。它不需要这样做，实际上除了第一个和最后一个元素之外的初始化可能由随机数组成，并且只是为了填写列表，以便它可以被索引而不是附加到一些非-tricksy 实现。因此，您可以盲目地模拟“参考实现”，代价是额外（浪费）malloc/free，或者您可以忽略特定于 Python 的实现细节并仅将参考实现用作可能正确答案的来源。然后你可以接受我的修复，然后通过删除thisrow数组的大部分初始化来节省时间。

更新 3：这是您的替代参考实现。它最初分配 3 行，以避免在外循环内创建列表的开销。它还避免了除了最后一个元素之外的所有不必要的初始化thisrow。这简化了转换为 C/Cython 的过程。

def damlevref2(seq1, seq2):
    # For Python 2.x as was the original.
    # Appears to work on Python 1.5.2 as well :-)
    seq2len = len(seq2)
    twoago = [-777] * (seq2len + 1) # pseudo-malloc; any old rubbish will do
    oneago = [-666] * (seq2len + 1) # ditto
    thisrow = range(1, seq2len + 1) + [0]
    for x in xrange(len(seq1)):
        twoago, oneago, thisrow = oneago, thisrow, twoago # circular "pointer" shuffle
        thisrow[-1] = x + 1
        for y in xrange(seq2len):
            delcost = oneago[y] + 1
            addcost = thisrow[y - 1] + 1
            subcost = oneago[y - 1] + (seq1[x] != seq2[y])
            thisrow[y] = min(delcost, addcost, subcost)
            if (x > 0 and y > 0 and seq1[x] == seq2[y - 1]
                and seq1[x-1] == seq2[y] and seq1[x] != seq2[y]):
                thisrow[y] = min(thisrow[y], twoago[y - 2] + 1)
    return thisrow[seq2len - 1]
于 2010-08-07T23:55:21.080 回答