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如何纠正Damerau Levenshtein实现中的错误?

edit-distance levenshtein-distance cython python-3.x python

3

flow · 技术社区 · 16 年前

我又问了一个很长的问题。试验了一些基于Python的Damerau Levenshtein 编辑距离实现, I finally found the one listed below 作为 editdistance_reference() . 它似乎提供了正确的结果,似乎有一个有效的实施。

所以我开始把代码转换成Cython。在我的测试数据中,reference方法能够提供结果对于11000个比较(对于12个字母长的单词对),而Cythonized方法每秒200000次比较。不幸的是,结果是不正确的:当你看变量 thisrow 我打印出来进行调试,我的版本里有很多,不管我向它扔了什么数据, 而参考输出显示另一张图片。例如,测试 'helo' 反对 'world' 产生以下输出( ED 标记我的功能, EDR 是正确的工作参考):

从 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]

从 编辑距离参考() :

#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]

我一定很笨,因为这个错误可能是非常明显的事情之一。但我好像找不到。

还有第二个问题:我 malloc 三个阵列的空间 twoago , oneago ,和 这排 , 然后他们以循环的方式交换。当我试图 free( twoago ) 等等,我得到一个 glibc抱怨的地方 double free or corruption . 我在网上搜索了一下,可能是指针交换业务使glibc有点头晕,因此无法正确释放内存?

下面我先列出 setup.py 这是编译所需要的 ( /path/to/python3.1 ./setup.py build_ext --inplace ),则编辑距离代码正确,所以感兴趣人们发现它更容易复制。

还有一件事:这是用Python3.1运行的;一件有趣的事是 *.pyx 我们有文件裸unicode字符串,但是 print 仍然是一个语句,而不是函数。

是的,我知道这里有很多代码要粘贴,但问题是当您减少太多。我相信所有的方法除了 编辑距离() 正确工作,但请放心指出你感觉到的任何问题。

安装程序.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 (一直滚动到最后可以看到有趣的东西):

############################################################################################################
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 还有Cython列表。

1 回复 | 直到 15 年前

1

2

John Machin Santi 16 年前

做一些初步的调试。你知道它在第二个输出行出错了 #ED B . 错误的值似乎表明它很早就找到了一个编辑,再也找不到了。这可能是因为 min() args不知怎的被限制在1。打印 deletion_cost , substitution_cost , addition_cost ... 哪个是错的?为什么错了?打印输入的文本值。暂时禁用换位部分,看看这是否会使问题消失。检查并重新检查 _warp 开玩笑(如果我见过的话,这是一个诡异的霍比特人的噱头)及其用法。如果你把“aaaaa”和“aaaaa”比较会怎么样“qwerty”和“qwerty”带“yyyy”的“xxxxx”?问题是否发生在 bytes , bytearray 和 str 输入?

自由的问题是:我会怀疑腐败,而不是头晕。打印三个数组;它们的内容是否符合预期?尝试启用 free() 一次一个阵列--都坏了?只有一个?哪一个?

关于内存管理的一些建议:您可以阅读 this 并考虑使用特定于Python的例程,而不是malloc/free。如果有代孕妈妈的话,缩小你的阵型似乎是最重要的。

更新: 遵循我自己的建议。删除成本已满。”“一个月前”和“这排”是一样的。问题导致了错误的答案和加倍的(-!没有腐败!-)自由: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,只是简单的普通调试铲作业,正如我所建议的。”把注意力集中在这个问题上“不是很容易理解的”。引用代码确实为每次传递创建了一个新列表,它可以这样做,因为 thisrow 指的是上一次通行证没有结转。它不需要这样做,事实上,除了第一个和最后一个元素之外的初始化可以由随机数组成,并且只在那里填写列表,以便它可以被索引到,而不是像某些非欺骗的实现那样附加到列表中。因此,您可以以额外(浪费的)malloc/free为代价,盲目地模拟“引用实现”,也可以忽略特定于Python的实现细节,将引用实现仅用作可能正确答案的来源。然后你就可以接受我的修复,然后继续节省时间,将 这排 阵列。

更新3: 这是一个替代的参考实现。它最初分配3行,以避免在外部循环中创建列表的开销。它还避免了不必要的初始化,除了 这排 . 这就简化了C/Cython的翻译。

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]