我带着另一个较长的问题回来了。尝试了一些基于 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并且似乎有一个有效的实现。
所以我着手将代码转换为 Cython。在我的测试数据上，引用方法设法提供了结果
进行 11,000 次比较(对于大约 12 个字母长的单词对)，而 Cythonized 方法完成
每秒 200,000 次比较。遗憾的是，结果不正确:当您查看变量 thisrow
我打印出来进行调试，无论我向它扔什么数据，我的版本都充满了它，
而引用输出显示另一张图片。例如，针对 'world' 测试 'helo'
产生以下输出(ED 标记我的函数，EDR 是正确工作的引用):
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非常明显的事情之一。但我似乎找不到它。
还有第二个问题:malloc 三个数组twoago、oneago 和thisrow 的空间,
然后他们以循环方式交换。当我尝试 free( twoago ) 等等时，我得到一个
glibc 提示 double free or corruption 的行。我用谷歌搜索；难道是那个
指针交换业务让 glibc 有点晕所以它变得无法正确释放内存？
下面我首先列出了编译所需的setup.py
(/path/to/python3.1 ./setup.py build_ext --inplace)，然后编辑距离代码合适，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；一件有趣的事是在 *.pyx 文件中我们确实有
裸 unicode 字符串，但 print 仍然是一个语句，而不是一个函数。
是的，我知道这里要粘贴很多代码，但问题是当您
削减太多了。我相信除 editdistance() 之外的所有方法都可以正常工作，但请放心
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(一直滚动到最后以查看有趣的内容):
############################################################################################################
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 我也将此文本发布到 pastebin和 Cython 列表。
最佳答案
做一些基本的调试。您知道在标记为 #ED B 的第二行输出中出错了。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。打印 deletion_cost, substitution_cost, addition_cost ... 哪个是错误的？为什么错了？打印输入文本值。暂时禁用移调部分，看看是否可以解决问题。检查并重新检查 _warp caper(如果我见过的话，这是一个棘手的霍比特人噱头)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”？所有 bytes、bytearray 和 str 输入都会出现问题吗？
免费问题:我怀疑是腐败，而不是头晕。打印三个数组；它们的内容是否符合预期？尝试一次启用一个数组的 free() —— 都坏了吗？只有一个？哪个？
关于内存管理的一些旁白:您可能喜欢阅读 this并考虑使用特定于 Python 的例程而不是 malloc/free。如果有代理人，缩小阵列规模似乎有点过头了。
更新:遵循我自己的建议。删除成本被塞满了。 “oneago”与“thisrow”相同。导致错误答案和双倍(-!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，只是普通的调试 spadework，正如我所建议的。 “专注于我的修复”不是“ super 可读”。引用代码确实为每个 channel 创建了一个新列表，它可以这样做，因为 thisrow 没有引用从上一个 channel 遗留下来的任何内容。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而不是附加到一些非-tricksy 实现。因此，您可以以执行额外(浪费的)malloc/free 为代价盲目地模仿“引用实现”，或者您可以忽略特定于 Python 的实现细节并仅将引用实现用作可能正确答案的来源。然后你可以接受我的修复，然后继续通过砍掉 thisrow 数组的大部分初始化来节省时间。
更新 3:这是为您提供的替换引用实现。它最初分配 3 行，以避免在外部循环内创建列表的开销。它还避免了对 thisrow 的最后一个元素以外的所有元素进行不必要的初始化。这简化了到 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]
关于python - 如何纠正此 Damerau-Levenshtein 实现中的错误？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/3431933/