问题
I'm back with another longish question. Having experimented with a number of Python-based Damerau-Levenshtein
edit distance implementations, I finally found the one listed below as editdistance_reference(). It
seems to deliver correct results and appears to have an efficient implementation.
So I set down to convert the code to Cython. on my test data, the reference method manages to deliver results
for 11,000 comparisons (for pairs of words aound 12 letters long), while the Cythonized method does over
200,000 comparisons per second. Sadly, the results are incorrect: when you look at the variable thisrow
which I print out for debugging, my version has it filled with ones no matter what data I throw at it,
while the reference output shows another picture. For example, testing 'helo' against 'world'
produces the following output (ED marks my function, EDR is the correctly working reference):
From 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]
from 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]
i must be very dumb as the error is probably one of those very very obvious things. but i can't seem to find it.
there is a second problem: i malloc space for the three arrays twoago, oneago, and thisrow,
then they get swapped around in a circular fashion. when i try to free( twoago ) and so on, i get a
line where glibc complains about double free or corruption. i googled for that; could it be that the
pointer-swapping business makes glibc a bit dizzy so it becomes unable to correctly free memory?
below i list first the setup.py that is needed to do the compilation
(/path/to/python3.1 ./setup.py build_ext --inplace), then the edit distance code proper, so interested
people find it easier to replicate.
one more thing: this is run with Python3.1; one funny thing is that inside the *.pyx file we do have
bare unicode strings, but print is still a statement, not a function.
and yes i know this is a lot of code to paste here but the thing is the code is simply not runnable when you
cut it down all too much. i believe all methods except editdistance() to work correctly, but feel free
to point out any problems you perceive.
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 (scroll all the way to the end to see the interesting stuff):
############################################################################################################
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 ]
edit i also posted this text to pastebin and the Cython list.
回答1:
Do some elementary debugging. You know that it is going wrong in the 2nd output line marked #ED B. The wrong values seem to indicate that it finds one edit early on and never finds any more. This is possibly because one of the min() args is somehow clamped at 1. Print deletion_cost, substitution_cost, addition_cost ... which is wrong? Why is it wrong? Print the input text values. Temporarily disable the transposition section to see if that makes the problem go away. Check and re-check the _warp caper (a tricksy hobbit gimmick if I ever saw one) and the usage thereof. What happens if you compare "aaaaa" with "aaaaa"? "qwerty" with "qwerty"? "xxxxx" with "yyyyy"? Does the problem happen with all of bytes, bytearray and str input?
The free problem: I'd suspect corruption, not dizzyness. Print the three arrays; are their contents as expected? Try enabling the free() one array at a time -- all broken? only one? which one?
Some asides on memory management: You may like to read this and consider using the Python-specific routines instead of malloc/free. Downsizing your array if there have been surrogates seems over the top.
Update: Followed my own suggestions. Deletion cost was stuffed. "oneago" was same as "thisrow". Problem causing both the wrong answer and the doubled (-! not corrupted !-) free: circular shuffle of pointers wasn't circular.
# twoago, oneago = oneago, thisrow ### BUG ###
twoago, oneago, thisrow = oneago, thisrow, twoago ### FIXED ###
Update 2: [comment capacity too small] No mojo, just plain ordinary debugging spadework, as I suggested. "concentrating on this for my fix" is not "super-readible". The reference code does create a new list for each pass, which it CAN do because thisrow refers to nothing carried over from the previous pass. It doesn't NEED to do this, and in fact the initialisation apart from the first and last elements could consist of random numbers, and are only there to fill out the list so that it can be indexed into instead of appended to as some non-tricksy implementations do. So you can slavishly emulate the "reference implementation", at the cost of doing an extra (wasted) malloc/free, or you could ignore the Python-specific implementation details and use the reference implementation solely as a source of presumably correct answers. Then you could accept my fix, and later go on to saving time by chopping out most of the initialisation of the thisrow array.
Update 3: Here's a replacement reference implementation for you. It allocates 3 rows initially, in order to avoid the overhead of list creation inside the outer loop. It also avoids the unnecessary initialisation of all but the last element of thisrow. This eases the translation into C/Cython.
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]
来源：https://stackoverflow.com/questions/3431933/how-to-correct-bugs-in-this-damerau-levenshtein-implementation