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			256 lines
		
	
	
		
			6.8 KiB
		
	
	
	
		
			Cython
		
	
	
	
	
	
			
		
		
	
	
			256 lines
		
	
	
		
			6.8 KiB
		
	
	
	
		
			Cython
		
	
	
	
	
	
# cython: infer_types=True
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'''Do Levenshtein alignment, for evaluation of tokenized input.
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Random notes:
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  r i n g
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  0 1 2 3 4
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r 1 0 1 2 3
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a 2 1 1 2 3
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n 3 2 2 1 2
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g 4 3 3 2 1
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0,0: (1,1)=min(0+0,1+1,1+1)=0 S
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1,0: (2,1)=min(1+1,0+1,2+1)=1 D
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2,0: (3,1)=min(2+1,3+1,1+1)=2 D
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3,0: (4,1)=min(3+1,4+1,2+1)=3 D
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0,1: (1,2)=min(1+1,2+1,0+1)=1 D
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1,1: (2,2)=min(0+1,1+1,1+1)=1 S
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2,1: (3,2)=min(1+1,1+1,2+1)=2 S or I
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3,1: (4,2)=min(2+1,2+1,3+1)=3 S or I
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0,2: (1,3)=min(2+1,3+1,1+1)=2 I
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1,2: (2,3)=min(1+1,2+1,1+1)=2 S or I
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2,2: (3,3)
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3,2: (4,3)
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At state (i, j) we're asking "How do I transform S[:i+1] to T[:j+1]?"
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We know the costs to transition:
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S[:i]   -> T[:j]   (at D[i,j])
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S[:i+1] -> T[:j]   (at D[i+1,j])
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S[:i]   -> T[:j+1] (at D[i,j+1])
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Further, we now we can tranform:
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S[:i+1] -> S[:i] (DEL) for 1,
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T[:j+1] -> T[:j] (INS) for 1.
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S[i+1]  -> T[j+1] (SUB) for 0 or 1
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Therefore we have the costs:
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SUB: Cost(S[:i]->T[:j])   + Cost(S[i]->S[j])
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i.e. D[i, j] + S[i+1] != T[j+1]
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INS: Cost(S[:i+1]->T[:j]) + Cost(T[:j+1]->T[:j])
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i.e. D[i+1,j] + 1
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DEL: Cost(S[:i]->T[:j+1]) + Cost(S[:i+1]->S[:i]) 
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i.e. D[i,j+1] + 1
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    Source string S has length m, with index i
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    Target string T has length n, with index j
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    Output two alignment vectors: i2j (length m) and j2i (length n)
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    # function LevenshteinDistance(char s[1..m], char t[1..n]):
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    # for all i and j, d[i,j] will hold the Levenshtein distance between
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    # the first i characters of s and the first j characters of t
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    # note that d has (m+1)*(n+1) values
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    # set each element in d to zero
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    ring rang
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      - r i n g
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    - 0 0 0 0 0
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    r 0 0 0 0 0
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    a 0 0 0 0 0
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    n 0 0 0 0 0
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    g 0 0 0 0 0
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    # source prefixes can be transformed into empty string by
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    # dropping all characters
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    # d[i, 0] := i
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    ring rang
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      - r i n g
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    - 0 0 0 0 0
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    r 1 0 0 0 0
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    a 2 0 0 0 0
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    n 3 0 0 0 0
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    g 4 0 0 0 0
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    # target prefixes can be reached from empty source prefix
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    # by inserting every character
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    # d[0, j] := j
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      - r i n g
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    - 0 1 2 3 4
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    r 1 0 0 0 0
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    a 2 0 0 0 0
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    n 3 0 0 0 0
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    g 4 0 0 0 0
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'''
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from __future__ import unicode_literals
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from libc.stdint cimport uint32_t
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import numpy
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cimport numpy as np
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from .compat import unicode_
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from murmurhash.mrmr cimport hash32
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def align(S, T):
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    cdef int m = len(S)
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    cdef int n = len(T)
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    cdef np.ndarray matrix = numpy.zeros((m+1, n+1), dtype='int32')
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    cdef np.ndarray i2j = numpy.zeros((m,), dtype='i')
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    cdef np.ndarray j2i = numpy.zeros((n,), dtype='i')
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    cdef np.ndarray S_arr = _convert_sequence(S)
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    cdef np.ndarray T_arr = _convert_sequence(T)
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    fill_matrix(<int*>matrix.data,
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        <const int*>S_arr.data, m, <const int*>T_arr.data, n)
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    fill_i2j(i2j, matrix)
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    fill_j2i(j2i, matrix)
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    for i in range(i2j.shape[0]):
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        if i2j[i] >= 0 and len(S[i]) != len(T[i2j[i]]):
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            i2j[i] = -1
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    for j in range(j2i.shape[0]):
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        if j2i[j] >= 0 and len(T[j]) != len(S[j2i[j]]):
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            j2i[j] = -1
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    return matrix[-1,-1], i2j, j2i, matrix
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def multi_align(np.ndarray i2j, np.ndarray j2i, i_lengths, j_lengths):
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    '''Let's say we had:
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    Guess: [aa bb cc dd]
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    Truth: [aa bbcc dd]
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    i2j: [0, None, -2, 2]
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    j2i: [0, -2, 3]
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    We want:
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    i2j_multi: {1: 1, 2: 1}
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    j2i_multi: {}
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    '''
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    i2j_miss = _get_regions(i2j, i_lengths)
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    j2i_miss = _get_regions(j2i, j_lengths)
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    i2j_multi, j2i_multi = _get_mapping(i2j_miss, j2i_miss, i_lengths, j_lengths)
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    return i2j_multi, j2i_multi
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def _get_regions(alignment, lengths):
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    regions = {}
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    start = None
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    offset = 0
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    for i in range(len(alignment)):
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        if alignment[i] < 0:
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            if start is None:
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                start = offset
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                regions.setdefault(start, [])
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            regions[start].append(i)
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        else:
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            start = None
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        offset += lengths[i]
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    return regions
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def _get_mapping(miss1, miss2, lengths1, lengths2):
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    i2j = {}
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    j2i = {}
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    for start, region1 in miss1.items():
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        if not region1 or start not in miss2:
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            continue
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        region2 = miss2[start]
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        if sum(lengths1[i] for i in region1) == sum(lengths2[i] for i in region2):
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            j = region2.pop(0)
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            buff = []
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            # Consume tokens from region 1, until we meet the length of the
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            # first token in region2. If we do, align the tokens. If
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            # we exceed the length, break.
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            while region1:
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                buff.append(region1.pop(0))
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                if sum(lengths1[i] for i in buff) == lengths2[j]:
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                    for i in buff:
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                        i2j[i] = j
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                    j2i[j] = buff[-1]
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                    j += 1
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                    buff = []
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                elif sum(lengths1[i] for i in buff) > lengths2[j]:
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                    break
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            else:
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                if buff and sum(lengths1[i] for i in buff) == lengths2[j]:
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                    for i in buff:
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                        i2j[i] = j
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                    j2i[j] = buff[-1]
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    return i2j, j2i
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def _convert_sequence(seq):
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    if isinstance(seq, numpy.ndarray):
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        return numpy.ascontiguousarray(seq, dtype='uint32_t')
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    cdef np.ndarray output = numpy.zeros((len(seq),), dtype='uint32')
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    cdef bytes item_bytes
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    for i, item in enumerate(seq):
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        if item == "``":
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            item = '"'
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        elif item == "''":
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            item = '"'
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        if isinstance(item, unicode):
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            item_bytes = item.encode('utf8')
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        else:
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            item_bytes = item
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        output[i] = hash32(<void*><char*>item_bytes, len(item_bytes), 0)
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    return output
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cdef void fill_matrix(int* D, 
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        const int* S, int m, const int* T, int n) nogil:
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    m1 = m+1
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    n1 = n+1
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    for i in range(m1*n1):
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        D[i] = 0
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    for i in range(m1):
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        D[i*n1] = i
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    for j in range(n1):
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        D[j] = j
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    cdef int sub_cost, ins_cost, del_cost
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    for j in range(n):
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        for i in range(m):
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            i_j = i*n1 + j
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            i1_j1 = (i+1)*n1 + j+1
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            i1_j = (i+1)*n1 + j
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            i_j1 = i*n1 + j+1
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            if S[i] != T[j]:
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                sub_cost = D[i_j] + 1
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            else:
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                sub_cost = D[i_j]
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            del_cost = D[i_j1] + 1
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            ins_cost = D[i1_j] + 1
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            best = min(min(sub_cost, ins_cost), del_cost)
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            D[i1_j1] = best
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cdef void fill_i2j(np.ndarray i2j, np.ndarray D) except *:
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    j = D.shape[1]-2
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    cdef int i = D.shape[0]-2
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    while i >= 0:
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        while D[i+1, j] < D[i+1, j+1]:
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            j -= 1
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        if D[i, j+1] < D[i+1, j+1]:
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            i2j[i] = -1
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        else:
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            i2j[i] = j
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            j -= 1
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        i -= 1
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cdef void fill_j2i(np.ndarray j2i, np.ndarray D) except *:
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    i = D.shape[0]-2
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    cdef int j = D.shape[1]-2
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    while j >= 0:
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        while D[i, j+1] < D[i+1, j+1]:
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            i -= 1
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        if D[i+1, j] < D[i+1, j+1]:
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            j2i[j] = -1
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        else:
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            j2i[j] = i
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            i -= 1
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        j -= 1
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