gpt4 book ai didi

python - Needleman-Wunsch 算法动态规划实现中的回溯

转载 作者:太空狗 更新时间:2023-10-29 17:33:52 28 4
gpt4 key购买 nike

我的 needleman-wunsch 实现几乎可以正常工作,但我对如何处理特定案例的回溯感到困惑。

想法是,为了重新构建序列(最长路径),我们重新计算以确定得分来自的矩阵。我遇到问题的边缘情况是右下角的分数不在匹配矩阵中,而是在插入列矩阵中(这意味着生成的追溯序列应该有一个插入。

这些序列以 a2m 格式记录,其中序列中的插入被记录为小写字符。所以在最终输出中,ZZAAAC 的对齐方式应该是 AAac。当我手动回溯时,我以 AAAc 结束,因为我只访问了 Ic 矩阵一次。 Here是我的白板的图片。如您所见,我有三个黑色箭头和一个绿色箭头,这就是为什么我的回溯给了我 AAAc。我应该数第一个单元格,然后停在位置 1,1 吗?我不确定我将如何改变我实现它的方式。

请注意,此处使用的替换矩阵是 BLOSUM62。递归关系是

M(i,j) = max(Ic(i-1,j-1)+subst, M(i-1,j-1)+subst, Ir(i-1,j-1)+subst)
Ic(i,j) = max(Ic(i,j-1)-extend, M(i,j-1)-open, Ir(i,j-1)-double)
Ir(i,j) = max(Ic(i-1,j)-double, M(i-1,j)-open, Ir(i-1,j)-extend)

编辑:这里重写了 traceback_col_seq 函数以使其更清晰。请注意,score_cell 现在返回 thisM、thisC、thisR 而不是其中的最大值。这个版本的比对得分为 AaAc,仍然有同样的问题,现在又出现了另一个问题,为什么它会在 1,2 处再次进入 Ic。然而,这个版本更清晰。

def traceback_col_seq(self):
i, j = self.maxI-1, self.maxJ-1
self.traceback = list()
matrixDict = {0:'M',1:'Ic',2:'Ir',3:'M',4:'Ic',5:'Ir',6:'M',7:'Ic',8:'Ir'}
while i > 0 or j > 0:
chars = self.col_seq[j-1] + self.row_seq[i-1]
thisM, thisC, thisR = self.score_cell(i, j, chars)
cell = thisM + thisC + thisR
prevMatrix = matrixDict[cell.index(max(cell))]
print(cell, prevMatrix,i,j)
if prevMatrix == 'M':
i -= 1; j -= 1
self.traceback.append(self.col_seq[j])
elif prevMatrix == 'Ic':
j -= 1
self.traceback.append(self.col_seq[j].lower())
elif prevMatrix == 'Ir':
i -= 1
self.traceback.append('-')
return ''.join(self.traceback[::-1])

这是生成动态规划矩阵并追溯对齐的python类。还有一个评分函数用于检查生成的比对是否正确。

class global_aligner():
def __init__(self, subst, open=12, extend=1, double=3):
self.extend, self.open, self.double, self.subst = extend, open, double, subst
def __call__(self, row_seq, col_seq):
#add alphabet error checking?
score_align(row_seq, col_seq)
return traceback_col_seq()
def init_array(self):
"""initialize three numpy arrays, set values in 1st column and row"""
self.M = zeros((self.maxI, self.maxJ), float)
self.Ic = zeros((self.maxI, self.maxJ), float)
self.Ir = zeros((self.maxI, self.maxJ), float)
for i in xrange(1,self.maxI):
self.M[i][0], self.Ic[i][0], self.Ir[i][0] = \
-float('inf'), -float('inf'), -(self.open+self.extend*(i-1))
for j in xrange(1,self.maxJ):
self.M[0][j], self.Ir[0][j], self.Ic[0][j] = \
-float('inf'), -float('inf'), -(self.open+self.extend*(j-1))
self.Ic[0][0] = self.Ir[0][0] = -float('inf')
def score_cell(self, i, j, chars):
"""score a matrix cell based on the 9 total neighbors (3 each direction)"""
thisM = [self.M[i-1][j-1]+self.subst[chars], self.Ic[i-1][j-1]+ \
self.subst[chars], self.Ir[i-1][j-1]+self.subst[chars]]
thisC = [self.M[i][j-1]-self.open, self.Ic[i][j-1]-self.extend, \
self.Ir[i][j-1]-self.double]
thisR = [self.M[i-1][j]-self.open, self.Ic[i-1][j]-self.double, \
self.Ir[i-1][j]-self.extend]
return max(thisM), max(thisC), max(thisR)
def score_align(self, row_seq, col_seq):
"""build dynamic programming matrices to align two sequences"""
self.row_seq, self.col_seq = list(row_seq), list(col_seq)
self.maxI, self.maxJ = len(self.row_seq)+1, len(self.col_seq)+1
self.init_array() #initialize arrays
for i in xrange(1, self.maxI):
row_char = self.row_seq[i-1]
for j in xrange(1, self.maxJ):
chars = row_char+self.col_seq[j-1]
self.M[i][j], self.Ic[i][j], self.Ir[i][j] = self.score_cell(i, j, chars)
def traceback_col_seq(self):
"""trace back column sequence in matrices in a2m format"""
i, j = self.maxI-1, self.maxJ-1
self.traceback = list()
#find which matrix to start in
#THIS IS WHERE THE PROBLEM LIES I THINK
cell = (self.M[i][j], self.Ic[i][j], self.Ir[i][j])
prevMatrix = cell.index(max(cell))
while i > 1 and j > 1:
if prevMatrix == 0: #M matrix
i -= 1; j -= 1 #step up diagonally
prevChars = self.row_seq[i-1]+self.col_seq[j-1]
diag = self.score_cell(i, j, prevChars) #re-score diagonal cell
prevMatrix = diag.index(max(diag)) #determine which matrix that was
self.traceback.append(self.col_seq[j])
elif prevMatrix == 1: #Ic matrix
j -= 1
prevChars = self.row_seq[i-1]+self.col_seq[j-1]
left = self.score_cell(i, j, prevChars)
prevMatrix = left.index(max(left))
self.traceback.append(self.col_seq[j].lower())
elif prevMatrix == 2: #Ir matrix
i -= 1
prevChars = self.row_seq[i-1]+self.col_seq[j-1]
up = self.score_cell(i, j, prevChars)
prevMatrix = up.index(max(up))
self.traceback.append('-')
for j in xrange(j,0,-1): #hit top of matrix before ending, add chars
self.traceback.append(self.col_seq[j-1])
for i in xrange(i,0,-1): #hit left of matrix before ending, add gaps
self.traceback.append('-')
print(''.join(self.row[::-1]))
return ''.join(self.traceback[::-1])
def score_a2m(self, s1, s2):
"""scores an a2m alignment of two sequences. I believe this function correctly
scores alignments and is used to test my alignments. The value produced by this
function should be the same as the largest value in the bottom right of the three
matrices"""
s1, s2 = list(s1.strip('.')), list(s2.strip('.'))
s1_pos, s2_pos = len(s1)-1, len(s2)-1
score, gap = 0, False
while s1_pos >= 0 and s2_pos >= 0:
if s1[s1_pos].islower() and gap is False:
score -= self.open; s1_pos -= 1; gap = True
elif s1[s1_pos].islower() and gap is True:
score -= self.extend; s1_pos -= 1
elif s2[s2_pos].islower() and gap is False:
score -= self.open; s2_pos -= 1; gap = True
elif s2[s2_pos].islower() and gap is True:
score -= self.extend; s2_pos -= 1
elif s1[s1_pos] == '-' and gap is False:
score -= self.open; s1_pos -= 1; s2_pos -= 1; gap = True
elif s1[s1_pos] == '-' and gap is True:
score -= self.extend; s1_pos -= 1; s2_pos -= 1
elif s2[s2_pos] == '-' and gap is False:
score -= self.open; s1_pos -= 1; s2_pos -= 1; gap = True
elif s2[s2_pos] == '-' and gap is True:
score -= self.extend; s1_pos -= 1; s2_pos -= 1
elif gap is True:
score += self.subst[s1[s1_pos].upper() + s2[s2_pos].upper()]
s1_pos -= 1; s2_pos -= 1; gap = False
else:
score += self.subst[s1[s1_pos].upper() + s2[s2_pos].upper()]
s1_pos -= 1; s2_pos -= 1
if s1_pos >= 0 and gap is True:
score -= self.extend*s1_pos
elif s1_pos >= 0 and gap is False:
score -= self.open+s1_pos*self.extend
if s2_pos >= 0 and gap is True:
score -= self.extend*s2_pos
elif s2_pos >= 0 and gap is False:
score -= self.open+s2_pos*self.extend
return score


test = global_aligner(blosumMatrix)
s1,s2 = 'ZZ','AAAC'
test.score_align(s1, s2)
align = test.traceback_col_seq()
print('This score: ', test.score_a2m(s1,align)
print('Correct score: ', test.score_a2m(s1,'AAac'))

Blosum 解析器

def parse_blosum(blosumFile):
blosumMatrix, commentFlag = dict(), False
for line in blosumFile:
if not line.startswith('#') and not commentFlag:
alphabet = line.rstrip().split()
commentFlag = True
elif commentFlag:
line = line.rstrip().split()
thisChar, line = line[0], line[1:]
for x in xrange(len(line)):
alphaChar, thisValue = alphabet[x], line[x]
blosumMatrix[thisChar+alphaChar] = int(thisValue)
return blosumMatrix

最佳答案

def traceback_col_seq(self):
"""
Traces back the column sequence to determine final global alignment.
Recalculates the score using score_cell.
"""
i, j = self.maxI-1, self.maxJ-1
self.traceback = list() #stores final sequence
matrixDict = {0:'M',1:'Ic',2:'Ir'} #mapping between numeric value and matrix
chars = self.col_seq[j-1] + self.row_seq[i-1] #store first characters
thisM, thisC, thisR = self.score_cell(i,j, chars)
cell = max(thisM), max(thisC), max(thisR) #determine where to start
prevMatrix = matrixDict[cell.index(max(cell))] #store where to go first
while i > 0 or j > 0:
#loop until the top left corner of the matrix is reached
if prevMatrix == 'M':
self.traceback.append(self.col_seq[j-1])
i -= 1; j -= 1
prevMatrix = matrixDict[thisM.index(max(thisM))]
elif prevMatrix == 'Ic':
self.traceback.append(self.col_seq[j-1].lower())
j -= 1
prevMatrix = matrixDict[thisC.index(max(thisC))]
elif prevMatrix == 'Ir':
self.traceback.append('-')
i -= 1
prevMatrix = matrixDict[thisR.index(max(thisR))]
chars = self.col_seq[j-1] + self.row_seq[i-1] #store next characters
thisM, thisC, thisR = self.score_cell(i,j, chars) #rescore next cell
return ''.join(self.traceback[::-1])

关于python - Needleman-Wunsch 算法动态规划实现中的回溯,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20229817/

28 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com