我的理解是，虽然基本上我能找到的所有关于动态规划的讨论都在填充矩阵时存储了指针，但在回溯步骤中重新计算先前的单元格会更快。
据我所知，我有自己的动态规划算法可以正确构建矩阵，但我对如何进行回溯计算感到困惑。我还被告知有必要重新计算这些值(而不是仅仅查找它们)，但我不知道如何得出不同的数字。
我正在实现的 SW 版本包括一个打开两个序列中的间隙的选项，因此每个矩阵的递归关系具有三个选项。下面是我的全局对齐类的当前版本。根据我的手工计算，我认为 score_align 正确生成了矩阵，但显然 traceback_col_seq 不起作用。
INF = 2147483647 #max size of int32
class global_aligner():
def __init__(self, subst, open=10, extend=2, 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):
self.M = zeros((self.maxI, self.maxJ), int)
self.Ic = zeros((self.maxI, self.maxJ), int)
self.Ir = zeros((self.maxI, self.maxJ), int)
for i in xrange(self.maxI):
self.M[i][0], self.Ir[i][0], self.Ic[i][0] = \
-INF, -INF, -(self.open+self.extend*i)
for j in xrange(self.maxJ):
self.M[0][j], self.Ic[0][j], self.Ir[0][j] = \
-INF, -INF, -(self.open+self.extend*j)
self.M[0][0] = 0
self.Ic[0][0] = -self.open
def score_cell(self, i, j, chars):
thisM = [self.Ic[i-1][j-1]+self.subst[chars], self.M[i-1][j-1]+\
self.subst[chars], self.Ir[i-1][j-1]+self.subst[chars]]
thisC = [self.Ic[i][j-1]-self.extend, self.M[i][j-1]-self.open, \
self.Ir[i][j-1]-self.double]
thisR = [self.M[i-1][j]-self.open, self.Ir[i-1][j]-self.extend, \
self.Ic[i-1][j]-self.double]
return max(thisM), max(thisC), max(thisR)
def score_align(self, row_seq, col_seq):
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()
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):
self.traceback = list()
i, j = self.maxI-1, self.maxJ-1
while i > 1 and j > 1:
cell = [self.M[i][j], self.Ic[i][j], self.Ir[i][j]]
cellMax = max(cell)
chars = self.row_seq[i-1]+self.col_seq[j-1]
if cell.index(cellMax) == 0: #M
diag = [diagM, diagC, diagR] = self.score_cell(i-1, j-1, chars)
diagMax = max(diag)
if diag.index(diagMax) == 0: #match
self.traceback.append(self.col_seq[j-1])
elif diag.index(diagMax) == 1: #insert column (open)
self.traceback.append('-')
elif diag.index(diagMax) == 2: #insert row (open other)
self.traceback.append(self.col_seq[j-1].lower())
i, j = i-1, j-1
elif cell.index(cellMax) == 1: #Ic
up = [upM, upC, upR] = self.score_cell(i-1, j, chars)
upMax = max(up)
if up.index(upMax) == 0: #match (close)
self.traceback.append(self.col_seq[j-1])
elif up.index(upMax) == 1: #insert column (extend)
self.traceback.append('-')
elif up.index(upMax) == 2: #insert row (double)
self.traceback.append('-')
i -= 1
elif cell.index(cellMax) == 2: #Ir
left = [leftM, leftC, leftR] = self.score_cell(i, j-1, chars)
leftMax = max(left)
if left.index(leftMax) == 0: #match (close)
self.traceback.append(self.col_seq[j-1])
elif left.index(leftMax) == 1: #insert column (double)
self.traceback.append('-')
elif left.index(leftMax) == 2: #insert row (extend other)
self.traceback.append(self.col_seq[j-1].lower())
j -= 1
for j in xrange(0,j,-1):
self.traceback.append(self.col_seq[j-1])
for i in xrange(0,i, -1):
self.traceback.append('-')
return ''.join(self.traceback[::-1])
test = global_aligner(blosumMatrix)
test.score_align('AA','AAA')
test.traceback_col_seq()
请您参考如下方法:
我认为主要问题是在生成可能来自的单元格时，您没有考虑当前使用的矩阵。 cell = [self.M[i][j], self.Ic[i][j], self.Ir[i][j]] 第一次通过while循环就对了, 但在那之后你不能只选择得分最高的矩阵。您的选择受到您来自哪里的限制。我在执行您的代码时遇到了一些麻烦，但我认为您在 while 循环的 if 语句中考虑到了这一点。如果是这样的话，那么我认为沿着这些方向的改变就足够了:
cell = [self.M[i][j], self.Ic[i][j], self.Ir[i][j]]
cellIndex = cell.index(max(cell))
while i > 1 and j > 1:
chars = self.row_seq[i-1]+self.col_seq[j-1]
if cellIndex == 0: #M
diag = [diagM, diagC, diagR] = self.score_cell(i-1, j-1, chars)
diagMax = max(diag)
...
cellIndex = diagMax
i, j = i-1, j-1
elif cell.index(cellMax) == 1: #Ic
up = [upM, upC, upR] = self.score_cell(i-1, j, chars)
upMax = max(up)
...
cellIndex = upMax
i -= 1
elif cell.index(cellMax) == 2: #Ir
left = [leftM, leftC, leftR] = self.score_cell(i, j-1, chars)
leftMax = max(left)
...
cellIndex = leftMax
j -= 1
就像我说的，我不确定我是否正确地遵循了您的代码，但看看这是否有帮助。
