最近在自学python，又用python实现了一下BLAST。

这次更新了打分函数如下，空位罚分改为-5，但不区分gap open 和 gap extend。

''''' @author: JiuYu ''' def score(a,b):#scoring function score=0 lst=['AC','GT','CA','TG'] if a==b: score +=2 elif a+b in lst: score += -5 else: score += -7 return score def BLAST(seq1,seq2):#Basic Local Alignment Search Tool l1 = len(seq1) l2 = len(seq2) GAP =-5 #-5 for any gap scores =[] point =[] for j in range(l2+1): if j == 0: line1=[0] line2=[0] for i in range(1,l1+1): line1.append(GAP*i) line2.append(2) else: line1=[] line2=[] line1.append(GAP*j) line2.append(3) scores.append(line1) point.append(line2) #fill the blank of scores and point for j in range(1,l2+1): letter2 = seq2[j-1] for i in range(1,l1+1): letter1 = seq1[i-1] diagonal_score = score(letter1, letter2) + scores[j-1][i-1] left_score = GAP + scores[j][i-1] up_score = GAP + scores[j-1][i] max_score = max(diagonal_score, left_score, up_score) scores[j].append(max_score) if scores[j][i] == diagonal_score: point[j].append(1) elif scores[j][i] == left_score: point[j].append(2) else: point[j].append(3) #trace back alignment1='' alignment2='' i = l2 j = l1 print 'scores =',scores[i][j] while True: if point[i][j] == 0: break elif point[i][j] == 1: alignment1 += seq1[j-1] alignment2 += seq2[i-1] i -= 1 j -= 1 elif point[i][j] == 2: alignment1 += seq1[j-1] alignment2 += '-' j -= 1 else: alignment1 += '-' alignment2 += seq2[i-1] i -= 1 #reverse alignment alignment1 = alignment1[::-1] alignment2 = alignment2[::-1] print 'The best alignment:' print alignment1 print alignment2 seq1=raw_input('Please input your first sequences:\n') seq2=raw_input('input second sequences:\n') BLAST(seq1, seq2)

运行结果：

无疑python对字符串的处理更加强大，语言也更加简单，优雅。比如最后逆序输出alignment，java我是单独写了一个逆序函数，而python只用一个语句就可以完成相同任务。

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