利用Needleman–Wunsch算法进行DNA序列全局比对

生物信息学原理作业第二弹：利用Needleman–Wunsch算法进行DNA序列全局比对。

具体原理：https://en.wikipedia.org/wiki/Needleman%E2%80%93Wunsch_algorithm。

利用Needleman–Wunsch算法进行DNA序列全局比对

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贴上python代码：

 # -*- coding: utf-8 -*-

 """

 Created on Sat Nov 25 18:20:01 2017

 @author: zxzhu

 后需修改：

 1.加命令行参数

 2.给出多种比对结果

 """

 import numpy as np

 import pandas as pd

 sequence1 = 'AACGTACTCA'

 sequence2 = 'TCGTACTCA'

 s1 = ''

 s2 = ''

 gap = -4

 score_matrix = pd.read_excel('score.xlsx')        #score matrix

 best_matrix = np.empty(shape= (len(sequence2)+1,len(sequence1)+1),dtype = int)

 def get_match_score(s1,s2):

     score = score_matrix[s1][s2]

     return score

 for i in range(len(sequence2)+1):

     for j in range(len(sequence1)+1):

         if i == 0:

             best_matrix[i][j] = gap * j

         elif j == 0:

             best_matrix[i][j] = gap *i

         else:

             match = get_match_score(sequence2[i-1],sequence1[j-1])

             gap1_score = best_matrix[i-1][j]+gap

             gap2_score = best_matrix[i][j-1]+gap

             match_score = best_matrix[i-1][j-1]+match

             best_matrix[i][j] = max(gap1_score,gap2_score,match_score)

 print(best_matrix)

 i,j = len(sequence2),len(sequence1)

 while(i>0 or j>0):

     match = get_match_score(sequence2[i-1],sequence1[j-1])

     if i>0 and j>0 and best_matrix[i][j] == best_matrix[i-1][j-1]+match:

         s1 += sequence1[j-1]

         s2 += sequence2[i-1]

         i-=1;j-=1

     elif i>0 and best_matrix[i,j] == best_matrix[i-1,j]+gap:

         s1+='-'

         s2+=sequence2[i-1]

         i-=1

     else:

         s1+=sequence1[j-1]

         s2+='-'

         j-=1

 print(s1[::-1]+'\n'+s2[::-1])

后面会加入命令行。

多种结果这里只取了一种，这个问题有待解决。

如果有其他的方法我会及时添加。