Needleman–Wunsch 算法的代碼實現
# -*- coding: utf-8 -*-
"""
:Author: huangsh
:Date: 19-7-28 下午19:17
:Description: 使用bidu Needleman–Wunsch 算法來計算兩條序列的最大相似得分
如果您對此算法不熟悉，可以去看看我寫的一篇拙文：https://www.jianshu.com/p/002bbebcaaef
"""
from collections import namedtuple
F = namedtuple('F', ('score', 'pointer'))
## 初始化二維矩陣， # 生成 x行，y列的二維矩陣，初始化第0行，0列的元素
def init_array(x, y):
array = [[0] * (y) for _ in range(x)]
array[0][0] = F(0, None)
for j in range(1, y):
array[0][j] = F((-5)*j, [0, j-1])
for i in range(1, x):
array[i][0] = F((-5)*i, [i-1, 0])
return array
## 一行一行的計算矩陣中的每個各自中的最優結果。當前格子中的最優結果由它的三個來源推出
def compute(array, seq1, seq2):
row, col = len(seq2), len(seq1)
for i in range(1, row+1):
for j in range(1, col+1):
if seq1[j-1] == seq2[i-1]: # 這里簡化了得分矩陣，完全匹配得10分，不完全得5分，有gap減5分
s = 10
else:
s = 5
lu = [array[i-1][j-1].score+s, [i-1, j-1]] # idx 0：最大得分，idx 1：來源坐標
left = [array[i-1][j].score-5, [i-1, j]]
up = [array[i][j-1].score-5, [i, j-1]]
max_choice = max([lu,left, up], key=lambda x: x[0])
score= max_choice[0]
pointer = max_choice[1]
array[i][j] = F(score, pointer) # 在當前保存最大得分，和來源坐標，方便回溯。
return array
## 回溯。從（m,n）一直回溯到（0，0）
def backtrack(array, seq1, seq2):
s1 = []
s2 = []
row, col = len(seq2), len(seq1)
while array[row][col].score != 0:
i, j = array[row][col].pointer # pointer 指向來源方的坐標
if i+1 == row and j+1 == col: # 左上方
s1.append(seq1[col-1])
s2.append(seq2[row-1])
row, col = i, j
elif row == i+1 and col == j: # 來源：上方
s1.append("-")
s2.append(seq2[i])
row, col = i, j
elif row == i and col == j+1: # 左方
s1.append(seq1[j])
s2.append("-")
row, col = i, j
s1 = ''.join(s1[::-1]) #因為是從最后往前回溯的，需要將逆轉一下list
s2 = ''.join(s2[::-1])
return s1, s2
def main(seq1, seq2):
x, y = len(seq2)+1 , len(seq1)+1 # x是矩陣行數，y是矩陣列數
array = init_array(x, y)
array = compute(array, seq1, seq2)
s1, s2 = backtrack(array, seq1, seq2)
max_score = array[x-1][y-1].score
print("最大得分：", max_score)
print(s1)
print(s2)
if __name__ == '__main__':
seq1 = "ATCGCGCAACTGCGCGC"
seq2 = "ACGCGCACTGCGGC"
main(seq1, seq2)