浅谈动态规划-NW算法

两个矩阵

一个是惩罚矩阵：

image.png

#生成惩罚矩阵
A=np.ones([5,5],dtype=np.int16)*-2+np.eye(5,dtype=np.int16)*5
A[0,0]=0
name_list=["_","A","C","G","T"]
punish_matrix=pd.DataFrame(A,columns=name_list,index=name_list)

这个的作用是为打分提供个参考，将每个碱基的对应情况进行打分

打分矩阵：
我们看下原理图：

image.png

s来自惩罚矩阵，w为惩罚分，H来自打分矩阵
假设我有两条序列：
str_one="ACTTTCA"
str_two="AGGCTTA"

str_one="ACTTTCA"
str_two="AGGCTTA"

index=["_"]+[i for i in str_one]
columns=["_"]+[i for i in str_two]
score_matrix=pd.DataFrame(np.zeros([len(str_one)+1,len(str_two)+1]),index=index,columns=columns)
# 设置行列名，以-，和你的序列为行列名，一条为行名，另一条为列名
punish=-3 #惩罚分
for i in range(len(str_one)+1):
    for j in range(len(str_two)+1):
        if i==0 or j ==0:
            score_matrix.iloc[i,j]=0+punish*i+punish*j
        else:
            insert=score_matrix.iloc[i,j-1]+punish
            delect=score_matrix.iloc[i-1,j]+punish
            match=score_matrix.iloc[i-1,j-1]+punish_matrix.loc[str_one[i-1],str_two[j-1]]
            score_matrix.iloc[i, j]=max(insert,delect,match) #选择最大的值作为填充

image.png

对于这个矩阵来说，我们要找到最底角的值开始回溯
对于这个矩阵来说是3.0

摘自知乎

那么我们先找到右下角的那个值，通过右下角这个值进行路径的回溯，即获得行列的全长即可：

i=len(str_one)
j=len(str_two)

while (i > 0 or j > 0) :
    if (i > 0 and j > 0 and score_matrix.iloc[i, j]==score_matrix.iloc[i-1,j-1]+punish_matrix.loc[str_one[i-1],str_two[j-1]]):
        s1 += str_one[i - 1]
        s2 += str_two[j - 1]
        i -= 1
        j -= 1
    elif (j > 0 and score_matrix.iloc[i, j]==score_matrix.iloc[i,j-1]+punish):
        s1 += '-'
        s2 += str_two[j - 1]
        j -= 1
    elif (i > 0 and score_matrix.iloc[i, j]==score_matrix.iloc[i-1,j]+punish):
        s1 += str_one[i - 1]
        s2 += '-'
        i -= 1

全部代码

import numpy as np
import pandas as pd

A=np.ones([5,5],dtype=np.int16)*-2+np.eye(5,dtype=np.int16)*5
A[0,0]=0
name_list=["_","A","C","G","T"]
punish_matrix=pd.DataFrame(A,columns=name_list,index=name_list)


str_one="ACTTTCA"
str_two="AGGCTTA"
s1 = ''
s2 = ''

index=["_"]+[i for i in str_one]
columns=["_"]+[i for i in str_two]
score_matrix=pd.DataFrame(np.zeros([len(str_one)+1,len(str_two)+1]),index=index,columns=columns)

punish=-3
for i in range(len(str_one)+1):
    for j in range(len(str_two)+1):
        if i==0 or j ==0:
            score_matrix.iloc[i,j]=0+punish*i+punish*j
        else:
            insert=score_matrix.iloc[i,j-1]+punish
            delect=score_matrix.iloc[i-1,j]+punish
            match=score_matrix.iloc[i-1,j-1]+punish_matrix.loc[str_one[i-1],str_two[j-1]]
            score_matrix.iloc[i, j]=max(insert,delect,match)

i=len(str_one) 
j=len(str_two)
# i，j 均为最大值，即满足矩阵右下角的元素的角标

while (i > 0 or j > 0) :
    if (i > 0 and j > 0 and score_matrix.iloc[i, j]==score_matrix.iloc[i-1,j-1]+punish_matrix.loc[str_one[i-1],str_two[j-1]]):
        s1 += str_one[i - 1]
        s2 += str_two[j - 1]
        i -= 1
        j -= 1
    elif (j > 0 and score_matrix.iloc[i, j]==score_matrix.iloc[i,j-1]+punish):
        s1 += '-'
        s2 += str_two[j - 1]
        j -= 1
    elif (i > 0 and score_matrix.iloc[i, j]==score_matrix.iloc[i-1,j]+punish):
        s1 += str_one[i - 1]
        s2 += '-'
        i -= 1

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

结果是：

image.png

考虑空间：

如果序列太长，那么空间复杂度会大大加大，我们可以采用中点法，将长序列分割成两段，然后进行比对

image.png

如上图，它的打分矩阵就不需要完全填充完，分成两段后只需要将蓝色部分的打分举证填充完毕即可

image.png

即绿色部分

import numpy as np
import pandas as pd
import argparse

A=np.ones([5,5],dtype=np.int16)*-2+np.eye(5,dtype=np.int16)*5
A[0,0]=0
name_list=["_","A","C","G","T"]
punish_matrix=pd.DataFrame(A,columns=name_list,index=name_list)

def mid_align(str_one,str_two):
    s1 = ''
    s2 = ''

    index = ["_"] + [i for i in str_one]
    columns = ["_"] + [i for i in str_two]
    score_matrix = pd.DataFrame(np.zeros([len(str_one) + 1, len(str_two) + 1]), index=index, columns=columns)

    punish = -3
    for i in range(len(str_one) + 1):
        for j in range(len(str_two) + 1):
            if i == 0 or j == 0:
                score_matrix.iloc[i, j] = 0 + punish * i + punish * j
            else:
                insert = score_matrix.iloc[i, j - 1] + punish
                delect = score_matrix.iloc[i - 1, j] + punish
                match = score_matrix.iloc[i - 1, j - 1] + punish_matrix.loc[str_one[i - 1], str_two[j - 1]]
                score_matrix.iloc[i, j] = max(insert, delect, match)

    i = len(str_one) 
    j = len(str_two)
# i，j 为矩阵最大的，即满足矩阵“右下角"元素

    while (i > 0 or j > 0):
        if (i > 0 and j > 0 and score_matrix.iloc[i, j] == score_matrix.iloc[i - 1, j - 1] + punish_matrix.loc[
            str_one[i - 1], str_two[j - 1]]):
            s1 += str_one[i - 1]
            s2 += str_two[j - 1]
            i -= 1
            j -= 1
        elif (j > 0 and score_matrix.iloc[i, j] == score_matrix.iloc[i, j - 1] + punish):
            s1 += '-'
            s2 += str_two[j - 1]
            j -= 1
        elif (i > 0 and score_matrix.iloc[i, j] == score_matrix.iloc[i - 1, j] + punish):
            s1 += str_one[i - 1]
            s2 += '-'
            i -= 1
    list = [s1[::-1],s2[::-1]]
    return list

if __name__ == "__main__":
    parser = argparse.ArgumentParser(description="Multi-value dictionary")
    parser.add_argument("-s1", "--s1", required=True, type=str, help="First sequence")
    parser.add_argument("-s2", "--s2", required=True, type=str, help="Second sequence")
    Args = parser.parse_args()
    A = np.ones([5, 5], dtype=np.int16) * -2 + np.eye(5, dtype=np.int16) * 5
    A[0, 0] = 0
    name_list = ["_", "A", "C", "G", "T"]
    punish_matrix = pd.DataFrame(A, columns=name_list, index=name_list)

    str_one = Args.s1
    str_two = Args.s2

    n = len(str_one)
    m = len(str_two)
    i_mid = n // 2  #取中点，然后将序列分为两段
    string_one = str_one[0:i_mid]  #分割序列
    string_two = str_two[0:i_mid]
    string_three = str_one[i_mid:len(str_one)]
    string_four = str_two[i_mid:len(str_two)]

    print(mid_align(string_one,string_two)[0] + mid_align(string_three,string_four)[0])
    print(mid_align(string_one,string_two)[1] + mid_align(string_three,string_four)[1])  #拼接序列

结果为：

image.png

https://www.jianshu.com/p/90bbd4ac447f