Question

我正在尝试使用仿射间隙成本来实现全局对齐算法。我首先用Java实现它，然后用Python实现它。但我在Java中的输出与我在python中得到的不同。我在Python中实现的代码与在Java中完全相同。

我在Python中的代码如下：

import numpy as np

def deletionMatrix(D,S,i,j):
    res = 0
    if i==0 and j>=0:
        res = S[i-1][j] - (gapOpen + gapExt)
        D[i][j] = res

    if i > 1 and j >= 0:
        res_S = S[i-1][j] - (gapOpen + gapExt)
        res_D = D[i-1][j] - gapOpen

        res = max(res_S, res_D)
        D[i][j] = res

    return(res)

def insertionMatrix(I, S, i, j):
    res = 0

    if i >= 0 and j == 1:
        res = S[i][j-1] - (gapOpen + gapExt)
        I[i][j] = res

    if i >= 0 and j > 1:
        res_S = S[i][j-1] - (gapOpen + gapExt)
        res_I = I[i][j-1] - gapOpen

        res = max(res_S, res_I)
        I[i][j] = res

    return(res)

def matrix(S,D,I,m,n,match,mismatch):
    for i in range(0,len(m)):
        for j in range(0,len(n)):
             if i == 0 and j == 0:
                 S[i][j] = 0

             if i == 0 and j > 0:
                 S[i][j] = insertionMatrix(I,S,i,j)

             if i > 0 and j == 0:
                 S[i][j] = deletionMatrix(D,S,i,j)

             if i > 0 and j > 0:
                 if m[i-1] == n[j-1]:
                    res_S = S[i-1][j-1] + match
                    res_D = deletionMatrix(D,S,i,j)
                    res_I = insertionMatrix(I,S,i,j)

                    S[i][j] = max(res_S,res_D,res_I)

                 elif m[i-1] != n[j-1]:
                    res_S = S[i-1][j-1] + mismatch
                    res_D = deletionMatrix(D,S,i,j)
                    res_I = insertionMatrix(I,S,i,j)

                    S[i][j] = max(res_S,res_D,res_I)

     return(S)


gapOpen = 5
gapExt = 2

m = "GAATTCAGTTA"
n = "GGATCGA"

mLen = len(m) + 1
nLen = len(n) + 1

S = np.zeros([mLen,nLen])
D = np.zeros([mLen,nLen])
I = np.zeros([mLen,nLen])

match = 1
mismatch = -3

S = matrix(S,D,I,m,n,match,mismatch)

for i in range(0,len(m)):
   for j in range(0,len(n)):
       print(S[i][j],)

   print()

我附上了我想要实现的算法的图像。有人能告诉我哪里出错了。 This is the algorithm that I'm trying to implement. The alpha is the gapOpen variable in my code and beta is the gapExt variable. And A[i],B[j] simply is match or mismatch variable. That is the two characters being read at a given time (from String m and n), if they are the same then its a match otherwise it is a mismatch

以下是我的java代码。



public class AffineGapCost {

    static int gapOpen = 5;
    static int gapExt = 2;

    public static int deletionMatrix(int[][] D, int[][] S ,int i, int j)
    {
        int res = 0;
        if(i == 1 && j>=0)
        {
            res = S[i-1][j] - (gapOpen + gapExt);
            D[i][j] = res;
        }

        if(i > 1 && j >= 0)
        {
            int res_S = S[i-1][j] - (gapOpen + gapExt);
            int res_D = D[i-1][j] - gapOpen;

            res = Math.max(res_S, res_D);
            D[i][j] = res;
        }

        return res;
    }

    public static int insertionMatrix(int[][] I, int[][] S, int i, int j)
    {
        int res = 0;
        if(i >= 0 && j==1)
        {
            res = S[i][j-1] - (gapOpen + gapExt);
            I[i][j] = res;
        }

        if(i >= 0 && j > 1)
        {
            int res_S = S[i][j-1] - (gapOpen + gapExt);
            int res_I = I[i][j-1] - gapOpen;

            res = Math.max(res_S,res_I);
            I[i][j] = res;
        }

        return res;
    }

    public static int[][] matrix(int[][] S, int[][] D, int[][] I,String m, String n,int match, int mismatch)
    {

        for(int i=0;i<=m.length();i++)
        {
            for(int j=0;j<=n.length();j++)
            {
                if(i == 0 && j == 0)
                {
                    S[i][j] = 0;
                }

                if(i == 0 && j > 0)
                {
                    S[i][j] = insertionMatrix(I, S, i, j);
                }

                if(i > 0 && j == 0)
                {
                    S[i][j] = deletionMatrix(D,S,i,j);
                }

                if(i >0 && j>0)
                {
                    int res_S = S[i-1][j-1] + (m.charAt(i-1) == n.charAt(j-1) ? match:mismatch);
                    int res_D = deletionMatrix(D,S,i,j);
                    int res_I = insertionMatrix(I,S,i,j);

                    S[i][j] = Math.max(Math.max(res_S,res_D),res_I);
                }
            }
        }

        return S;
    }
<pre><code>
public static void main(String[] args)
    {
        // Initializing two strings
        String m = "GAATTCAGTTA";
        String n = "GGATCGA";

        //String m = "ctaca";
        //String n = "cttca";

        //String m = "ACGGCT";
        //String n = "ACGT";

        int[][] S = new int[m.length() + 1][n.length() + 1];
        int[][] D = new int[m.length() + 1][n.length() + 1];
        int[][] I = new int[m.length() + 1][n.length() + 1];

        int match = 1;
        int mismatch = -3;

        S = matrix(S,D,I,m,n,match,mismatch);

        for(int i=0;i<=m.length();i++)
        {
            for(int j=0;j<=n.length();j++)
            {
                System.out.print(S[i][j] + "\t");
            }
            System.out.println();
        }

        traceBack(S, D, I, m, n,match,mismatch);




    }

    }

如果有人能告诉我哪里出错了，我真的很感激。过去两天我一直试图找到问题，但似乎无法弄清楚我做错了什么。

Answer 1

这里有一些潜在的问题：

在deletionMatrix中，Java说：

if(i == 1 && j>=0)

但是Python说：

if i==0 and j>=0:

在matrix中，Java说：

for(int i=0;i<=m.length();i++) {
    for(int j=0;j<=n.length();j++) {

但Python说：

for i in range(0,len(m)):
    for j in range(0,len(n)):

每个循环的缩短一个。 range(m,n)从m转到n-1，因此您可能需要：

for i in range(len(m) + 1):
    for j in range(len(n) + 1):

在main代码的嵌套打印循环中再次发生这种情况。

下面是对Python代码的返工，它产生与Java代码相同的输出（如果在Java代码中注释掉对未定义方法traceBack()的调用）：

import numpy as np

gapOpen = 5
gapExt = 2

def deletionMatrix(D, S, i, j):
    res = 0
    if i == 1 and j >= 0:
        res = S[i - 1][j] - (gapOpen + gapExt)
        D[i][j] = res

    if i > 1 and j >= 0:
        res_S = S[i - 1][j] - (gapOpen + gapExt)
        res_D = D[i - 1][j] - gapOpen

        res = max(res_S, res_D)
        D[i][j] = res

    return res

def insertionMatrix(I, S, i, j):
    res = 0

    if i >= 0 and j == 1:
        res = S[i][j-1] - (gapOpen + gapExt)
        I[i][j] = res

    if i >= 0 and j > 1:
        res_S = S[i][j-1] - (gapOpen + gapExt)
        res_I = I[i][j-1] - gapOpen

        res = max(res_S, res_I)
        I[i][j] = res

    return res

def matrix(S, D, I, m, n, match, mismatch):

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

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

            if i == 0 and j == 0:
                S[i][j] = 0

            if i == 0 and j > 0:
                S[i][j] = insertionMatrix(I, S, i, j)

            if i > 0 and j == 0:
                S[i][j] = deletionMatrix(D, S, i, j)

            if i > 0 and j > 0:
                res_S = S[i-1][j-1] + (match if m[i-1] == n[j-1] else mismatch)
                res_D = deletionMatrix(D, S, i, j)
                res_I = insertionMatrix(I, S, i, j)

                S[i][j] = max(res_S, res_D, res_I)

    return S

if __name__ == '__main__':

    # Initializing two strings
    m = "GAATTCAGTTA"
    n = "GGATCGA"

    mLen = len(m) + 1
    nLen = len(n) + 1

    S = np.zeros([mLen, nLen])
    D = np.zeros([mLen, nLen])
    I = np.zeros([mLen, nLen])

    match = 1
    mismatch = -3

    S = matrix(S, D, I, m, n, match, mismatch)

    for i in range(mLen):
        for j in range(nLen):
            print(S[i][j], end='\t')
        print()

仿射间隙序列对齐

1 个答案: