python 中的 Smith-Waterman 实施

Question

我想在 python 中用基本函数编写 Smith-Waterman 算法的第一部分。

我找到了这个 example，但它没有给我想要的东西。

def zeros(X: int, Y: int):
    #        ^       ^  incorrect type annotations. should be str
    lenX = len(X) + 1
    lenY = len(Y) + 1
    matrix = []
    for i in range(lenX):
        matrix.append([0] * lenY)
    # A more "pythonic" way of expressing the above would be:
    # matrix = [[0] * len(Y) + 1 for _ in range(len(x) + 1)]
    
    def score(X, Y):
        #     ^  ^ shadowing variables from outer scope. this is not a bug per se but it's considered bad practice
        if X[n] == Y[m]: return 4
        #    ^       ^  variables not defined in scope
        if X[n] == '-' or Y[m] == '-': return -4
        #    ^              ^  variables not defined in scope
        else: return -2

    def SmithWaterman(X, Y, score): # this function is never called
        #                   ^ unnecessary function passed as parameter. function is defined in scope
        for n in range(1, len(X) + 1):
            for m in range(1, len(Y) + 1):
                align = matrix[n-1, m-1] + (score(X[n-1], Y[m-1]))
                #                 ^ invalid list lookup. should be: matrix[n-1][m-1]
                indelX = matrix[n-1, m] + (score(X[n-1], Y[m]))
                #                                          ^ out of bounds error when m == len(Y)
                indelY = matrix[n, m-1] + (score(X[n], Y[m-1]))
                #                                  ^ out of bounds error when n == len(X)
        matrix[n, m] = max(align, indelX, indelY, 0)
        # this should be nested in the inner for-loop. m, n, indelX, and indelY are not defined in scope here
    print(matrix)

zeros("ACGT", "ACGT")

在书上找到了这个算法，但是我无法正确实现。

input: sequences s and t, with |s| =n, |t| = m, score function, penality InDel

匹配 +1，不匹配 -2，InDel -1

M = matrix of size n+1 * m+1
M[i,j] = 0
i=j=0

请帮忙谢谢

Answer 1

图像算法实现：

M = []
for i in range(n):
    M.append([])
    for j in range(m):
        first = max(M[i - 1][j - 1] + score(s[i], t[j])
        second = M[i - 1][j] + penal
        third = M[i][j - 1] + penal

        M[i].append(first, second, third, 0))

但是您必须修复边缘情况（超出范围）并添加一些默认值。

Answer 2

你提供的代码的问题在那段代码的注释中有很好的描述。

假设你想要一个2分的线性gap-penalty，并且你只寻找第一阶段算法（因此不包括回溯过程），代码可以固定如下：

def score(x, y):
    return 4 if x == y else (
        -4 if '-' in (x, y) else -2
    )

def zeros(a, b):
    penalty = 2  # linear penalty (see Wikipedia)
    nextrow = [0] * (len(b) + 1)
    matrix = [nextrow]
    for valA in a:
        row, nextrow = nextrow, [0]
        for m, valB in enumerate(b):
            nextrow.append(max(
                row[m] + score(valA, valB),
                row[m+1] - penalty,
                nextrow[m] - penalty,
                0
            ))
        matrix.append(nextrow)
    return matrix

# Example run:
result = zeros("ACGT", "AC-GT")
print(result)

python 中的 Smith-Waterman 实施

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