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我需要一些帮助是这段代码。我知道应该递归的部分，或者至少我认为我知道但不确定如何实现它。我正在尝试从对齐矩阵中实现寻路程序，该程序将找到返回零值的多条路线。例如，如果您执行我的代码并插入 CGCA 作为第一个序列，将 CACGTAT 作为第二个序列，以及 1、0 和 -1 作为匹配、不匹配和差距分数。该程序给出的路径为 HDHHDD，对齐方式为

CACGTAT

CGC--A-。

然而，除了我不知道有多少之外，还有更多可能的路径和路线。我想要做的是让我的一段代码自行循环并找到其他路径和对齐方式，使用与第一次相同的代码，直到它用完可能的对齐方式。我在网上找到的最好的方法是递归，除了没有人能解释如何做到这一点。在这种情况下，应该有另外两条路径和对齐方式 HDDDHHD 和 CACGTAT 以及 C--GCA- 和。HDDDDHH、CACGTAT 和--CGCA-。我只是不知道如何编写代码来执行此任务。

# Implementation of Needleman and Wunsch Algorithm

my($seq1, $len1, $seq2, $len2, $data, @matrix, $i, $j, $x, $y, $val1, $val2);
my($val3, $pathrow, $pathcol, $seq1loc, $seq2loc, $gapscore, $matchscore, $mismatchscore);

#first obtain the data from the user. 
print "Please enter the first sequence for comaprsion\n";
$seq1=<STDIN>;
chomp $seq1;

print "Please enter the second sequence for comparsion\n";
$seq2=<STDIN>;
chomp $seq2;


# adding extra characters so sequences align with matrix
# saves some calculations later on
$seq1 = " " . $seq1;
$seq2 = " " . $seq2;
$len1 = length($seq1);
$len2 = length($seq2);

print "Enter the match score: ";
$matchscore=<STDIN>;
chomp $matchscore;

print "Enter the mismatch score: ";
$mismatchscore=<STDIN>;
chomp $mismatchscore;

print "Enter the gap score: ";
$gapscore=<STDIN>;
chomp $gapscore;

# declare a two dimensional array and initialize to spaces
# array must contain one extra row and one extra column
@matrix = ();  
for($i = 0; $i < $len1; $i++){  
   for($j = 0; $j < $len2; $j++){  
      $matrix[$i][$j] = ' ';  
   }  
}  

# initialize 1st row and 1st column of matrix  
$matrix[0][0] = 0;  
for ($i = 1; $i < $len1; $i ++){  
    $matrix[$i][0] = $matrix[$i-1][0] + $gapscore;  
}  
for ($i = 1; $i < $len2; $i ++){  
    $matrix[0][$i] = $matrix[0][$i-1] + $gapscore;  
}  


# STEP 1:
# Fill in rest of matrix using the following rules:
# determine three possible values for matrix[x][y]
# value 1 = add gap score to matrix[x][y-1]
# value 2 = add gap score to matrix[x-1][y]
# value 3 = add match score or mismatch score to 
#           matrix[x-1][y-1] depending on nucleotide 
#           match for position x of $seq1 and position y
#           of seq2
# place the largest of the three values in matrix[x][y]
#
# Best alignment score appears in matrix[$len1][$len2].

for($x = 1; $x < $len1; $x++){  
   for($y = 1; $y < $len2; $y++){  
 $val1 = $matrix[$x][$y-1] + $gapscore;  
 $val2 = $matrix[$x-1][$y] + $gapscore;  
 if (substr($seq1, $x, 1) eq substr($seq2, $y, 1)){  
           $val3 = $matrix[$x-1][$y-1] + $matchscore;  
 }  
 else{  
    $val3 = $matrix[$x-1][$y-1] + $mismatchscore;  
 }  
 if (($val1 >= $val2) && ($val1 >= $val3)){  
    $matrix[$x][$y] = $val1;  
 }  
 elsif (($val2 >= $val1) && ($val2 >= $val3)){  
    $matrix[$x][$y] = $val2;  
 }  
 else{  
    $matrix[$x][$y] = $val3;  
 }  
   }  
}  

# Display scoring matrix  
print "MATRIX:\n";   
for($x = 0; $x < $len1; $x++){  
   for($y = 0; $y < $len2; $y++){  
 print "$matrix[$x][$y] ";  
   }  
   print "\n";  
}  
print "\n";    

# STEP 2:  
# Begin at matrix[$len1][$len2] and find a path to   
# matrix[0][0].  
# Build string to hold path pattern by concatenating either   
# 'H' (current cell produced by cell horizontally to left),   
# 'D' (current cell produced by cell on diagonal),   
# 'V' (current cell produced by cell vertically above)
# ***This is were I need help I need this code to be recursive, so I can find more then one path***

$pathrow = $len1-1;  
$pathcol = $len2-1;  

while (($pathrow != 0) || ($pathcol != 0)){  
    if ($pathrow == 0){  
       # must be from cell to left  
       $path = $path . 'H';  
       $pathcol = $pathcol - 1;  
    }  
    elsif ($pathcol == 0){  
       # must be from cell above  
       $path = $path . 'V';  
       $pathrow = $pathrow - 1;  
    }  
    # could be from any direction  
    elsif  (($matrix[$pathrow][$pathcol-1] + $gapscore) == $matrix[$pathrow][$pathcol]){  
       # from left  
       $path = $path . 'H';  
       $pathcol = $pathcol - 1;  
    }  
    elsif (($matrix[$pathrow-1][$pathcol] + $gapscore) == $matrix[$pathrow][$pathcol]){  
       # from above  
       $path = $path . 'V';  
       $pathrow = $pathrow - 1;  
    }   
    else{  
       # must be from diagonal  
       $path = $path . 'D';  
       $pathrow = $pathrow - 1;  
       $pathcol = $pathcol - 1;  
    }  


}   



print "Path is $path\n";   

# STEP 3:
# Determine alignment pattern by reading path string 
# created in step 2.
# Create two string variables ($alignseq1 and $alignseq2) to hold 
# the sequences for alignment.
# Reading backwards from $seq1, $seq2 and path string, 
#   if string character is 'D', then 
#      concatenate to front of $alignseq1, last char in $seq1 
#      and to the front of $alignseq2, last char in $seq2
#   if string character is 'V', then 
#      concatenate to front of $alignseq1, last char in $seq1 
#      and to the front of $alignseq2, the gap char
#   if string character is 'H', then 
#      concatenate to front of $alignseq1 the gap char
#      and to the front of $alignseq2, last char in $seq2
# Continue process until path string has been traversed.
# Result appears in string $alignseq1 and $seq2
***#I need this code to be recursive as well.***

$seq1loc = $len1-1;  
$seq2loc = $len2-1;  
$pathloc = 0;  
print length($path);  
while ($pathloc < length($path)){  
   if (substr($path, $pathloc, 1) eq 'D'){  
      $alignseq1 = substr($seq1, $seq1loc, 1) . $alignseq1;  
      $alignseq2 = substr($seq2, $seq2loc, 1) . $alignseq2;  
      $seq1loc--;  
      $seq2loc--;  
   }  
   elsif (substr($path, $pathloc, 1) eq 'V'){  
      $alignseq1 = substr($seq1, $seq1loc, 1) . $alignseq1;  
      $alignseq2 = '-' . $alignseq2;  
      $seq1loc--;  
   }    
   else{  # must be an H  
      $alignseq1 = '-' . $alignseq1;  
      $alignseq2 = substr($seq2, $seq2loc, 1) . $alignseq2;  
      $seq2loc--;  
   }    
   $pathloc++;  
}  

print "\nAligned Sequences:\n";  
print "$alignseq2 \n";  
print "$alignseq1 \n";  

# statement may be needed to hold output screen  
print "Press any key to exit program";  
$x = <STDIN>;

如果有人想知道这是一种needleman-wunsch 算法。这里的任何帮助都会非常感激。

4

3 回答 3

7

我无法提供答案，因为我不完全了解您要做什么，但我可以提供一些一般性建议。

开始将您的代码组织成执行狭义任务的离散子例程。此外，实现中央算法的子程序不应面向从键盘接收输入并在屏幕上产生输出；相反，他们应该接收输入作为参数并返回他们的结果。如果需要用户输入或屏幕输出，这些任务应该在单独的子程序中，而不是与您的主要算法混合。

沿着该路径的第一步（也是部分）是获取整个程序，将其包含在子例程定义中，然后使用所需的参数调用子例程。子例程不应该打印它的关键结果，而是应该返回它们——具体来说，是对 的引用以及, ,@matrix的值。$path$alignseq1$alignseq2

sub NW_algo {
    my ($seq1, $seq2, $matchscore, $mismatchscore, $gapscore) = @_;

    # The rest of your code here, but with all print
    # statements and <STDIN> inputs commented out.

    return \@matrix, $path, $alignseq1, $alignseq2;
}

my(@return_values) = NW_algo('CGCA', 'CACGTAT', 1, 0, -1);

Print_matrix($return_values[0]);

sub Print_matrix {
    for my $m ( @{$_[0]} ){
        print join(' ', @$m), "\n";
    }
}

此时，您将拥有一个可由其他代码调用的算法，从而更容易测试和调试您的程序。例如，您可以定义各种输入数据集并NW_algo()在每组上运行。只有这样，才有可能考虑递归或其他技术。

于 2009-09-18T14:32:09.133 回答
5

由于 Needleman-Wunsch 是一种动态规划算法，所以大部分工作在您计算 DP 矩阵时已经完成。一旦你有了你的 DP 矩阵，你应该通过矩阵回溯以找到最佳对齐方式。这个问题有点像出租车的几何形状，只是你可以沿对角线移动。本质上，当您需要回溯矩阵时，不是在向上、向左或对角线之间进行选择，而是通过进行三个递归调用来完成这三个操作，并且每个向上、向左或对角线调用自己，直到你到达你的起点。每条递归所追踪的路径将绘制出每个对齐。

编辑：所以基本上你需要把第 2 步放在一个子过程中（它需要位置和到目前为止跟踪的路径），所以它可以一遍又一遍地调用自己。当然，在定义过程之后，您需要对其进行一次调用才能真正开始跟踪过程：

sub tracePaths {
    $x = shift;
    $y = shift;
    $pathSoFar = shift; # assuming you're storing your path as a string
    #
    # ... do some work ...
    #
    tracePaths($x - 1, $y, $pathSoFar . something);
    tracePaths($x, $y - 1, $pathSoFar . somethingelse);
    tracePaths($x - 1, $y - 1, $pathSoFar . somethingelselse);
    #
    #
    #
    if(reached the end) return $pathSoFar;
}
# launch the recursion
tracePaths(beginningx, beginningy, "");
于 2009-09-18T14:42:08.023 回答
4

这并没有具体说明您的问题，但您也许应该查看《高阶 Perl 》一书。它讨论了如何使用许多更高级别的技术（例如递归）。

于 2009-09-18T17:59:14.650 回答