问题
Summary:
I'm beginning with some details about alignment algorithms, and at the end, I ask my question. If you know about alignment algorithm pass the beginning.
Consider we have two strings like:
ACCGAATCGA
ACCGGTATTAAC
There is some algorithms like: Smith-Waterman Or Needleman–Wunsch, that align this two sequence and create a matrix. take a look at the result in the following section:
Smith-Waterman Matrix
§ § A C C G A A T C G A
§ 0 0 0 0 0 0 0 0 0 0 0
A 0 4 0 0 0 4 4 0 0 0 4
C 0 0 13 9 4 0 4 3 9 4 0
C 0 0 9 22 17 12 7 3 12 7 4
G 0 0 4 17 28 23 18 13 8 18 13
G 0 0 0 12 23 28 23 18 13 14 18
T 0 0 0 7 18 23 28 28 23 18 14
A 0 4 0 2 13 22 27 28 28 23 22
T 0 0 3 0 8 17 22 32 27 26 23
T 0 0 0 2 3 12 17 27 31 26 26
A 0 4 0 0 2 7 16 22 27 31 30
A 0 4 4 0 0 6 11 17 22 27 35
C 0 0 13 13 8 3 6 12 26 22 30
Optimal Alignments
A C C G A - A T C G A
A C C G G A A T T A A
Question:
My question is simple, but maybe the answer is not easy as it looks. I want to use a group of character as a single one like: [A0][C0][A1][B1]. But in these algorithms, we have to use individual characters. How can we achieve that?
P.S. Consider we have this sequence: #read #write #add #write. Then I convert this to something like that: #read to A .... #write to B.... #add to C. Then my sequence become to: ABCB. But I have a lot of different words that start with #. And the ASCII table is not enough to convert all of them. Then I need more characters. the only way is to use something like [A0] ... [Z9] for each word. OR to use numbers.
P.S: some sample code for Smith-Waterman is exist in this link
P.S: there is another post that want something like that, but what I want is different. In this question, we have a group of character that begins with a [ and ends with ]. And no need to use semantic like ee is equal to i.
回答1:
Imagine we have a log file with alphabetic sequences. Like something you said, I converted sequences to A0A1... . For example, if there was a sequence like #read #write #add #write, it converted to A0A1A2A1. Every time, I read two character and compare them but keep score matrix like before. Here is my code in C# for smith-waterman string alignment.
Notice that Cell is a user defined class.
private void alignment()
{
string strSeq1;
string strSeq2;
string strTemp1;
string strTemp2;
scoreMatrix = new int[Log.Length, Log.Length];
// Lists That Holds Alignments
List<char> SeqAlign1 = new List<char>();
List<char> SeqAlign2 = new List<char>();
for (int i = 0; i<Log.Length; i++ )
{
for (int j=i+1 ; j<Log.Length; j++)
{
strSeq1 = "--" + logFile.Sequence(i);
strSeq2 = "--" + logFile.Sequence(j);
//prepare Matrix for Computing optimal alignment
Cell[,] Matrix = DynamicProgramming.Intialization_Step(strSeq1, strSeq2, intSim, intNonsim, intGap);
// Trace back matrix from end cell that contains max score
DynamicProgramming.Traceback_Step(Matrix, strSeq1, strSeq2, SeqAlign1, SeqAlign2);
this.scoreMatrix[i, j] = DynamicProgramming.intMaxScore;
strTemp1 = Reverse(string.Join("", SeqAlign1));
strTemp2 = Reverse(string.Join("", SeqAlign2));
}
}
}
class DynamicProgramming
{
public static Cell[,] Intialization_Step(string Seq1, string Seq2,int Sim,int NonSimilar,int Gap)
{
int M = Seq1.Length / 2 ;//Length+1//-AAA //Changed: /2
int N = Seq2.Length / 2 ;//Length+1//-AAA
Cell[,] Matrix = new Cell[N, M];
//Intialize the first Row With Gap Penality Equal To Zero
for (int i = 0; i < Matrix.GetLength(1); i++)
{
Matrix[0, i] = new Cell(0, i, 0);
}
//Intialize the first Column With Gap Penality Equal To Zero
for (int i = 0; i < Matrix.GetLength(0); i++)
{
Matrix[i, 0] = new Cell(i, 0, 0);
}
// Fill Matrix with each cell has a value result from method Get_Max
for (int j = 1; j < Matrix.GetLength(0); j++)
{
for (int i = 1; i < Matrix.GetLength(1); i++)
{
Matrix[j, i] = Get_Max(i, j, Seq1, Seq2, Matrix,Sim,NonSimilar,Gap);
}
}
return Matrix;
}
public static Cell Get_Max(int i, int j, string Seq1, string Seq2, Cell[,] Matrix,int Similar,int NonSimilar,int GapPenality)
{
Cell Temp = new Cell();
int intDiagonal_score;
int intUp_Score;
int intLeft_Score;
int Gap = GapPenality;
//string temp1, temp2;
//temp1 = Seq1[i*2].ToString() + Seq1[i*2 + 1]; temp2 = Seq2[j*2] + Seq2[j*2 + 1].ToString();
if ((Seq1[i * 2] + Seq1[i * 2 + 1]) == (Seq2[j * 2] + Seq2[j * 2 + 1])) //Changed: +
{
intDiagonal_score = Matrix[j - 1, i - 1].CellScore + Similar;
}
else
{
intDiagonal_score = Matrix[j - 1, i - 1].CellScore + NonSimilar;
}
//Calculate gap score
intUp_Score = Matrix[j - 1, i].CellScore + GapPenality;
intLeft_Score = Matrix[j, i - 1].CellScore + GapPenality;
if (intDiagonal_score<=0 && intUp_Score<=0 && intLeft_Score <= 0)
{
return Temp = new Cell(j, i, 0);
}
if (intDiagonal_score >= intUp_Score)
{
if (intDiagonal_score>= intLeft_Score)
{
Temp = new Cell(j, i, intDiagonal_score, Matrix[j - 1, i - 1], Cell.PrevcellType.Diagonal);
}
else
{
Temp = new Cell(j, i, intDiagonal_score, Matrix[j , i - 1], Cell.PrevcellType.Left);
}
}
else
{
if (intUp_Score >= intLeft_Score)
{
Temp = new Cell(j, i, intDiagonal_score, Matrix[j - 1, i], Cell.PrevcellType.Above);
}
else
{
Temp = new Cell(j, i, intDiagonal_score, Matrix[j , i - 1], Cell.PrevcellType.Left);
}
}
if (MaxScore.CellScore <= Temp.CellScore)
{
MaxScore = Temp;
}
return Temp;
}
public static void Traceback_Step(Cell[,] Matrix, string Sq1, string Sq2, List<char> Seq1, List<char> Seq2)
{
intMaxScore = MaxScore.CellScore;
while (MaxScore.CellPointer != null)
{
if (MaxScore.Type == Cell.PrevcellType.Diagonal)
{
Seq1.Add(Sq1[MaxScore.CellColumn * 2 + 1]); //Changed: All of the following lines with *2 and +1
Seq1.Add(Sq1[MaxScore.CellColumn * 2]);
Seq2.Add(Sq2[MaxScore.CellRow * 2 + 1]);
Seq2.Add(Sq2[MaxScore.CellRow * 2]);
}
if (MaxScore.Type == Cell.PrevcellType.Left)
{
Seq1.Add(Sq1[MaxScore.CellColumn * 2 + 1]);
Seq1.Add(Sq1[MaxScore.CellColumn * 2]);
Seq2.Add('-');
}
if (MaxScore.Type == Cell.PrevcellType.Above)
{
Seq1.Add('-');
Seq2.Add(Sq2[MaxScore.CellRow * 2 + 1]);
Seq2.Add(Sq2[MaxScore.CellRow * 2]);
}
MaxScore = MaxScore.CellPointer;
}
}
}
回答2:
I adapted this Python implementation (GPL version 3 licensed) of both the Smith-Waterman and the Needleman-Wunsch algorithms to support sequences with multiple character groups:
#This software is a free software. Thus, it is licensed under GNU General Public License.
#Python implementation to Smith-Waterman Algorithm for Homework 1 of Bioinformatics class.
#Forrest Bao, Sept. 26 <http://fsbao.net> <forrest.bao aT gmail.com>
# zeros() was origianlly from NumPy.
# This version is implemented by alevchuk 2011-04-10
def zeros(shape):
retval = []
for x in range(shape[0]):
retval.append([])
for y in range(shape[1]):
retval[-1].append(0)
return retval
match_award = 10
mismatch_penalty = -5
gap_penalty = -5 # both for opening and extanding
gap = '----' # should be as long as your group of characters
space = ' ' # should be as long as your group of characters
def match_score(alpha, beta):
if alpha == beta:
return match_award
elif alpha == gap or beta == gap:
return gap_penalty
else:
return mismatch_penalty
def finalize(align1, align2):
align1 = align1[::-1] #reverse sequence 1
align2 = align2[::-1] #reverse sequence 2
i,j = 0,0
#calcuate identity, score and aligned sequeces
symbol = []
found = 0
score = 0
identity = 0
for i in range(0,len(align1)):
# if two AAs are the same, then output the letter
if align1[i] == align2[i]:
symbol.append(align1[i])
identity = identity + 1
score += match_score(align1[i], align2[i])
# if they are not identical and none of them is gap
elif align1[i] != align2[i] and align1[i] != gap and align2[i] != gap:
score += match_score(align1[i], align2[i])
symbol.append(space)
found = 0
#if one of them is a gap, output a space
elif align1[i] == gap or align2[i] == gap:
symbol.append(space)
score += gap_penalty
identity = float(identity) / len(align1) * 100
print 'Identity =', "%3.3f" % identity, 'percent'
print 'Score =', score
print ''.join(align1)
# print ''.join(symbol)
print ''.join(align2)
def needle(seq1, seq2):
m, n = len(seq1), len(seq2) # length of two sequences
# Generate DP table and traceback path pointer matrix
score = zeros((m+1, n+1)) # the DP table
# Calculate DP table
for i in range(0, m + 1):
score[i][0] = gap_penalty * i
for j in range(0, n + 1):
score[0][j] = gap_penalty * j
for i in range(1, m + 1):
for j in range(1, n + 1):
match = score[i - 1][j - 1] + match_score(seq1[i-1], seq2[j-1])
delete = score[i - 1][j] + gap_penalty
insert = score[i][j - 1] + gap_penalty
score[i][j] = max(match, delete, insert)
# Traceback and compute the alignment
align1, align2 = [], []
i,j = m,n # start from the bottom right cell
while i > 0 and j > 0: # end toching the top or the left edge
score_current = score[i][j]
score_diagonal = score[i-1][j-1]
score_up = score[i][j-1]
score_left = score[i-1][j]
if score_current == score_diagonal + match_score(seq1[i-1], seq2[j-1]):
align1.append(seq1[i-1])
align2.append(seq2[j-1])
i -= 1
j -= 1
elif score_current == score_left + gap_penalty:
align1.append(seq1[i-1])
align2.append(gap)
i -= 1
elif score_current == score_up + gap_penalty:
align1.append(gap)
align2.append(seq2[j-1])
j -= 1
# Finish tracing up to the top left cell
while i > 0:
align1.append(seq1[i-1])
align2.append(gap)
i -= 1
while j > 0:
align1.append(gap)
align2.append(seq2[j-1])
j -= 1
finalize(align1, align2)
def water(seq1, seq2):
m, n = len(seq1), len(seq2) # length of two sequences
# Generate DP table and traceback path pointer matrix
score = zeros((m+1, n+1)) # the DP table
pointer = zeros((m+1, n+1)) # to store the traceback path
max_score = 0 # initial maximum score in DP table
# Calculate DP table and mark pointers
for i in range(1, m + 1):
for j in range(1, n + 1):
score_diagonal = score[i-1][j-1] + match_score(seq1[i-1], seq2[j-1])
score_up = score[i][j-1] + gap_penalty
score_left = score[i-1][j] + gap_penalty
score[i][j] = max(0,score_left, score_up, score_diagonal)
if score[i][j] == 0:
pointer[i][j] = 0 # 0 means end of the path
if score[i][j] == score_left:
pointer[i][j] = 1 # 1 means trace up
if score[i][j] == score_up:
pointer[i][j] = 2 # 2 means trace left
if score[i][j] == score_diagonal:
pointer[i][j] = 3 # 3 means trace diagonal
if score[i][j] >= max_score:
max_i = i
max_j = j
max_score = score[i][j];
align1, align2 = [], [] # initial sequences
i,j = max_i,max_j # indices of path starting point
#traceback, follow pointers
while pointer[i][j] != 0:
if pointer[i][j] == 3:
align1.append(seq1[i-1])
align2.append(seq2[j-1])
i -= 1
j -= 1
elif pointer[i][j] == 2:
align1.append(gap)
align2.append(seq2[j-1])
j -= 1
elif pointer[i][j] == 1:
align1.append(seq1[i-1])
align2.append(gap)
i -= 1
finalize(align1, align2)
If we run this with the following input:
seq1 = ['[A0]', '[C0]', '[A1]', '[B1]']
seq2 = ['[A0]', '[A1]', '[B1]', '[C1]']
print "Needleman-Wunsch"
needle(seq1, seq2)
print
print "Smith-Waterman"
water(seq1, seq2)
We get this output:
Needleman-Wunsch
Identity = 60.000 percent
Score = 20
[A0][C0][A1][B1]----
[A0]----[A1][B1][C1]
Smith-Waterman
Identity = 75.000 percent
Score = 25
[A0][C0][A1][B1]
[A0]----[A1][B1]
For the specific changes I made, see: this GitHub repository.
来源：https://stackoverflow.com/questions/36596389/sequence-alignment-algorithm-with-a-group-of-characters-instead-of-one-character