Permalink
Cannot retrieve contributors at this time
Name already in use
A tag already exists with the provided branch name. Many Git commands accept both tag and branch names, so creating this branch may cause unexpected behavior. Are you sure you want to create this branch?
Genomics/lcs.py
Go to fileThis commit does not belong to any branch on this repository, and may belong to a fork outside of the repository.
99 lines (82 sloc)
2.52 KB
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
from fasta import readFASTA | |
import numpy as np | |
import string | |
import operator as op | |
# The key to this algorithm is that the LCS of A[0..n] and B[0..m] | |
# can be computed from the LCS of A[0..(n-1)] and B[0..(m-1)] | |
# if A[n]==B[m] then the LCS of A[0..n] and B[0..m] is the LCS | |
# of A[0..(n-1)] and B[0..(m-1)] with A[n] (B[m]) appended -- then length | |
# goes up by one. If A[n]!=B[m] then the LCS of the two is the longer | |
# the lcs of A[0..(n-1)],B[m] and A[0..n] and B[0..(m-1)]. | |
# note that this algorithm is very close to Needleman-Wunsch and Smith-Waterman, | |
# just the scoring part is different. | |
s=readFASTA("rosalind_lcsq.txt") | |
A=s.values()[0] | |
B=s.values()[1] | |
#A='AACCTTGG' | |
#B='ACACTGTGA' | |
nA=len(A) | |
nB = len(B) | |
dp=np.zeros((nB,nA)) | |
pointerx=np.zeros((nB,nA),dtype=int) | |
pointery=np.zeros((nB,nA),dtype=int) | |
savedp=0 | |
savedi=0 | |
savedj=0 | |
if A[0]==B[0]: | |
dp[0][0]=1 | |
for i in range(1,nA): | |
pointerx[0][i]=-1 | |
if B[0]==A[i]: | |
dp[0][i]=1 | |
savedi=i | |
savedp=1 | |
pointerx[0][i]=-1 | |
pointery[0][i]=-1 | |
else: | |
dp[0][i]=dp[0][i-1] | |
pointerx[0][i]=-1 | |
for j in range(1,nB): | |
pointery[j][0]=-1 | |
if B[j]==A[0]: | |
dp[j][0]=1 | |
savedj=j | |
savedp=1 | |
pointerx[j][0]=-1 | |
pointery[j][0]=-1 | |
else: | |
dp[j][0]=dp[j-1][0] | |
pointery[j][0]=-1 | |
for i in range(1,nA): | |
for j in range(1,nB): | |
if A[i]==B[j]: | |
dp[j][i]=dp[j-1][i-1]+1 | |
(dx,dy)=(-1,-1) | |
else: | |
t=[(dp[j-1][i],(-1,0)),(dp[j][i-1],(0,-1))] | |
(t,(dx,dy))=max(t,key=lambda x: op.getitem(x,0)) | |
dp[j][i]=dp[j+dx][i+dy] | |
pointerx[j][i],pointery[j][i]=dx,dy | |
if dp[j][i]>savedp: | |
savedp=dp[j][i] | |
savedi,savedj=i,j | |
answer=[] | |
i,j = savedi,savedj | |
while True: | |
## print i,j,pointerx[j][i],pointery[j][i],dp[j][i] | |
if pointerx[j][i]==pointery[j][i] and pointerx[j][i]==-1: | |
answer.append(A[i]) | |
i,j=i+pointery[j][i],j+pointerx[j][i] | |
if i<0 or j<0: | |
break | |
print ''.join(reversed(answer)) | |
##### This version came from the Rosalind Web Site | |
##### it preserves only the last row of the table, and the | |
##### current row; but it has to keep the corresponding longest subsequences | |
#S, T = open('rosalind_lcsq.txt').read().splitlines() | |
#cur = [''] * (len(T) + 1) #dummy entries as per wiki | |
#for s in A: | |
# last, cur = cur, [''] | |
# for i, t in enumerate(B): | |
# cur.append(last[i] + s if s==t else max(last[i+1], cur[-1], key=len)) | |
#print cur[-1] | |