[BioPython] Informative content problem, SOLVED!!!

[Sebastian Bassi]

Hi,

Now I know what's going on. The formulae used by biopython is ONLY for 
aligments (since it uses information from every sequence on the aligment).
My formula is LCC (local content complexity), so I implemented here:

Before submiting my code, I know it sucks, so it would be nice to have 
it as a module, like lcc(STRING, STARTPOSITION, ENDPOSITION)

Now, my code:


from Bio import Fasta
import string
import math


parser=Fasta.RecordParser()
entrada=open("C:\\bioinfo-adv\\blast\\data\\vector.nn","r")
cur_record=1
iterator=Fasta.Iterator(entrada,parser)

while cur_record:
     cur_record=iterator.next()
     if cur_record is None:
         break
     tamseq=len(cur_record.sequence)
     print tamseq

     for ini in range(tamseq-18):
         fin=ini+18
         primer=cur_record.sequence[ini:fin]

         if string.count(primer,'A')==0:
             term_a=0
         else:
 
term_a=(string.count(primer,'A')/float(len(primer)))*((math.log(string.count(primer,'A')/float(len(primer))))/math.log(2))

         if string.count(primer,'C')==0:
             term_c=0
         else:
 
term_c=(string.count(primer,'C')/float(len(primer)))*((math.log(string.count(primer,'C')/float(len(primer))))/math.log(2))

         if string.count(primer,'T')==0:
             term_t=0
         else:
 
term_t=(string.count(primer,'T')/float(len(primer)))*((math.log(string.count(primer,'T')/float(len(primer))))/math.log(2))

         if string.count(primer,'G')==0:
             term_g=0
         else:
 
term_g=(string.count(primer,'G')/float(len(primer)))*((math.log(string.count(primer,'G')/float(len(primer))))/math.log(2))

         lcc=-(term_a+term_c+term_t+term_g)
         print lcc
     print 'Cambio crom'
     print ''
     print ''

entrada.close()


-- 
Best regards,

//=\ Sebastian Bassi - Diplomado en Ciencia y Tecnologia, UNQ   //=\
\=// IT Manager Advanta Seeds - Balcarce Research Center -      \=//
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