[Python] Iterare una sequenza

[Giuseppe Costanzi]

ciao giorgio,
temo che a breve dovro' studiarmelo....
grazie per la segnalazione

On Sun, Nov 26, 2017 at 11:34 AM, Giorgio Zoppi <giorgio.zoppi a gmail.com> wrote:
> Io userei el automa di aho corasick.
> In Bioinformatics si usa tanto.
> http://carshen.github.io/data-structures/algorithms/2014/04/07/aho-corasick-implementation-in-python.html
>
>
> 2017-11-26 10:20 GMT+01:00 Giuseppe Costanzi <giuseppecostanzi a gmail.com>:
>>
>> salve a tutti,
>>
>> ho una sequenza del tipo
>>
>> ACTGATCGATTACGTATAGTAGAATTCTATCATACATATATATCGATGCGTTCAT
>>
>> scorrendola devo trovare una sequenza target GAATTC
>>
>> ACTGATCGATTACGTATAGTA  "GAATTC"  TATCATACATATATATCGATGCGTTCAT
>>
>>
>> quindi dividere la sequenza da G, la prima lettera della sequenza target,
>> e calcolarmi la lunghezza dei due frammenti risultanti
>>
>> ACTGATCGATTACGTATAGTAG
>>
>> e di questa
>>
>> GAATTCTATCATACATATATATCGATGCGTTCAT
>>
>> io avrei fatto
>>
>> seq = "ACTGATCGATTACGTATAGTAGAATTCTATCATACATATATATCGATGCGTTCAT"
>> target = "GAATTCT"
>>
>> s = []
>>
>> for i,base in enumerate(seq):
>>
>>     s.append(base)
>>
>>     if len(s)==8:
>>
>>         s.pop(0)
>>
>>         if ''.join(s) == target:
>>              print i-5
>>              print len(seq)-(i-5)
>>              break
>>
>> funziona ma non e' che ne sia proprio convinto,
>> avete suggerimenti su come iterare la sequenza?
>>
>> saluti
>> beppe
>>
>>
>> p.s.
>>
>> e' un esercizio che ho trovato su p4b, python for biologist
>>
>>
>> "Let's start this exercise by solving the problem manually. If we look
>> through the
>> DNA sequence we can spot the EcoRI site at position 21. Here's the
>> sequence with
>> the base positions labelled above and the EcoRI motif in bold:
>>
>> in bold sarebbe questa GAATTCT
>>
>> 0123456789012345678901234567890123456789012345678901234
>> ACTGATCGATTACGTATAGTAGAATTCTATCATACATATATATCGATGCGTTCAT
>> Since the EcoRI enzyme cuts the DNA between the G and first A, we can
>> figure out
>> that the first fragment will run from position 0 to position 21, and the
>> second
>> fragment from position 22 to the last position, 54. Therefore the
>> lengths of the two
>> fragments are 22 and 33."
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