[Python] Iterare una sequenza

[Christian Barra]

In pseudo code

seq = "ACTGATCGATTACGTATAGTAGAATTCTATCATACATATATATCGATGCGTTCAT"
    ...:
    ...: def finder(seq, target):
    ...:     len_target = len(target)
    ...:     i = 0
    ...:     while i+len_target < len(seq):
    ...:         yield seq[i:i+len_target], i
    ...:         i += 1
		
Il check per == lo puoi inserire all’interno del finder o fuori.		

——
Christian Barra
Python Freelancer // Consultant // Trainer
Board member of the EuroPython Society
www.chrisbarra.xyz

> On 26 Nov 2017, at 10:42, Giuseppe Costanzi <giuseppecostanzi a gmail.com> wrote:
> 
> ciao carlo,
> 
> si, interessante ma vorrei iterare la sequenza mano a mano,
> 
> stavo pensando a qualcosa tipo
> 
> next(iterator, default)
> 
> grazie comunque
> 
> 
> 
> On Sun, Nov 26, 2017 at 10:30 AM, Carlo Miron <miron a python.it> wrote:
>> On Sun, Nov 26, 2017 at 10:20 AM, Giuseppe Costanzi
>> <giuseppecostanzi a gmail.com> wrote:
>> 
>>> 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
>> 
>> qualcosa tipo
>> 
>>>>> seq = "ACTGATCGATTACGTATAGTAGAATTCTATCATACATATATATCGATGCGTTCAT"
>>>>> target = "GAATTCT"
>>>>> first, second = seq.split(target, 1)
>>>>> first += target[1]
>>>>> second = target[1:] + second
>>>>> len(first), len(second)
>> (22, 33)
>> 
>> ?
>> ㎝
>> 
>> --
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