clarification

[Michael Bentley]

On Aug 16, 2007, at 2:42 AM, Beema shafreen wrote:

> hi every body,
> i have compared two files:
> code:
>
> fh = open('HPRD_MAIN_20.txt','r')
> for line in fh.readlines():
>         data = line.strip().split('#')
>         fh1 = open('NOMENCLATURE_MAIN_20.txt','r')
>         for line1 in fh1.readlines():
>                 data1 = line1.strip().split('#')
>                 if  data1[0] == data[0]:
>                         result = data[0] +'#'+data[3]+'|'+ data[4] 
> +'|'+data[9]+'|'+ data1[3]
>                         print result
> the result was as given below:
>
>
> 00017#ACTG1|actin, gamma 1|Actin gamma 1|ACTG
> 00017#ACTG1|actin, gamma 1|Actin gamma 1|Actin gamma
> 00017#ACTG1|actin, gamma 1|Actin gamma 1|Cytoskeletal gamma actin
>
>
> but i need the result to be like this :
>
>
> 00017#ACTG1|actin, gamma 1|Actin gamma 1|ACTG,Actin  
> gamma,Cytoskeletal gamma, actin
>
>
> with out redundancy and the name in the same line separated by  
> commas..
> please suggest what should i do for this to get the result like this.

# untested
fh = open('HPRD_MAIN_20.txt','r')
for line in fh.readlines():
         data = line.strip().split('#')
         hprd = '%s#%s|%s|%s|' % (data[0], data[3], data[4], data[9])
         nomenclature = []
         fh1 = open('NOMENCLATURE_MAIN_20.txt','r')
         for line1 in fh1.readlines():
                 data1 = line1.strip().split('#')
                 if  data1[0] == data[0]:
                         nomenclature.append(data1[3])
         print '%s%s' % (hprd, ','.join(nomenclature))

hth,
Michael

---
"I would rather use Java than Perl. And I'd rather be eaten by a  
crocodile than use Java." — Trouser


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