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Awk Assistance

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# 8  
Old 11-14-2009
I see... I think.... A sequence includes all the lines between the headers?
# 9  
Old 11-14-2009
Yes,

I think he just needs a count of the sequences that do not contain the letter J. It seems a bit more simpler now.

So a script to read in each sequence and print how many sequences with the letter J.

The example is 1. Due to 1 J sequence.

THanks again for your time and assistance.

Abacus
# 10  
Old 11-14-2009
Something like this then?
Code:
awk '/^>/{if(s==1)t++;s=1;next} /J/{s=0} END{if(s==1)t++;print "J-less sequences: "t}' infile

# 11  
Old 11-15-2009
Hmm,
OSX Snow Leopard tested.


Code:
machine1:~ jasonralph$ awk '/^>/{if(s==1)t++;s=1;next} /J/{s=0} END{if(s==1)t++;print "J-less sequences: "t}' infile
J-less sequences: 
c-machine1:~ jasonralph$

Thans for your time as always.

Abacus
# 12  
Old 11-15-2009
you can know the number of sequences that have J on it on each description, then you can grep the zero or write if condition grep the "0" values counter.

Code:-

Code:
gawk '
/^>/{ t=$0 ; a[t]=0 ;next}
/J/{a[t]++  }
END{ for ( i in a ) { printf "J exist in %s sequence(s) of header %s\n",a[i],i } }
'  input_file



Code:
Output

J exist in 2 sequence(s) of header >description 1-blah blah blah
J exist in 1 sequence(s) of header >description 2-blah blah blah


SmilieSmilieSmilieSmilie
# 13  
Old 11-15-2009
[solved]
Thanks to ahmad.diab, scruitinizer and jsmithstl. The code examples that you have all provided have done the trick.

My friend and I are very satisfied with the level of professionalism found on this site by the users.

Abacus
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