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Delete rows with conditions

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# 1  
Old 12-03-2014
Delete rows with conditions
Hi everyone,

I will appreciate a lot if anyone can help me about a simple issue.
I have a data file, and I need to remove some rows with a given condition.
So here is a part of my data file:

Code:
5,14,1,3,3,0,0,-0.29977188269E+01
5,16,1,4,4,0,0,0.30394279900E+02
5,13,1,4,4,0,0,0.15108184243E+02
5,-14,1,4,4,0,0,0.38275921713E+01
6,-2,-1,0,0,0,501,0.00000000000E+00
6,2,-1,0,0,501,0,0.00000000000E+00
6,-15,2,1,2,0,0,0.43485959710E+02
6,15,2,1,2,0,0,-0.43485959710E+02
7,4,-1,0,0,501,0,0.00000000000E+00
7,-4,-1,0,0,0,501,0.00000000000E+00
7,23,2,1,2,0,0,0.14210854715E-13
7,-15,2,3,3,0,0,0.57831183575E+02

The first column is the data set ID, I need to remove data sets in which any of its rows contains 23 in the second column. According to given data above, after the operation it should look like:

Code:
5,14,1,3,3,0,0,-0.29977188269E+01
5,16,1,4,4,0,0,0.30394279900E+02
5,13,1,4,4,0,0,0.15108184243E+02
5,-14,1,4,4,0,0,0.38275921713E+01
6,-2,-1,0,0,0,501,0.00000000000E+00
6,2,-1,0,0,501,0,0.00000000000E+00
6,-15,2,1,2,0,0,0.43485959710E+02
6,15,2,1,2,0,0,-0.43485959710E+02

the number 23 in the second column appears only in data set 7, so the entire data set must be removed from the file.

thanks in advance for any help.
# 2  
Old 12-03-2014
I didn't consider too many possible scenarios beyond the sample you gave above; try
Code:
awk     '$1 != L        {if (!D) for (i=1; i<=CNT; i++) print O[i]; delete O; CNT=D=0}
                        {O[++CNT]=$0; L=$1}
         $2==23         {D=1}
         END            {if (!D) for (i=1; i<=CNT; i++) print O[i]}
        ' FS=, file

This User Gave Thanks to RudiC For This Post:
hayreter
# 3  
Old 12-03-2014
or alternatively:
Code:
awk -F, 'FNR==NR{if ($2==23) no[$1]; next} !($1 in no)' myfile myfile

These 2 Users Gave Thanks to vgersh99 For This Post:
hayreter junior-helper
# 4  
Old 12-03-2014
Hi vgersh99,

your code works quite fine, can you briefly explain your code if you don't mind please?

thanks again.
# 5  
Old 12-03-2014
And the inevitable:
Code:
awk '
  $1!=p {
    if(NR>1 && m) printf "%s", s
    s=x; p=$1; m=1
  }
  { 
    if($2==23) m=0
    s=s $0 ORS
  } 
  END {
    if(m)printf "%s", s
  }
' FS=, OFS=, file

Last edited by Scrutinizer; 12-03-2014 at 09:04 PM..
This User Gave Thanks to Scrutinizer For This Post:
hayreter
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