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Split a file based on number sum at the second column and the third column.

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Top Forums UNIX for Beginners Questions & Answers Split a file based on number sum at the second column and the third column.
# 8  
Old 03-02-2020
Try
Code:
awk '
#NR == 1          ||
$3-ST>2999      {FN = "file" ++FCNT
                 ST = $2
                }
                {print  >  FN
                }
 ' file


This assumes the values in $3 start at values higher than 2999; if not so, remove the # before the NR == 1 line
This User Gave Thanks to RudiC For This Post:
nezabudka
# 9  
Old 03-02-2020
Quote:
Originally Posted by nezabudka
Hi
Maybe just like that?
Code:
awk '
/^\S+\s+12109/,/^(\S+\s+){2}14678\s/ {print > "file1"}
/^\S+\s+15573/,/^(\S+\s+){2}15612\s/ {print > "file2"}
/^\S+\s+20498/,/^(\S+\s+){2}21668\s/ {print > "file3"}
' file


Thank you so much. This is what I need.


Quote:
Originally Posted by nezabudka
maybe so?
Code:
#!/bin/bash

step=2999
declare -i start=12109 end=start+step count=1
stop=$(awk '{if($3>max) max=$3} END {print max}' file)

while [ $end -le $stop ]; do
        awk -vA=$start -vZ=$end -vf="file$count" '
                $2>=A && $3<=Z {print > f}
        ' file
        start+=step
        end+=step
        count+=1
done

# 10  
Old 03-02-2020
Sorry about my delay in gettng back think timezone differences are involved here. looks like we have some nice solutions coming together in this thread now.

This is the pseudo code I had in mind when I first read your requirements:
Code:
set startnum=0
set fileext = 1
loop:
    read line from input
    if column#3 - startnum > 2999 then
        startnum = column#2
        fileext = fileext + 1
    endif
    append line to filename("file" + fileext)
end loop

And the awk coded solution:
Code:
awk '
BEGIN { start=0 ; filenum=1 }
!start { start=$2 }
($3 - start) > 2999 {
   close("file" filenum)
   filenum++
   start=$2
}
{ print > "file" filenum }' infile

This is very similar to RudiC's proposal. The main difference being in the close statement. awk has a limited number of output buffers and using close will become necessary when dealing with a larger input files which can generate too many output files.

And in the spirit of this site this reduced solution could be derived from above:

Code:
awk '
!start || ($3 - start) > 2999 {
   close("file" filenum++)
   start=$2
}
{ print > "file" filenum }' infile

Last edited by Chubler_XL; 03-02-2020 at 04:09 PM..
This User Gave Thanks to Chubler_XL For This Post:
nezabudka
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