Delete several lines if the first line contain numbers > 200

Shell Programming and Scripting
1

I have a file of the following format:

$data1 size 1278
dataw
datat
datau
datai
$data2 size 456
datak
dataf
datat
datay
datal
$data3 size 154
datag
datas
datat
datar
datas
datap
$data4 size 234
datat
datat
datap
datai

so that certain line begin with $ and have the parameter "size". I want to delete the entire block of data of the lines for which "size" > 200.

In my case the output would be:

$data1 size 1278
dataw
datat
datau
datai
$data2 size 456
datak
dataf
datat
datay
datal
$data4 size 234
datat
datat
datap
datai

Any ideas? Thanks

2

Assuming you want the blocks with a size > 200:

awk '/^\$/{f=$NF>200?1:0}f' file
3

No, the size of the blocks does not match the parameter "size". What I want is to delete the block of data from one $ to the next, IF the parameter "size" is >200

4

But your desired output contains only blocks of > 200:

$data1 size 1278
dataw
datat
datau
datai
$data2 size 456
datak
dataf
datat
datay
datal
$data4 size 234
datat
datat
datap
datai
5

Sorry, misunderstanding.
I only want to keep the blocks for which the first line contain "size 'number' " where number >200.

But the number of rows for each block could also be less than 200

6

Have you tried the provided code?

awk '/^\$/{f=$NF>200?1:0}f' file
7

Hi,

sed -e '/size \(1\{,1\}[0-9]\{,2\}\|200\)$/,/size \(20[1-9]\|2[1-9][0-9]\|[3-9][0-9][0-9]\|[1-9][0-9]\{3,\}\)/{/size \(20[1-9]\|2[1-9][0-9]\|[3-9][0-9][0-9]\|[1-9][0-9]\{3,\}\)/!d;}'  file

Regards.

8

It works.

Thanks really a lot

9

I would take a simpler script

#! /bin/ksh
 
flag=0
while read line
do
   lin=`print $line | sed -n '/.data.*size/p`
   if [[ -n $lin ]]
   then
      num=`print $lin | sed 's/.data.*size //'`
      if [[ $num -gt 200 ]]
      then
         flag=1
      else
         flag=0
      fi
   fi
 
   if [[ $flag -eq 1 ]]
   then
      print $line
   fi
done < file1