sed Command to replace particular value.

Shell Programming and Scripting
1

Hi ,

My input file contain :

list = 3 14 15 10 9 11 12 18 19 20 21 22 23 24 25 26 6 1 2 3 4 5 7 8 16 17 27 28 30 29

Expected output :

list = 0 0 0 0 0 0 0 18 0 20 0 0 0 0 0 0 6 0 0 3 4 0 0 0 0 0 0 0 0 0

I want to keep the 8,10,16,17,22 value from the list and put 0 on rest of the positions.

command using :

set list=`3 14 15 10 9 11 12 18 19 20 21 22 23 24 25 26 6 1 2 3 4 5 7 8 16 17 27 28 30 29'
set new_list='0 0 0 0 0 0 0 18 0 20 0 0 0 0 0 0 6 0 0 3 4 0 0 0 0 0 0 0 0 0'
sed -i 's/'$list'/'$exp_list'/g'

but it is not working...

Thanks in advance....

2

Hi,
If you want to keep the 8,10,16,17,22 positions value, expect output will:

list = 0 0 0 0 0 0 0 18 0 20 0 0 0 0 0 26 6 0 0 0 0 5 0 0 0 0 0 0 0 0

Why do you do these with sed ?
It will most easy to do with awk.

Regards.

3

Note that you can't use a backquote in your assignment to list ; the opening quote type much either be a double-quote or a single-quote and must be the same type of quote as the ending quote.

And, you have to use the correct quoting in your sed command so the substitute command passed to it will be a single quoted string. And, you have to tell sed what file you want it to edit:

set list='3 14 15 10 9 11 12 18 19 20 21 22 23 24 25 26 6 1 2 3 4 5 7 8 16 17 27 28 30 29'
set new_list='0 0 0 0 0 0 0 18 0 20 0 0 0 0 0 0 6 0 0 3 4 0 0 0 0 0 0 0 0 0'
sed -i "s/$list/$exp_list/" pathname

Where pathname is the pathname of the file you want to edit.

But, note that this preserves the values of fields 8, 10, 17, 20, and 21; not 8, 10, 16, 17, and 22???

4

Just for the fun of it - can't be done with sed alone; but, with list as defined above, 8,10,16,17,22 in a file called POS, and a recent bash , try

echo $list | sed 's/[0-9]\+/0/g' | sed -f<(echo $list | cut -d" " -f$(< POS) | tr ' ' '\n' | paste - <(tr ',' '\n' < POS) | tac | sed 's/^/s\/0\//; s/\t/\//')
0 0 0 0 0 0 0 18 0 20 0 0 0 0 0 26 6 0 0 0 0 5 0 0 0 0 0 0 0 0
5 
awk '{for (i=3; i<=NF; i++) $i=(keep ~ ("," (i-2) ",")) ? $i : 0; print}' keep=,8,10,16,17,22, infile
6

input file contain :

list=1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

expected output :
keep values at 8,10,16,17,22 and put 0 at all other places in the same file.

list=0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 0 0 0 0

what command we can use for this changes?

previous post gives separate script command,but i want to change in the same input file.

7

Hi,
with perl:

perl -pe 'BEGIN{%idx=map { $_ => 1 } (8,10,16,17,22)};print;@x=split / /,(split /=/)[1];$i=0;s/([0-9]+)/$idx{$i++} == 1 ? "$x[$i-1]":"0"/ge' file

And if ok, you can just use '-i'.

Regards.

8

awk suggestion:

awk -v fields=8,10,16,17,22 -F' *= *|[ \t]*' '
  BEGIN {
    split(fields,F,/,/)
    for(i in F)
      C[F]
  }

  {
    for(i=1; i<NF; i++)
      if(!(i in C))
        $(i+1)=0
    sub(/ /,"=")
    print
  }
' file
9

Another way in (gnu) sed:

echo "8,10,16,17,22" | sed 's/,/;/g;s/[0-9]\+/s\/[0-9]\\+\/A\&A\/&/g' | sed -f - -e 's/^\|$/ /g;:B;s/\([^A]\)[0-9]\+\([^A]\)/\1A0A\2/;tB;s/A//g;s/^ \| $//g' file

Regards.

10

thank you all..