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sed Command to replace particular value.

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# 1  
Old 04-25-2017
sed Command to replace particular value.
Hi ,

My input file contain :
Code:
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 :
Code:
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 :
Code:
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  
Old 04-25-2017
Hi,
If you want to keep the 8,10,16,17,22 positions value, expect output will:
Code:
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  
Old 04-25-2017
Quote:
Originally Posted by Preeti Chandra
Hi ,

My input file contain :
Code:
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 :
Code:
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 :
Code:
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....
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:
Code:
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  
Old 04-26-2017
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

Code:
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

Last edited by RudiC; 04-26-2017 at 07:07 PM..
# 5  
Old 04-26-2017
Code:
awk '{for (i=3; i<=NF; i++) $i=(keep ~ ("," (i-2) ",")) ? $i : 0; print}' keep=,8,10,16,17,22, infile

# 6  
Old 04-27-2017
input file contain :
Code:
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.
Code:
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  
Old 04-29-2017
Hi,
with perl:
Code:
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.
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