remove single elements in a text file in bash - regex

Question

Basically what I have is a text file (file.txt), which contains lines of numbers (lines aren't necessarily the same length) e.g.
1 2 3 4
5 6 7 8
9 10 11 12 13
What I need to do is write new files with each of these numbers deleted, one at a time, with replacement e.g. the first new file will contain
2 3 4 <--- 1st element removed
5 6 7 8
9 10 11 12 13
and the 7th file will contain
1 2 3 4
5 6 8 <--- 7th element removed here
9 10 11 12 13
To generate these, I'm looping through each line, and then each element in each line. E.g. for the 7th file, where I remove the third element of the second line, I'm trying to do this by reading in the line, removing the appropriate element, then reinserting this new line
$lineNo is 2 (second line)
$line is 5 6 7 8
with cut, I remove the third number, making $newline 5 6 8
Then I try to replace the line $lineNo in file.txt with $newline using sed:
sed -n '$lineNo s/.*/'$newline'/' > file.txt
This is totally not working. I get an error
sed: can't read 25.780000: No such file or directory
(where 25.780000 is a number in my text file. It looks like it's trying to use $newline to read files or something)
I have reason to suspect my way of stating which line to replace isn't working either :(
My question is, a) is there a better way to do this rather than sed, and b) if sed is the way to go, what am I doing wrong?
Thanks!!

Answer 1

filename=file.txt
i=1
while [[ -s $filename ]]; do
new=file_$i.txt
awk 'NR==1 {if (NF==1) next; else sub(/^[^ ]+ /, "")} 1' $filename > $new
((i++))
filename=$new
done
This leaves a space at the beginning the the first line for each new file, and when a line becomes empty the line is removed. The loop ends when the last generated file is empty.
Update due to requirement clarification:
words=$(wc -w < file.txt)
for ((i=1; i<=words; i++)); do
awk -v n=$i '
words < n && n <= words+NF {$(n-words) = "" }
{words += NF; print}
' file.txt > file_$i.txt
done

Answer 2

Unless I misunderstood the question, the following should work, although it will be pretty slow if your files are large:
#! /bin/bash
remove_by_value()
{
local TO_REMOVE=$1
while read line; do
out=
for word in $line; do [ "$word" = "$TO_REMOVE" ] || out="$out $word"; done
echo "${out/ }"
done < $2
}
remove_by_position()
{
local NTH=$1
while read line; do
out=
for word in $line; do
((--NTH == 0)) || out="$out $word"
done
echo "${out/ }"
done < $2
}
FILE=$1
shift
for number; do
echo "Removing $number"
remove_by_position $number "$FILE"
done
This will dump all the output to stdout, but it should be trivial to change it so the output for each removed number is redirected (e.g. with remove_by_position $number $FILE > $FILE.$$ && mv $FILE.$$ $FILE.$number and proper quoting). Run it as, say,
$ bash script.sh file.txt $(seq 11)

Answer 3

I have to admit, that I'm a bit surprised how short the other solutions are.
#!/bin/bash
#
file=$1
lines=$(cat $file | wc -l)
out=0
dropFromLine () {
file=$1
row=$2
to=$((row-1))
from=$((row+1))
linecontent=($(sed -n "${row}p" $file))
# echo " linecontent: " ${linecontent[#]}
linelen=${#linecontent[#]}
# echo " linelength: " $linelen
for n in $(seq 0 $linelen)
do
(
if [[ $row > 1 ]] ; then sed -n "1,${to}p" $file ;fi
for i in $(seq 0 $linelen)
do
if [[ $n != $i ]]
then
echo -n ${linecontent[$i]}" "
fi
done
echo
# echo "mod - drop " ${linecontent[$n]}
sed -n "$from,${lines}p" $file
) > outfile-${out}.txt
out=$((out+1))
done
}
for row in $(seq 1 $lines)
do
dropFromLine $file $row
done
invocation:
./dropFromRow.sh num.dat
num.dat:
1 2 3 4
5 6 7 8
9 10 11
result:
outfile-0 outfile-10 outfile-12 outfile-2 outfile-4 outfile-6 outfile-8
outfile-1 outfile-11 outfile-13 outfile-3 outfile-5 outfile-7 outfile-9
samples:
asux:~/proj/mini/forum > cat outfile-0
2 3 4
5 6 7 8
9 10 11
asux:~/proj/mini/forum > cat outfile-1
1 3 4
5 6 7 8
9 10 11

Answer 4

One way using perl:
Content of file.txt:
1 2 3 4
5 6 7 8
9 10 11 12 13
Content of script.pl:
use warnings;
use strict;
## Read all input to a scalar variable as a single string.
my $str;
{
local $/ = undef;
$str = <>;
}
## Loop for each number found.
while ( $str =~ m/(\d+)(?:\h*)?/g ) {
## Open file for writing. The name of the file will be
## the number matched in previous regexp.
open my $fh, q[>], ($1 . q[.txt]) or
die qq[Couldn't create file $1.txt\n];
## Print everything prior to matched string plus everything
## after matched string.
printf $fh qq[%s%s], $`, $';
## Close file.
close $fh;
}
Run it like:
perl script.pl file.txt
Show files created:
ls [0-9]*.txt
With output:
10.txt 11.txt 12.txt 13.txt 1.txt 2.txt 3.txt 4.txt 5.txt 6.txt 7.txt 8.txt 9.txt
Show content of one of them:
cat 9.txt
Output:
1 2 3 4
5 6 7 8
10 11 12 13