how to find a sequence of numbers - bash

Question

I have a data file formatted like this:
0.00 0.00 0.00
1 10 1.0
2 12 1.0
3 15 1.0
4 20 0.0
5 23 0.0
0.20 0.15 0.6
1 12 1.0
2 15 1.0
3 20 0.0
4 18 0.0
5 20 0.0
0.001 0.33 0.15
1 8 1.0
2 14 1.0
3 17 0.0
4 25 0.0
5 15 0.0
I need to remove some data and reorder line like this:
1 10
1 12
1 8
2 12
2 15
2 14
3 15
3 20
3 17
4 20
4 18
4 25
5 23
5 20
5 15
My code do not show anything. The problem might be in the grep command. Could you please help me out?
touch extract_file.txt
for (( i=1; i<=band; i++))
do
sed -e '1, 7d' data_file | grep -w " '$(echo $i)' " | awk '{print $2}' > extract(echo $i).txt
paste -s extract_file.txt extract$(echo $i).txt > data
done
#rm eigen*.txt

Answer 1

The following code with comments:
cat <<EOF |
0.00 0.00 0.00
1 10 1.0
2 12 1.0
3 15 1.0
4 20 0.0
5 23 0.0
0.20 0.15 0.6
1 12 1.0
2 15 1.0
3 20 0.0
4 18 0.0
5 20 0.0
0.001 0.33 0.15
1 8 1.0
2 14 1.0
3 17 0.0
4 25 0.0
5 15 0.0
EOF
# remove lines not starting with a space
grep -v '^[^ ]' |
# remove leading space
sed 's/^[[:space:]]*//' |
# remove third arg
sed 's/[[:space:]]*[^[:space:]]*$//' |
# stable sort on first number
sort -s -n -k1 |
# each time first number changes, print additional newline
awk '{ if(length(last) != 0 && last != $1) printf "\n"; print; last=$1}'
outputs:
1 10
1 12
1 8
2 12
2 15
2 14
3 15
3 20
3 17
4 20
4 18
4 25
5 23
5 20
5 15
Tested on repl.

Answer 2

perl one-liner:
$ perl -lane 'push #{$nums{$F[0]}}, "#F[0,1]" if /^ /;
END { for $n (sort { $a <=> $b } keys %nums) {
print for #{$nums{$n}};
print "" }}' input.txt
1 10
1 12
1 8
2 12
2 15
2 14
3 15
3 20
3 17
4 20
4 18
4 25
5 23
5 20
5 15
Basically, for each line starting with a space, use the first number as a key to a hash table that stores lists of the first two numbers, and print them out sorted by first number.