grep: how to find ALL the lines between to expressions - regex

Question

We have a HUGE file (numbers), we want to get ALL the lines between two expressions, e.g.,
232445 -9998.01 xxxxxxxxxx
234566 -9998.02 xxxxxxxxx
.
.
324444 -8000.012 xxxxxxx
344444 -8000.0 xxxx
and the expressions are -9998.01 and -8000.0, so tried:
$ grep -A100000 '[0-9] -9998.[0-9]' mf.in | grep -B100000 '[0-9] -8000.[0-9]' mf.in > mfile.out
And this is OK ...ALL the lines between are get it... of course, 100000 is so big as to keept ALL the lines between... but if we are wrong? i.e., if there are more than 100000 between? How we can take ALL between without numeric specification after A and B ...
PD: I was unable to use sed with similar "[ ...]" expressions
PD2: the columns has more digits (here only 4 columns)
-1931076.0 -9998.96235 1.0002741998076021 0.0191476198569163
-1931075.0 -9998.95962 1.0000742544770280 0.0192495084654059
-1931074.0 -9998.95688 0.9998778097258081 0.0193725608470694

Answer 1

With awk:
awk '$2 ~ /^-9998.01$/{p=1} p{print} $2 ~ /^-8000.0$/{p=0}' file
Test:
$ cat file
232445 -9998.00 xxxxxxxxxx
232445 -9998.01 xxxxxxxxxx
234566 -9998.02 xxxxxxxxx
234566 -9998.03 xxxxxxxxx
234566 -9998.05 xxxxxxxxx
....
....
324444 -8000.011 xxxxxxx
324444 -8000.012 xxxxxxx
344444 -8000.0 xxxx
344444 -8000.1 xxxx
$ awk '$2 ~ /^-9998.01$/{p=1} p{print} $2 ~ /^-8000.0$/{p=0}' file
232445 -9998.01 xxxxxxxxxx
234566 -9998.02 xxxxxxxxx
234566 -9998.03 xxxxxxxxx
234566 -9998.05 xxxxxxxxx
....
....
324444 -8000.011 xxxxxxx
324444 -8000.012 xxxxxxx
344444 -8000.0 xxxx

Answer 2

sed already has this functionality builtin using this expression:
/regex1/,/regex2/ p=>p command prints all lines that are present in between 2 lines(start line having regex1 and end line having regex2(both inclusive in output)).
Here is an example wrt your file format:
$ cat file
124235 -69768.77 xxx
232445 -9998.01 xxxxxxxxxx
234566 -9998.02 xxxxxxxxx
12345 -124.66 xxxx
324444 -8000.012 xxxxxxx
344444 -8000.0 xxxx
344444 -7000.0 xxxx
$ sed -nr '/^[0-9]+\s-9998.[0-9]+\s/,/^[0-9]+\s-8000.[0-9]+\s/ p' file
232445 -9998.01 xxxxxxxxxx
234566 -9998.02 xxxxxxxxx
12345 -124.66 xxxx
324444 -8000.012 xxxxxxx
344444 -8000.0 xxxx
$

Answer 3

Well it might not be the best answer, but the easy fix for your command would be to use the file's number of lines as argument to -A and -B, so you're sure you cannot miss any lines:
NB_LINES=$(wc -l main.c | awk '{print $1}')
grep -A$NB_LINES '[0-9] -9998.[0-9]' mf.in | grep -B$NB_LINES '[0-9] -8000.[0-9]' mf.in > mfile.out
Though, tbh, in pure shell it's very likely I'd do something similar. Or I'd write a small python script, that would look like:
import re
LINE_RE = re.compile(r'[^ ]+ (-[0-9]+\.[0-9]+) .*')
with open('mf.in', 'r') as fin:
with open('mf.out', 'w') as fout:
for line in f:
match = LINE_RE.match(line)
if match:
if float(match.groups()[0]) > -9998.0:
fout.write(line)
elif float(match.groups()[0]) < -8000.0:
break
N.B.: this script is just to expose the algorithmic idea, and being blindly coded and untested, it might need some tweaking to actually work.
HTH