Matching bash variables as number literals with grep - regex

Question

I have a (GNU) bash script which establishes two variables to be matched in a file.
hour=$(head -n 1 sensorstest.log | cut -f5 | cut -d"-" -f1)
dom=$(head -n 1 sensorstest.log | cut -f5 | cut -d"-" -f4)
...and matches them to other occurrences in the file
grep -E [^0-9]+"$hour"-[0-9]+-[0-9]+-"$dom"-[0-9]+-[0-9]{4} sensorstest.log
Here is an example of the script calculating the mean for all values in field 2 of the input file for the given hour of day.
hMean=$(grep -E [^0-9]+"$hour"-[0-9]+-[0-9]+-"$dom"-[0-9]+-[0-9]{4} sensorstest.log | cut -f2 | awk ' {sum+=$
1}{count++}{mean=sum/count} END {printf("%.2f",mean) } ' );
Here is an example of the cleanup of the input file.
echo "removing: "$hour"th hour of the "$dom"th day of the "$month"th month"
sed -i -r '/'"$hour"'-[0-9]+-[0-9]+-'"$dom"'-'"$month"'-[0-9]{4}/d' sensorstest.log
And finally... Here is an example line in the file:
The format is:
status<tab>humidity<tab>temperature<tab>unix timestamp<tab>time/date
OK 94.4 16.9 1443058486 1-34-46-24-9-2015
I am attempting to match all instances of the hour on the day of the first entry in the file.
This works fine for numbers below 9, however;
Problem: Numbers over 9 are being matched as two single digit numbers, resulting in 12 matching 1, 2, 12, 21...etc.
Here is an example of where is trips up:
OK 100 17.2 1442570381 9-59-41-18-9-2015
OK 100 17.1 1442570397 9-59-57-18-9-2015
Moistening 100 17.6 1442574014 11-0-14-18-9-2015
Moistening 100 17.6 1442574030 11-0-30-18-9-2015
Here the output skips to 0-0-0-19-9-2015 (and yes I am missing an hour of entries from the log)
$ sudo statanhourtest.sh
100,1.4,1.40,-98.6 16.5,17.2,16.90,.7 1442566811 9-0-0-18-9-2015
removing: 9th hour of the 18th day of the 9th month
$ sudo statanhourtest.sh
100,1.4,1.40,-98.6 18.3,18.8,18.57,.5 1442620804 0-0-0-19-9-2015
removing: 0th hour of the 19th day of the 9th month
The problem is only happening with the hours. the day ($dom) is matching fine.
I have tried using the -w option with grep, but I think this only returns the exact match where I need the whole line.
There's not much online about matching numbers literally in grep. And I found nothing on using bash variables as a number literal.
Any help or relevant links would be greatly appreciated.
EDIT:
I have solved the problem after a night of dredging through the script.
The problem lay with my sed expression right at the end.
The problem being in single quoting parts of the sed expression and double quoting variables for expansion by the shell.
I took this from a suggestion on another thread.
Double quoting the whole expression solved the problem.
The awk suggestion has greatly increased the efficiency and accuracy of the script. Thanks again.

Answer 1

awk to the rescue!
I think you can combine everything to a simple awk script without needing any regex. For example,
awk 'NR==1{split($NF,h,"-")} {split($NF,t,"-")} t[1]==h[1] && t[4]==h[4]'
will parse the time stamp on the first row of the file and filters only the hour and day matching records.
This will take the average of field 2
awk 'NR==1
{
split($NF,h,"-")
}
{
split($NF,t,"-")
}
t[1]==h[1] && t[4]==h[4]
{
sum+=$2;
c++
}
END
{
print "Average: " sum/c
}'