select lines with n identical fields, after a sentinel character - regex

Question

In my file, each line includes five numerical fields delimited by spaces (preceded and followed by more fields). Via a shell script I need to be able to select lines with exactly 3, 4 and 5 identical entries among those five numeric fields (i.e. three separate searches, such that the search for lines with 3 matches in those fields does not also return lines with 4 or 5 matches in those fields).
To find the relevant fields, my search will have to locate the first open and closed parenthesis pair on a line. After the parenthesis is closed, the immediately following five fields are the ones I'm interested in. One potential complication: sometimes one or more of the numeric fields is replaced by a single dash/hyphen instead of a number. One potential means of simplification: the five fields will be in (non-strictly) ascending order and any hyphen entries will always precede the remaining numeric fields.
I would be grateful for some sed/awk suggestions with this. Many thanks!
[EDIT]: I can extract the relevant fields (as detailed in the comment below), thus the strike-through paragraph above is unnecessary. Here is sample data once the relevant fields are extracted:
109 110 111 111 112
110 110 111 111 112
99 99 99 112 112
99 99 99 112 112
100 101 101 112 112
102 102 102 112 112
102 102 103 112 112
102 103 103 112 112
102 104 104 112 112
102 104 104 112 112
103 104 104 112 112
102 105 105 112 112
102 105 105 112 112
103 105 105 112 112
102 106 106 112 112
102 106 107 112 112
103 106 107 112 112
104 106 107 112 112
102 107 107 112 112
104 107 107 112 112
104 107 107 112 112
106 107 108 112 112
107 107 108 112 112
107 107 108 112 112
102 109 109 112 112
102 109 109 112 112
104 109 109 112 112
102 109 110 112 112
103 109 110 112 112
104 109 110 112 112
102 110 110 112 112
104 110 110 112 112
104 110 110 112 112
107 109 111 112 112
107 109 111 112 112
106 110 111 112 112
107 110 111 112 112
107 110 111 112 112
109 110 112 112 112
110 110 112 112 112
107 112 112 112 112
112 112 112 112 112
This should produce hits when n=3 on these lines:
99 99 99 112 112
99 99 99 112 112
102 102 102 112 112
109 110 112 112 112
110 110 112 112 112
A hit when n=4 on this line:
107 112 112 112 112
and a hit when n=5 on this line:
112 112 112 112 112

Answer 1

Here is a Bash script solution using awk. It reads the file line by line and uses an AWK associative array to count how many times a number appeared on the line. Change filename.txt to your file that contains the numbers.
n=3
while read line
do
echo "$line" | awk -v n="$n" '
{
for(i=1; i <= NF; i++) {
a[$i]++
}
}
{
for(o in a) {
if (a[o] == n) {
print
}
}
}
'
done < filename.txt

Answer 2

You can do it using sed as well.
You can create a script:
n=$(($1-1))
sed -n "/\([0-9]*\)\( \1\)\{$n\}/p" filename
And run it like this, just supply n as a script argument:
./script.sh 3
Output:
99 99 99 112 112
99 99 99 112 112
102 102 102 112 112
109 110 112 112 112
110 110 112 112 112

Answer 3

Awk-only solution as a one-liner:
awk -v n=3 '{for(i=1;i<=NF;i++)a[$i]++;for(o in a)if(a[o]==n)p=1} p; {p=0;delete a}' inputfile
Split out for easier reading, this slightly resembles badjr's solution. (I've used his variables for easier comparison.)
{
for (i=1;i<=NF;i++) # populate an array with counts of unique elements
a[$i]++
for (o in a) # check the array for a matching count & set flag
if (a[o]==n)
p=1
}
p; # if we've set our flag, print the current line.
{ # clear our workspace for the next line.
p=0
delete a
}
If you're interested in a bash-only solution, the following implements the same awk logic, only without awk:
#!/usr/bin/env bash
n=5
while read -a a; do
unset b
for i in "${!a[#]}"; do
(( b[${a[$i]}]++ ))
done
for i in "${b[#]}"; do
[ "$i" -eq "$n" ] && echo "${a[#]}"
done
done < inputfile
Note that because the output here is printed using array elements, whitespace in the input file will not be maintained.
This solution is bash-only because of its use of arrays.

Answer 4

Another sed example, didn't want my work to go to waste ;)
#!/bin/bash
while (($1 > 0))
do
n="${n} \1"
set ${1}-1
done
sed -nr "\_\<([0-9]+)${n}\>_ p"
EDIT:
On BSD sed (OS X) you need to replace \< and \> with the fascinating [[:<:]] and [[:>:]] respectively.