awk: delete first and last entry of comma-separated field - awk

Question

I have a 4 column data that looks something like the following:
a 1 g 1,2,3,4,5,6,7
b 2 g 3,5,3,2,6,4,3,2
c 3 g 5,2,6,3,4
d 4 g 1,5,3,6,4,7
I am trying to delete first two numbers and the last two numbers on entire fourth column so the output looks like the following
a 1 g 3,4,5
b 2 g 3,2,6,4
c 3 g 6
d 4 g 3,6
Can someone give me a help? I would appreciate it.

Answer 1

You can use this:
$ awk '{n=split($4, a, ","); for (i=3; i<=n-2; i++) t=t""a[i](i==n-2?"":","); print $1, $2, $3, t; t=""}' file
a 1 g 3,4,5
b 2 g 3,2,6,4
c 3 g 6
d 4 g 3,6
Explanation
n=split($4, a, ",") slices the 4th field in pieces, based on comma as delimiter. As split() returns the number of pieces we got, we store it in n to work with it later on.
for (i=3; i<=n-2; i++) t=t""a[i](i==n-2?"":",") stores in t the last field, looping through all the slices.
print $1, $2, $3, t; t="" prints the new output and blanks the variable t.

Answer 2

This will work for your posted sample input:
$ awk '{gsub(/^([^,]+,){2}|(,[^,]+){2}$/,"",$NF)}1' file
a 1 g 3,4,5
b 2 g 3,2,6,4
c 3 g 6
d 4 g 3,6
If you have cases where there's less than 4 commas in your 4th field then update your question to show how those should be handled.

Answer 3

This uses bash array manipulation. It may be a little ... gnarly:
while read -a fields; do # read the fields for each line
IFS=, read -a values <<< "${fields[3]}" # split the last field on comma
new=("${values[#]:2:${#values[#]}-4}") # drop the first 2 and last fields
fields[3]=$(IFS=,; echo "${new[*]}") # join the new list on comma
printf "%s\t" "${fields[#]}"; echo # print the new line
done <<END
a 1 g 1,2,3,4,5,6,7
b 2 g 3,5,3,2,6,4,3,2
c 3 g 5,2,6,3,4
d 4 g 1,5,3,6,4,7
END
a 1 g 3,4,5
b 2 g 3,2,6,4
c 3 g 6
d 4 g 3,6