Duplicate elements of an array except the first and last elements

Question

I have an array, and I would like to repeat each element except the first and last elements.

For example if the array has five elements 1 2 3 4 5, then after repeating, its elements should be 1 2 2 3 3 4 4 5.

My bash commands are

$ newarr=("${myarr[0]}")
$ for i in $(seq 1 $((${#myarr[@]}-2))) ; do newarr+=( "${myarr[i]}" "${myarr[i]}"); done
$ newarr+=("${myarr[-1]}")

Is there a more clear way than mine?

I am also wondering about how to wrap that into a function which takes myarr as argument, and returns newarr. (After creating such a function, I will read an array from a file, so that each element stores a line in the file, and then call the function on the array. If creating such a function is not a good approach, let me know.)

Thanks.

Answer 1

You can get the number of operation down ever so slightly, and skip the call to seq:

for (( i = 1; i < ${#myarr[@]} - 1; ++i )); do
    newarr+=( "${myarr[i]}" "${myarr[i]}" )
done
newarr=( "${myarr[0]}" "${newarr[@]}" "${myarr[-1]}" )

This assumes that newarr is empty to start with. Do unset newarry first if it's not.

As a function (this modifies the array that is passed):

dup_interal_items () {
    typeset -n arr=$1
    local tmparr

    for (( i = 1; i < ${#arr[@]} - 1; ++i )); do
        tmparr+=( "${arr[i]}" "${arr[i]}" )
    done
    arr=( "${arr[0]}" "${tmparr[@]}" "${arr[-1]}" )
}

The name of the array is passed into the function and the name-reference variable arr is used to access the elements in the array. At the end, the original array is updated to contain the result.

I did it this way rather than returning an array as you can only return an exit status from a function. The other approach would have been to pass both the names of an input array and an output array and use two name-reference variables in the function.

Or, you could possibly echo or printf the values in the function (in which case you don't have to construct a new array at all) and then parse that data in the main code. The function would then be required to be called inside a command substitution.

Note that you can't call this function with an array called arr due to the particular name scoping rules used by bash. You may want to rename the arr variable in the function if this is an issue.

Testing:

$ myarr=( 1 2 3 "here we go" )
$ dup_interal_items myarr
$ printf 'Element: %s\n' "${myarr[@]}"
Element: 1
Element: 2
Element: 2
Element: 3
Element: 3
Element: here we go

---

To duplicate all lines of a file except for the first and last line:

sed -e '1b' -e '$b' -e 'p' <file

The sed script branches to the end (where there is an implicit print statement) if it's on the first or last line, but prints all other lines (all other lines are therefore both explicitly printed by that last p and implicitly printed).

Or, as contributed by don_crissti,

sed 'p;1d;$d' <file

which explicitly prints each line, then ends the cycle for the first and last line, but prints all other lines implicitly (a second time).

An equivalent awk program that does not store more than a single line in memory would be non-trivial to write.

Answer 2

Try also - no function needed -

$ ARR=(1 2 3 4 5)
$ IFS=$'\n\t '
$ readarray NEWARR < <(echo "${ARR[*]}" | sed '1!p; $d')
$ echo ${NEWARR[@]}
1 2 2 3 3 4 4 5

or even

NEWARR=($(printf "%s\n" ${ARR[@]} | sed '1!p; $d'))

Answer 3

A solution that accepts newlines:

#!/bin/bash
arr=( 1 2 "3 3" $'41\n42' 5 )

readarray -t -d $'\0' newarray < <(printf '%s\0' "${arr[@]}" | sed -z '1!p; $d')

printf '<%s>\n' "${newarr[@]}"

Runs as:

$ ./script
<1>
<2>
<2>
<3 3>
<3 3>
<41
42>
<41
42>
<5>