bash brace expansion: is it possible to associate another list(s) with the initial list?

Question

Example:

$ eval echo "{x,y,z}\ --opt\; "
x --opt; y --opt; z --opt;

Assume that the 2nd list is {1,2,3} and its length is equal to the length of the 1st (initial) list.

Questions:

1. How to make bash produce:

x --opt 1; y --opt 2; z --opt 3;

2. How to make bash produce (i.e. to refer the elements from the {x,y,z} list):

x --opt x; y --opt y; z --opt z;

One-liners are preferred.

Answer 1

Brace expansion would create all possible pairs, it doesn't walk both the lists in parallel:

$ echo {x,y,z}' --opt; '{1,2,3}
x --opt; 1 x --opt; 2 x --opt; 3 y --opt; 1 y --opt; 2 y --opt; 3 z --opt; 1 z --opt; 2 z --opt; 3

To produce the desired output, you need to use something else. For example, loop over the indices of an array:

#! /bin/bash
opt1=(x y z)
opt2=(1 2 3)
for i in "${!opt1[@]}" ; do
    printf '%s --opt %s; ' "${opt1[i]}" "${opt2[i]}"
done
echo

Or, use associative arrays:

#! /bin/bash
declare -A opts=([x]=1 [y]=2 [z]=3)
for i in "${!opts[@]}" ; do
    printf '%s --opt %s; ' "$i" "${opts[$i]}"
done
echo

Answer 2

As choroba pointed out, using {x..z}' --opt '{1..3}';' would generate all possible combinations of strings.

From the list of all possible combinations, 1 through to 9, we want every fourth combination, 1, 5, and 9.

set -- {x..z}' --opt '{1..3}';'
eval echo '${'{1..9..4}'}'

Or, using arrays, fetching elements 0, 4, and 8.

strings=( {x..z}' --opt '{1..3}';' )
eval echo '${strings['{0..8..4}']}'

Note that I would never ever write real code like this and that the code above generates strings that are virtually useless for use as real commands/arguments as they would need to be explicitly split by the shell.

The code is purely presented as an interesting tidbit.