How to declare 2D array in bash

Solution 1:

You can simulate them for example with hashes, but need care about the leading zeroes and many other things. The next demonstration works, but it is far from optimal solution.

#!/bin/bash
declare -A matrix
num_rows=4
num_columns=5

for ((i=1;i<=num_rows;i++)) do
    for ((j=1;j<=num_columns;j++)) do
        matrix[$i,$j]=$RANDOM
    done
done

f1="%$((${#num_rows}+1))s"
f2=" %9s"

printf "$f1" ''
for ((i=1;i<=num_rows;i++)) do
    printf "$f2" $i
done
echo

for ((j=1;j<=num_columns;j++)) do
    printf "$f1" $j
    for ((i=1;i<=num_rows;i++)) do
        printf "$f2" ${matrix[$i,$j]}
    done
    echo
done

the above example creates a 4x5 matrix with random numbers and print it transposed, with the example result

           1         2         3         4
 1     18006     31193     16110     23297
 2     26229     19869      1140     19837
 3      8192      2181     25512      2318
 4      3269     25516     18701      7977
 5     31775     17358      4468     30345

The principle is: Creating one associative array where the index is an string like 3,4. The benefits:

• it's possible to use for any-dimension arrays ;) like: 30,40,2 for 3 dimensional.
• the syntax is close to "C" like arrays ${matrix[2,3]}

Solution 2:

Bash doesn't have multi-dimensional array. But you can simulate a somewhat similar effect with associative arrays. The following is an example of associative array pretending to be used as multi-dimensional array:

declare -A arr
arr[0,0]=0
arr[0,1]=1
arr[1,0]=2
arr[1,1]=3
echo "${arr[0,0]} ${arr[0,1]}" # will print 0 1

If you don't declare the array as associative (with -A), the above won't work. For example, if you omit the declare -A arr line, the echo will print 2 3 instead of 0 1, because 0,0, 1,0 and such will be taken as arithmetic expression and evaluated to 0 (the value to the right of the comma operator).

Solution 3:

Bash does not support multidimensional arrays.

You can simulate it though by using indirect expansion:

#!/bin/bash
declare -a a0=(1 2 3 4)
declare -a a1=(5 6 7 8)
var="a1[1]"
echo ${!var}  # outputs 6

Assignments are also possible with this method:

let $var=55
echo ${a1[1]}  # outputs 55

Edit 1: To read such an array from a file, with each row on a line, and values delimited by space, use this:

idx=0
while read -a a$idx; do
    let idx++;
done </tmp/some_file

Edit 2: To declare and initialize a0..a3[0..4] to 0, you could run:

for i in {0..3}; do
    eval "declare -a a$i=( $(for j in {0..4}; do echo 0; done) )"
done

Solution 4:

You can also approach this in a much less smarter fashion

q=()
q+=( 1-2 )
q+=( a-b )

for set in ${q[@]};
do
echo ${set%%-*}
echo ${set##*-}
done

of course a 22 line solution or indirection is probably the better way to go and why not sprinkle eval every where to .