Looping over arrays, printing both index and value

Question

I want to do something like this:

foo=( )
foo[0]="bar"
foo[35]="baz"
for((i=0;i<${#foo[@]};i++))
do
    echo "$i: ${foo[$i]}"
done
# Output:
# 0: bar
# 1: 

Then i tried to loop through it using for in:

foo=( )
foo[0]="bar"
foo[35]="baz"
for i in ${foo[@]}
do
    echo "?: $i"
done
# Output:
# ?: bar
# ?: naz

but here I don't know the index value.

I know you could something like

foo=( )
foo[0]="bar"
foo[35]="baz"
declare -p foo
# Output:
# declare -a foo='([0]="bar" [35]="baz")'

but, can't you do it in another way?

Solution 1

You would find the array keys with "${!foo[@]}" (reference), so:

for i in "${!foo[@]}"; do 
  printf "%s\t%s\n" "$i" "${foo[$i]}"
done

Which means that indices will be in $i while the elements themselves have to be accessed via ${foo[$i]}

Solution 2

you can always use iteration param:

ITER=0
for I in ${FOO[@]}
do  
    echo ${I} ${ITER}
    ITER=$(expr $ITER + 1)
done

Solution 3

INDEX=0
for i in $list; do 
    echo ${INDEX}_$i
    let INDEX=${INDEX}+1
done

Solution 4

In bash 4, you can use associative arrays:

declare -A foo
foo[0]="bar"
foo[35]="baz"

# for Zsh, change this to: for key in "${(k)foo[@]}"
for key in "${!foo[@]}"
do
    echo "key: $key, value: ${foo[$key]}"
done

# output
# $ key: 0, value bar.
# $ key: 35, value baz.

In bash 3, this works (also works in zsh):

map=( )
map+=("0:bar")
map+=("35:baz")

for keyvalue in "${map[@]}" ; do
    key=${keyvalue%%:*}
    value=${keyvalue#*:}
    echo "key: $key, value $value."
done

Solution 5

Simple one line trick for dumping array

I've added one value with spaces:

foo=([12]="bar" [42]="foo bar baz" [35]="baz")

For a quick dump of bash arrays or associative arrays I use

This one line command:

paste <(printf "%s\n" "${!foo[@]}") <(printf "%s\n" "${foo[@]}")

Will render:

12  bar
35  baz
42  foo bar baz

Explained

• printf "%s\n" "${!foo[@]}" will print all keys separated by a newline,
• printf "%s\n" "${foo[@]}" will print all values separated by a newline,
• paste <(cmd1) <(cmd2) will merge output of cmd1 and cmd2 line by line.

Tunning

This could be tunned by -d switch:

paste -d : <(printf "%s\n" "${!foo[@]}") <(printf "%s\n" "${foo[@]}")
12:bar
35:baz
42:foo bar baz

or even:

paste -d = <(printf "foo[%s]\n" "${!foo[@]}") <(printf "'%s'\n" "${foo[@]}")
foo[12]='bar'
foo[35]='baz'
foo[42]='foo bar baz'

Associative array will work same:

declare -A bar=([foo]=snoopy [bar]=nice [baz]=cool [foo bar]='Hello world!')
paste -d = <(printf "bar[%s]\n" "${!bar[@]}") <(printf '"%s"\n' "${bar[@]}")
bar[foo bar]="Hello world!"
bar[foo]="snoopy"
bar[bar]="nice"
bar[baz]="cool"

Issue with newlines or special chars

Unfortunely, there is at least one condition making this not work anymore: when variable do contain newline:

foo[17]=$'There is one\nnewline'

Command paste will merge line-by-line, so output will become wrong:

paste -d = <(printf "foo[%s]\n" "${!foo[@]}") <(printf "'%s'\n" "${foo[@]}")
foo[12]='bar'
foo[17]='There is one
foo[35]=newline'
foo[42]='baz'
='foo bar baz'

For this work, you could use %q instead of %s in second printf command (and whipe quoting):

paste -d = <(printf "foo[%s]\n" "${!foo[@]}") <(printf "%q\n" "${foo[@]}")

Will render perfect ( and reusable! ):

foo[12]=bar
foo[17]=$'There is one\nnewline'
foo[35]=baz
foo[42]=foo\ bar\ baz

From man bash:

          %q     causes  printf  to output the corresponding argument in a
                 format that can be reused as shell input.

Or by using a function:

dumpArray() {
    local -n _ary=$1
    local _idx
    local -i _idlen=0
    for _idx in "${!_ary[@]}"; do
        _idlen=" ${#_idx} >_idlen ? ${#_idx} : _idlen "
    done
    for _idx in "${!_ary[@]}"; do
        printf "%-*s: %s\n" "$_idlen" "$_idx" \
            "${_ary["$_idx"]//$'\n'/$'\n\e['${_idlen}C: }"
    done
}

Then now:

dumpArray foo
12: bar
17: There is one
  : newline
35: baz
42: foo bar baz

dumpArray bar
foo    : snoopy
bar    : nice
baz    : cool
foo bar: Hello world!

About UTF-8 format output

From UTF-8 string length, adding:

strU8DiffLen() { local chLen=${#1} LANG=C LC_ALL=C;return $((${#1}-chLen));}

Then

dumpArray() {
    local -n _ary=$1
    local _idx
    local -i _idlen=0
    for _idx in "${!_ary[@]}"; do
        _idlen=" ${#_idx} >_idlen ? ${#_idx} : _idlen "
    done
    for _idx in "${!_ary[@]}"; do
        strU8DiffLen "$_idx"
        printf "%-*s: %s\n" $(($?+$_idlen)) "$_idx" \
            "${_ary["$_idx"]//$'\n'/$'\n\e['${_idlen}C: }"
    done
}

Demo:

foo=([12]="bar" [42]="foo bar baz" [35]="baz")
declare -A bar=([foo]=snoopy [bar]=nice [baz]=cool [foo bar]='Hello world!')

foo[17]=$'There is one\nnewline'
LANG=fr.UTF-8 printf -v bar[déjà]  $'%(%a %d %b\n%Y\n%T)T' -1

dumpArray bar
déjà   : ven 24 déc
       : 2021
       : 08:36:05
foo    : snoopy
bar    : nice
baz    : cool
foo bar: Hello world!

dumpArray foo
12: bar
17: There is one
  : newline
35: baz
42: foo bar baz

View more solutions

Comments

[David Tonhofer]

((ITER++)) in modern bash

[MikeW]

Why post-increment ? You only want the value incremented, hence ((++ITER)) is more directly a statement of what you want done ....

[Marco]

No, not "ALWAYS". A hash can have "holes", which means not all numbers are an index. In your example the ${ITER} is not always the index of ${I}.

[F. Hauri - Give Up GitHub]

This is wrong! Inner loop is useless and could drive to wrong results!

[techno]

My up vote for printf "%q\n" "${var[@]}" newline was my problem!

[Alexander D'Attore]

I liked the conciseness of the answer and strategy. For anyone wondering this is bash and the strat is you just keep track of the array index on your own.