unset array[0] removes the element but still if I do echo ${array[0]} I get a null value moreover there are other ways of doing this but if an element of an array contains spaces like below

array[0]='james young'
array[2]='randy orton'

but these also fail to do the job

array=${array[@]:1} #removed the 1st element

now I want the new array to be like

array[1]='randy orton'

The spaces cause the trouble after assignment and the actual array becomes like with substitution.

array=(mary randy orton)
  • 4
    No, not the spaces cause trouble, but the lack of quoting.
    – manatwork
    Mar 18, 2013 at 12:33

4 Answers 4


Just use array syntax on the assignment and quote your variable:

array=("${array[@]:1}") #removed the 1st element

Edit according to question in comment. For $@ you can use it like this:

set -- "${@:2}" #removed the 1st parameter
  • 8
    Note that it doesn't remove the 1st element but the element of indice 0 and reassigns the indices. If the first element was on indice 12, it won't remove anything but will reassign the indices so that what once was on indice 12 will now be on indice 0. It's probably not a concern in the OP's case but should probably be noted for future reference. The behaviour is different with zsh whose arrays are not sparse contrary to ksh or bash. Mar 18, 2013 at 13:05
  • 4
    Hi @StephaneChazelas. The singular of "indices" is "index". Thanks for your comment!
    – Steven Lu
    Aug 10, 2013 at 23:09
  • 3
    @manatwork - re: your edit - why not using shift ? Sep 8, 2016 at 9:53
  • 1
    @don_crissti, good point. I focused on the indexing difference and not thought further. Also had in mind the situation when you need to discard variable amount of items, for example to keep exactly the last 3: array=("${array[@]: -3}") and set -- "${@: -3}". So stuck at indices.
    – manatwork
    Sep 8, 2016 at 10:16
  • 1
    shift $[$#-3] for the last 3 is probably much faster for $@
    – Tino
    Dec 11, 2016 at 10:21

This had me thinking. The problem with sed/awk/tail is that they're line by line. After you delete the first line you have to write every other line from pattern space to the file.

  • You can use the following commands to do what you want in seconds.
  • This will write the entire file to an array.
  • Remove the first line as it dumps it back into the file.

    readarray -t aLargeFile < <(cat largefile)
    echo "${aLargeFile[@]:1}" >largefile

Just change the largefile to the name of your file.

  • Why not use sed -i 1d largefile instead? This even works for files bigger than RAM+swap
    – Tino
    Dec 11, 2016 at 10:15

To remove an element at particular index, we can use unset and then do copy to another array. Only just unset is not required in this case. Because unset does not remove the element it just sets null string to the particular index in array.

declare -a arr=('aa' 'bb' 'cc' 'dd' 'ee')
unset 'arr[1]'
declare -a arr2=()
for element in ${arr[@]}
echo ${arr[@]}
echo "1st val is ${arr[1]}, 2nd val is ${arr[2]}"
echo ${arr2[@]}
echo "1st val is ${arr2[1]}, 2nd val is ${arr2[2]}"

Output is

aa cc dd ee
1st val is , 2nd val is cc
aa cc dd ee
1st val is cc, 2nd val is dd
  • NO it does not replace element with a null, or empty string. it removes the element and creates a sparse array. That is an array with 'holes' in it. However looking up ANY undefined element in a sparse array returns a empty string! Sparse Arrays is perfectly valid for bash arrays! You can see it was removed using declare -p arr You can test if it is defined using [[ -v arr[1] ]] || echo not-defined
    – anthony
    Jun 19, 2020 at 11:35

q=( one two three four five )

echo -e "
  (remove) { [:range:] } <- [:list:]
                | [:range:] => return list with range removed range is in the form of [:digit:]-[:digit:]

function remove {
  if [[ $1 =~ ([[:digit:]])(-([[:digit:]]))?   ]]; then
    echo bad range
  array=( ${@} )
  local start=${array[@]::${from}}
  local rest
  [ -n "$to" ] && rest=${array[@]:((${to}+1))}  || rest=${array[@]:((${from}+1))}
  echo ${start[@]} ${rest[@]}

q=( `remove 1 ${q[*]}` )
echo ${q[@]}
  • 4
    This would be much better if there was something to explain how it works and not just a blob of code. And what's with the tildes at the bottom?
    – user
    Oct 8, 2013 at 20:16
  • 4
    Seriously, you are correct. This does look like it was written by a hooligan, but thank you. I really only get to sneak this in between hamburger serving days. Oct 9, 2013 at 6:54
  • If any element of q has spaces in it, this will break it up into multiple elements. Aug 21, 2014 at 20:40

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