How can I remove an element from an array completely?

Question

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[1]='mary'
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[0]='mary'
array[1]='randy orton'

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

array=(mary randy orton)

Answer 1

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

Answer 2

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.

Answer 3

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=()
i=0
for element in ${arr[@]}
do
    arr2[$i]=$element
    ((++i))
done
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

Answer 4

#!/bin/bash

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 [[ $1 =~ ([[:digit:]])(-([[:digit:]]))?   ]]; then
    from=${BASH_REMATCH[1]}
    to=${BASH_REMATCH[3]}
  else
    echo bad range
  fi;shift
  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[@]}
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