I need to remove an element from an array in bash shell.
Generally I'd simply do:
array=("${(#)array:#<element to remove>}")
Unfortunately the element I want to remove is a variable so I can't use the previous command.
Down here an example:
array+=(pluto)
array+=(pippo)
delete=(pluto)
array( ${array[#]/$delete} ) -> but clearly doesn't work because of {}
Any idea?
The following works as you would like in bash and zsh:
$ array=(pluto pippo)
$ delete=pluto
$ echo ${array[#]/$delete}
pippo
$ array=( "${array[#]/$delete}" ) #Quotes when working with strings
If need to delete more than one element:
...
$ delete=(pluto pippo)
for del in ${delete[#]}
do
array=("${array[#]/$del}") #Quotes when working with strings
done
Caveat
This technique actually removes prefixes matching $delete from the elements, not necessarily whole elements.
Update
To really remove an exact item, you need to walk through the array, comparing the target to each element, and using unset to delete an exact match.
array=(pluto pippo bob)
delete=(pippo)
for target in "${delete[#]}"; do
for i in "${!array[#]}"; do
if [[ ${array[i]} = $target ]]; then
unset 'array[i]'
fi
done
done
Note that if you do this, and one or more elements is removed, the indices will no longer be a continuous sequence of integers.
$ declare -p array
declare -a array=([0]="pluto" [2]="bob")
The simple fact is, arrays were not designed for use as mutable data structures. They are primarily used for storing lists of items in a single variable without needing to waste a character as a delimiter (e.g., to store a list of strings which can contain whitespace).
If gaps are a problem, then you need to rebuild the array to fill the gaps:
for i in "${!array[#]}"; do
new_array+=( "${array[i]}" )
done
array=("${new_array[#]}")
unset new_array
You could build up a new array without the undesired element, then assign it back to the old array. This works in bash:
array=(pluto pippo)
new_array=()
for value in "${array[#]}"
do
[[ $value != pluto ]] && new_array+=($value)
done
array=("${new_array[#]}")
unset new_array
This yields:
echo "${array[#]}"
pippo
This is the most direct way to unset a value if you know it's position.
$ array=(one two three)
$ echo ${#array[#]}
3
$ unset 'array[1]'
$ echo ${array[#]}
one three
$ echo ${#array[#]}
2
This answer is specific to the case of deleting multiple values from large arrays, where performance is important.
The most voted solutions are (1) pattern substitution on an array, or (2) iterating over the array elements. The first is fast, but can only deal with elements that have distinct prefix, the second has O(n*k), n=array size, k=elements to remove. Associative array are relative new feature, and might not have been common when the question was originally posted.
For the exact match case, with large n and k, possible to improve performance from O(nk) to O(n+klog(k)). In practice, O(n) assuming k much lower than n. Most of the speed up is based on using associative array to identify items to be removed.
Performance (n-array size, k-values to delete). Performance measure seconds of user time
N K New(seconds) Current(seconds) Speedup
1000 10 0.005 0.033 6X
10000 10 0.070 0.348 5X
10000 20 0.070 0.656 9X
10000 1 0.043 0.050 -7%
As expected, the current solution is linear to N*K, and the fast solution is practically linear to K, with much lower constant. The fast solution is slightly slower vs the current solution when k=1, due to additional setup.
The 'Fast' solution: array=list of input, delete=list of values to remove.
declare -A delk
for del in "${delete[#]}" ; do delk[$del]=1 ; done
# Tag items to remove, based on
for k in "${!array[#]}" ; do
[ "${delk[${array[$k]}]-}" ] && unset 'array[k]'
done
# Compaction
array=("${array[#]}")
Benchmarked against current solution, from the most-voted answer.
for target in "${delete[#]}"; do
for i in "${!array[#]}"; do
if [[ ${array[i]} = $target ]]; then
unset 'array[i]'
fi
done
done
array=("${array[#]}")
Here's a one-line solution with mapfile:
$ mapfile -d $'\0' -t arr < <(printf '%s\0' "${arr[#]}" | grep -Pzv "<regexp>")
Example:
$ arr=("Adam" "Bob" "Claire"$'\n'"Smith" "David" "Eve" "Fred")
$ echo "Size: ${#arr[*]} Contents: ${arr[*]}"
Size: 6 Contents: Adam Bob Claire
Smith David Eve Fred
$ mapfile -d $'\0' -t arr < <(printf '%s\0' "${arr[#]}" | grep -Pzv "^Claire\nSmith$")
$ echo "Size: ${#arr[*]} Contents: ${arr[*]}"
Size: 5 Contents: Adam Bob David Eve Fred
This method allows for great flexibility by modifying/exchanging the grep command and doesn't leave any empty strings in the array.
Partial answer only
To delete the first item in the array
unset 'array[0]'
To delete the last item in the array
unset 'array[-1]'
To expand on the above answers, the following can be used to remove multiple elements from an array, without partial matching:
ARRAY=(one two onetwo three four threefour "one six")
TO_REMOVE=(one four)
TEMP_ARRAY=()
for pkg in "${ARRAY[#]}"; do
for remove in "${TO_REMOVE[#]}"; do
KEEP=true
if [[ ${pkg} == ${remove} ]]; then
KEEP=false
break
fi
done
if ${KEEP}; then
TEMP_ARRAY+=(${pkg})
fi
done
ARRAY=("${TEMP_ARRAY[#]}")
unset TEMP_ARRAY
This will result in an array containing:
(two onetwo three threefour "one six")
Here's a (probably very bash-specific) little function involving bash variable indirection and unset; it's a general solution that does not involve text substitution or discarding empty elements and has no problems with quoting/whitespace etc.
delete_ary_elmt() {
local word=$1 # the element to search for & delete
local aryref="$2[#]" # a necessary step since '${!$2[#]}' is a syntax error
local arycopy=("${!aryref}") # create a copy of the input array
local status=1
for (( i = ${#arycopy[#]} - 1; i >= 0; i-- )); do # iterate over indices backwards
elmt=${arycopy[$i]}
[[ $elmt == $word ]] && unset "$2[$i]" && status=0 # unset matching elmts in orig. ary
done
return $status # return 0 if something was deleted; 1 if not
}
array=(a 0 0 b 0 0 0 c 0 d e 0 0 0)
delete_ary_elmt 0 array
for e in "${array[#]}"; do
echo "$e"
done
# prints "a" "b" "c" "d" in lines
Use it like delete_ary_elmt ELEMENT ARRAYNAME without any $ sigil. Switch the == $word for == $word* for prefix matches; use ${elmt,,} == ${word,,} for case-insensitive matches; etc., whatever bash [[ supports.
It works by determining the indices of the input array and iterating over them backwards (so deleting elements doesn't screw up iteration order). To get the indices you need to access the input array by name, which can be done via bash variable indirection x=1; varname=x; echo ${!varname} # prints "1".
You can't access arrays by name like aryname=a; echo "${$aryname[#]}, this gives you an error. You can't do aryname=a; echo "${!aryname[#]}", this gives you the indices of the variable aryname (although it is not an array). What DOES work is aryref="a[#]"; echo "${!aryref}", which will print the elements of the array a, preserving shell-word quoting and whitespace exactly like echo "${a[#]}". But this only works for printing the elements of an array, not for printing its length or indices (aryref="!a[#]" or aryref="#a[#]" or "${!!aryref}" or "${#!aryref}", they all fail).
So I copy the original array by its name via bash indirection and get the indices from the copy. To iterate over the indices in reverse I use a C-style for loop. I could also do it by accessing the indices via ${!arycopy[#]} and reversing them with tac, which is a cat that turns around the input line order.
A function solution without variable indirection would probably have to involve eval, which may or may not be safe to use in that situation (I can't tell).
Using unset
To remove an element at particular index, we can use unset and then do copy to another array. Only just unset is not required in this case. Because unset does not remove the element it just sets null string to the particular index in array.
declare -a arr=('aa' 'bb' 'cc' 'dd' 'ee')
unset 'arr[1]'
declare -a arr2=()
i=0
for element in "${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 ${arr2[2]}"
Output is
aa cc dd ee
1st val is , 2nd val is cc
aa cc dd ee
1st val is cc, 2nd val is dd
Using :<idx>
We can remove some set of elements using :<idx> also. For example if we want to remove 1st element we can use :1 as mentioned below.
declare -a arr=('aa' 'bb' 'cc' 'dd' 'ee')
arr2=("${arr[#]:1}")
echo "${arr2[#]}"
echo "1st val is ${arr2[1]}, 2nd val is ${arr2[2]}"
Output is
bb cc dd ee
1st val is cc, 2nd val is dd
http://wiki.bash-hackers.org/syntax/pe#substring_removal
${PARAMETER#PATTERN} # remove from beginning
${PARAMETER##PATTERN} # remove from the beginning, greedy match
${PARAMETER%PATTERN} # remove from the end
${PARAMETER%%PATTERN} # remove from the end, greedy match
In order to do a full remove element, you have to do an unset command with an if statement. If you don't care about removing prefixes from other variables or about supporting whitespace in the array, then you can just drop the quotes and forget about for loops.
See example below for a few different ways to clean up an array.
options=("foo" "bar" "foo" "foobar" "foo bar" "bars" "bar")
# remove bar from the start of each element
options=("${options[#]/#"bar"}")
# options=("foo" "" "foo" "foobar" "foo bar" "s" "")
# remove the complete string "foo" in a for loop
count=${#options[#]}
for ((i = 0; i < count; i++)); do
if [ "${options[i]}" = "foo" ] ; then
unset 'options[i]'
fi
done
# options=( "" "foobar" "foo bar" "s" "")
# remove empty options
# note the count variable can't be recalculated easily on a sparse array
for ((i = 0; i < count; i++)); do
# echo "Element $i: '${options[i]}'"
if [ -z "${options[i]}" ] ; then
unset 'options[i]'
fi
done
# options=("foobar" "foo bar" "s")
# list them with select
echo "Choose an option:"
PS3='Option? '
select i in "${options[#]}" Quit
do
case $i in
Quit) break ;;
*) echo "You selected \"$i\"" ;;
esac
done
Output
Choose an option:
1) foobar
2) foo bar
3) s
4) Quit
Option?
Hope that helps.
There is also this syntax, e.g. if you want to delete the 2nd element :
array=("${array[#]:0:1}" "${array[#]:2}")
which is in fact the concatenation of 2 tabs. The first from the index 0 to the index 1 (exclusive) and the 2nd from the index 2 to the end.
POSIX shell script does not have arrays.
So most probably you are using a specific dialect such as bash, korn shells or zsh.
Therefore, your question as of now cannot be answered.
Maybe this works for you:
unset array[$delete]
What I do is:
array="$(echo $array | tr ' ' '\n' | sed "/itemtodelete/d")"
BAM, that item is removed.
This is a quick-and-dirty solution that will work in simple cases but will break if (a) there are regex special characters in $delete, or (b) there are any spaces at all in any items. Starting with:
array+=(pluto)
array+=(pippo)
delete=(pluto)
Delete all entries exactly matching $delete:
array=(`echo $array | fmt -1 | grep -v "^${delete}$" | fmt -999999`)
resulting in
echo $array -> pippo, and making sure it's an array:
echo $array[1] -> pippo
fmt is a little obscure: fmt -1 wraps at the first column (to put each item on its own line. That's where the problem arises with items in spaces.) fmt -999999 unwraps it back to one line, putting back the spaces between items. There are other ways to do that, such as xargs.
Addendum: If you want to delete just the first match, use sed, as described here:
array=(`echo $array | fmt -1 | sed "0,/^${delete}$/{//d;}" | fmt -999999`)
Actually, I just noticed that the shell syntax somewhat has a behavior built-in that allows for easy reconstruction of the array when, as posed in the question, an item should be removed.
# let's set up an array of items to consume:
x=()
for (( i=0; i<10; i++ )); do
x+=("$i")
done
# here, we consume that array:
while (( ${#x[#]} )); do
i=$(( $RANDOM % ${#x[#]} ))
echo "${x[i]} / ${x[#]}"
x=("${x[#]:0:i}" "${x[#]:i+1}")
done
Notice how we constructed the array using bash's x+=() syntax?
You could actually add more than one item with that, the content of a whole other array at once.
In ZSH this is dead easy (note this uses more bash compatible syntax than necessary where possible for ease of understanding):
# I always include an edge case to make sure each element
# is not being word split.
start=(one two three 'four 4' five)
work=(${(#)start})
idx=2
val=${work[idx]}
# How to remove a single element easily.
# Also works for associative arrays (at least in zsh)
work[$idx]=()
echo "Array size went down by one: "
[[ $#work -eq $(($#start - 1)) ]] && echo "OK"
echo "Array item "$val" is now gone: "
[[ -z ${work[(r)$val]} ]] && echo OK
echo "Array contents are as expected: "
wanted=("${start[#]:0:1}" "${start[#]:2}")
[[ "${(j.:.)wanted[#]}" == "${(j.:.)work[#]}" ]] && echo "OK"
echo "-- array contents: start --"
print -l -r -- "-- $#start elements" ${(#)start}
echo "-- array contents: work --"
print -l -r -- "-- $#work elements" "${work[#]}"
Results:
Array size went down by one:
OK
Array item two is now gone:
OK
Array contents are as expected:
OK
-- array contents: start --
-- 5 elements
one
two
three
four 4
five
-- array contents: work --
-- 4 elements
one
three
four 4
five
To avoid conflicts with array index using unset - see https://stackoverflow.com/a/49626928/3223785 and https://stackoverflow.com/a/47798640/3223785 for more information - reassign the array to itself: ARRAY_VAR=(${ARRAY_VAR[#]}).
#!/bin/bash
ARRAY_VAR=(0 1 2 3 4 5 6 7 8 9)
unset ARRAY_VAR[5]
unset ARRAY_VAR[4]
ARRAY_VAR=(${ARRAY_VAR[#]})
echo ${ARRAY_VAR[#]}
A_LENGTH=${#ARRAY_VAR[*]}
for (( i=0; i<=$(( $A_LENGTH -1 )); i++ )) ; do
echo ""
echo "INDEX - $i"
echo "VALUE - ${ARRAY_VAR[$i]}"
done
exit 0
[Ref.: https://tecadmin.net/working-with-array-bash-script/ ]
How about something like:
array=(one two three)
array_t=" ${array[#]} "
delete=one
array=(${array_t// $delete / })
unset array_t
#/bin/bash
echo "# define array with six elements"
arr=(zero one two three 'four 4' five)
echo "# unset by index: 0"
unset -v 'arr[0]'
for i in ${!arr[*]}; do echo "arr[$i]=${arr[$i]}"; done
arr_delete_by_content() { # value to delete
for i in ${!arr[*]}; do
[ "${arr[$i]}" = "$1" ] && unset -v 'arr[$i]'
done
}
echo "# unset in global variable where value: three"
arr_delete_by_content three
for i in ${!arr[*]}; do echo "arr[$i]=${arr[$i]}"; done
echo "# rearrange indices"
arr=( "${arr[#]}" )
for i in ${!arr[*]}; do echo "arr[$i]=${arr[$i]}"; done
delete_value() { # value arrayelements..., returns array decl.
local e val=$1; new=(); shift
for e in "${#}"; do [ "$val" != "$e" ] && new+=("$e"); done
declare -p new|sed 's,^[^=]*=,,'
}
echo "# new array without value: two"
declare -a arr="$(delete_value two "${arr[#]}")"
for i in ${!arr[*]}; do echo "arr[$i]=${arr[$i]}"; done
delete_values() { # arraydecl values..., returns array decl. (keeps indices)
declare -a arr="$1"; local i v; shift
for v in "${#}"; do
for i in ${!arr[*]}; do
[ "$v" = "${arr[$i]}" ] && unset -v 'arr[$i]'
done
done
declare -p arr|sed 's,^[^=]*=,,'
}
echo "# new array without values: one five (keep indices)"
declare -a arr="$(delete_values "$(declare -p arr|sed 's,^[^=]*=,,')" one five)"
for i in ${!arr[*]}; do echo "arr[$i]=${arr[$i]}"; done
# new array without multiple values and rearranged indices is left to the reader
Imagine I created an array like this:
IFS="|" read -ra ARR <<< "zero|one|||four"
now
echo ${#ARR[#]}
> 5
echo "${ARR[#]}"
> zero one four
echo "${ARR[0]}"
> zero
echo "${ARR[2]}"
> # Nothing, because it is empty
The question is how can I replace the empty elements with another string?
I have tried
${ARR[#]///other}
${ARR[#]//""/other}
none of them worked.
I want this as output:
zero one other other four
To have the shell expansion behave, you need to loop through its elements and perform the replacement on each one of them:
$ IFS="|" read -ra ARR <<< "zero|one|||four"
$ for i in "${ARR[#]}"; do echo "${i:-other}"; done
zero
one
other
other
four
Where:
${parameter:-word}
If parameter is unset or null, the expansion of word is substituted. Otherwise, the value of parameter is substituted.
To store them in a new array, just do so by appending with +=( element ):
$ new=()
$ for i in "${ARR[#]}"; do new+=("${i:-other}"); done
$ printf "%s\n" "${new[#]}"
zero
one
other
other
four
If you want to replace all empty values (actually modifying the list), you could do this :
for i in "${!ARR[#]}" ; do ARR[$i]="${ARR[$i]:-other}"; done
Which looks like this when indented (more readable I would say) :
for i in "${!ARR[#]}"
do
ARR[$i]="${ARR[$i]:-other}"
done
# Temporary array initialization
NEW=()
# Loop over the array, add only non-empty values to the new array
for i in "${ARR[#]}"; do
# Skip null items
if [ -z "$i" ]; then
continue
fi
# Add the rest of the elements to an array
NEW+=("${i}")
done
# Reinitialize your array
ARR=(${NEW[#]})
I have the following sample code for my shell script:
#!/bin/bash
x[1,1]=0
x[2,1]=1
echo "x[1,1]=${x[1,1]}"
echo "x[2,1]=${x[2,1]}"
for i in {1..2}; do
x[$i,1]=${i}
echo "loop$i x[$i,1]=${i}"
done
echo "x[1,1]=${x[1,1]}"
echo "x[2,1]=${x[2,1]}"
and I am expecting for x[1,1] to have the value of 1 and x[2,2] to have the value of 2.
But when I run the script the result is:
$ ./test3.sh
x[1,1]=1
x[2,1]=1
loop1 x[1,1]=1
loop2 x[2,1]=2
x[1,1]=2
x[2,1]=2
I expect x[1,1] to retain the value of 1 but it happens to be 2 now. Is there something wrong with my script?
Bash does not have 2-D arrays. The best you can do is emulate them with associative arrays.
Add the following line to the beginning of your script:
declare -A x
This makes x into an associative array. When that is done, the script produces the output that you expect:
$ bash script
x[1,1]=0
x[2,1]=1
loop1 x[1,1]=1
loop2 x[2,1]=2
x[1,1]=1
x[2,1]=2
Bash indexed arrays
Unless declare -A is used, a bash array is just an indexed array. Let's define y as an indexed array:
$ y=()
Now, let's assign two values:
$ y[2,3]=1
$ y[22,3]=2
Now, let's use declare -p to find out what the contents of the array really are:
$ declare -p y
declare -a y='([3]="2")'
As you can see, there is only y[3]. The reason is that the index in an indexed array is subject to arithmetic expansion and, when given a list of comma-separated values, arithmetic expansion returns just the last one.
In other words, as far as bash is concerned, assignments to y[2,3] and y[22,3] are both just assignments to y[3]. The second assignment overwrites the first.
We can see this directly if we echo the results of arithmetic expansion:
$ echo $((3))
3
$ echo $((2,3))
3
$ echo $((22,3))
3
When given a list of comma-separated values, arithmetic expansion returns the last one. This is true even if the comma-separated list is a long one:
$ echo $((1+2,3*4,5,6,7,8))
8
It is always the last value which is returned.
Bash associative arrays
Let's examine what happens with associative arrays. Let's define z as an associative array and assign some values to it:
$ declare -A z
$ z[1,2]=1
$ z[3,4]=2
$ z["Jim Bob"]=3
Now, let's see what was stored in z:
$ declare -p z
declare -A z='([3,4]="2" ["Jim Bob"]="3" [1,2]="1" )'
This seems to be what you need.
The code
SourceFolder[0]=""
SourceFolder[1]="inbound2"
SourceFolder[2]="inbound3"
for i in "${!SourceFolder[#]}"
do
if [ -z "${SourceFolder[$i]}"]; then
${SourceFolder[$i]} = "TEST"
fi
done
${SourceFolder[$i]} = "TEST" - doesn't work
it says
=: command not found
How to change value in current loop index in an array?
Because of the first space = is not interpreted as an assignment. There is a full explanation on So.
Btw ${SourceFolder[$i]} evaluate the array element, which is not what you want to do. For instance for the first one it is the empty string.
Replaces with SourceFolder[$i]=
You must change indexnumber in the your array:
ARRAYNAME[indexnumber]=value
ok, you have array is:
array=(one two three)
you can add count to your script for initialize and change element for array indexnumber, example:
#!/bin/bash
count=0
array=(one two three)
for i in ${array[#]}
do
echo "$i"
array[$count]="$i-indexnumber-is-$count"
count=$((count + 1))
echo $count
done
echo ${array[*]}
Result:
bash test-arr.sh
one
1
two
2
three
3
one-indexnumber-is-0 two-indexnumber-is-1 three-indexnumber-is-2
Assigning arrays to variables in Bash script seems rather complicated:
a=("a" "b" "c")
b=$a
echo ${a[0]}
echo ${a[1]}
echo ${b[0]}
echo ${b[1]}
leads to
a
b
a
instead of
a
b
a
b
Why? How can I fix it?
If you want to copy a variable that holds an array to another name, you do it like this:
a=('a' 'b' 'c')
b=( "${a[#]}" )
Why?
If a is an array, $a expands to the first element in the array. That is why b in your example only has one value. In bash, variables that refer to arrays aren't assignable like pointers would be in C++ or Java. Instead variables expand (as in Parameter Expansion) into strings and those strings are copied and associated with the variable being assigned.
How can I fix it?
To copy a sparse array that contains values with spaces, the array must be copied one element at a time by the indices - which can be obtained with ${!a[#]}.
declare -a b=()
for i in ${!a[#]}; do
b[$i]="${a[$i]}"
done
From the bash man page:
It is possible to obtain the keys (indices) of an array as well as the values.
${!name[#]} and ${!name[*]} expand to the indices assigned in array variable name.
The treatment when in double quotes is similar to the expansion of the special
parameters # and * within double quotes.
Here's a script you can test on your own:
#!/bin/bash
declare -a a=();
a[1]='red hat'
a[3]='fedora core'
declare -a b=();
# Copy method that works for sparse arrays with spaces in the values.
for i in ${!a[#]}; do
b[$i]="${a[$i]}"
done
# does not work, but as LeVar Burton says ...
#b=("${a[#]}")
echo a indicies: ${!a[#]}
echo b indicies: ${!b[#]}
echo "values in b:"
for u in "${b[#]}"; do
echo $u
done
Prints:
a indicies: 1 3
b indicies: 1 3 # or 0 1 with line uncommented
values in b:
red hat
fedora core
This also works for associative arrays in bash 4, if you use declare -A (with capital A instead of lower case) when declaring the arrays.