bash array leave elements containing string - arrays

I will be short, what i have is
array=( one.a two.b tree.c four.b five_b_abc)
I want this
array=( two.b four.b five_b_abc )
from here i found this
# replace any array item matching "b*" with "foo"
array=( foo bar baz )
array=( "${array[#]/%b*/foo}" )
echo "${orig[#]}"$'\n'"${array[#]}"
how ever this does not work
array2=( ${array[#]//%^.p/})
result array2=array
this deletes all with p
array2=(${array[#]/*p*/})
result array2=( one.a tree.c )
I need an idea how to do add ^p (all accept p), and get my solution
array2=(${array[#]/*^p*/}
Its a rather large array about 10k elements where i need to do this i need it to be as fast as possible with data so please no loop solutions.

Edit: added timing comparisons (at end) and got rid of the tr
It is possible to replace the content of array elements using bash Parameter expansion, ie. ${var[#]....} but you cannot actually delete the elements. The best you can get with parameter expansion is to make null ("") those elements you do not want.
Instead, you can use printf and sed, and IFS. This has the advantabe of allowing you to use full regular expression syntax (not just the shell globbing expressions)... and it is much faster than using a loop...
This example leaves array elements which contain c
Note: This method caters for spaces in data. This is achieved via IFS=\n
IFS=$'\n'; a=($(printf '%s\n' "${a[#]}" |sed '/c/!d'))
Here it is again with before/after dumps:
#!/bin/bash
a=(one.ac two.b tree.c four.b "five b abcdefg" )
echo "======== Original array ===="
printf '%s\n' "${a[#]}"
echo "======== Array containing only the matched elements 'c' ===="
IFS=$'\n'; a=($(printf '%s\n' "${a[#]}" |sed '/c/!d'))
printf '%s\n' "${a[#]}"
echo "========"
Output:
======== Original array ====
one.ac
two.b
tree.c
four.b
five b abcdefg
======== Array containing only the matched elements 'c' ====
one.ac
tree.c
five b abcdefg
========
For general reference: Testing an array containing 10k elements. selecting 5k:
a=( a\ \ \ \ b{0..9999} )
 The printf method took: 0m0.226s (results in sequential index values)
    The first loop method: 0m4.007s (it leaves gaps in index values)
The second loop method: 0m7.862s (results in sequential index values)
printf method:
IFS=$'\n'; a=($(printf '%s\n' "${a[#]}" |sed '/.*[5-9]...$/!d'))
1st loop method:
iz=${#a[#]}; j=0
for ((i=0; i<iz; i++)) ;do
[[ ! "${a[i]}" =~ .*[5-9]...$ ]] && unset a[$i]
done
2nd loop method:
iz=${#a[#]}; j=0
for ((i=0; i<iz; i++)) ;do
if [[ ! "${a[i]}" =~ .*[5-9]...$ ]] ;then
unset a[$i]
else
a[$j]="${a[i]}=$i=$j"; ((j!=i)) && unset a[$i]; ((j+=1));
fi
done

You can try this:
array2=(`echo ${array[#]} | sed 's/ /\n/g' | grep b`)

If you set IFS to $'\n', then you could use the echo command on the array with an '*' as the subscript, instead of printf with an '#':
IFS=$'\n'; a=($(echo "${a[*]}" | sed '/.*[5-9]...$/!d'))
The '*' subscript concatenates the array elements together, separated by the IFS.
(I have only recently learned this myself, by the way.)

I know that it is a bash question, but I used a python script to make it one liner and easy to read.
$ array=( one.a two.b tree.c four.b five_b_abc)
$ printf '%s\n' "${array[#]}" | python3 -c "import sys; print(' '.join([ s.strip() for s in sys.stdin if 'b' in s ]));"
two.b four.b five_b_abc

Related

Does array 1 contain any of the strings in array 2

I am trying to make an if statement where if array1 contains any of the strings in array2 it should print "match" else print "no match"
So far I have the following. Not sure how to complete it. Both loops should break as soon as the first match is found.
#!/bin/bash
array1=(a b c 1 2 3)
array2=(b 1)
for a in "${array1[#]}"
do
for b in "${array2[#]}"
do
if [ "$a" == "$b" ]; then
echo "Match!"
break
fi
done
done
Maybe this isn't even the best way to do it?
This illustrates the desired result
if [ array1 contains strings in array2 ]
then
echo "match"
else
echo "no match"
fi
To check whether array1 contains any entry from array2 you can use grep. This will be way faster and shorter than loops in bash.
The following commands exit with status code 0 if and only if there is a match. Use them as ...
if COMMAND FROM BELOW; then
echo match
else
echo no match
fi
Single-Line Array Entries
The simple version for strings without linebreaks is
printf %s\\n "${array1[#]}" | grep -qFxf <(printf %s\\n "${array2[#]}")
Multiline Array Entries
Sadly there doesn't seem to be a straightforward way to make this work for array entries with linebreaks. GNU grep has the option -z to set the "line" delimiters in the input to null, but apparently no option to do the same for the file provided to -f. Listing the entries from array2 as -e arguments to grep is not working either -- grep -F seems to be unable to match multiline patterns. However, we can use the following hack:
printf %q\\n "${array1[#]}" | grep -qFxf <(printf %q\\n "${array2[#]}")
Here we assume that bash's built-in printf %q always prints a unique single line -- which it currently does. However, future implementations of bash may change this. The documentation help printf only states that the output thas to be correctly quoted for bash.
For a fast solution, you're better off using an external tool that can process the entire array as a whole (such as the grep-based answers). Doing nested loops in pure bash is likely to be slower for any substantial amount of data (where the item-by-item processing in bash is likely to be more expensive than the external process start-up time).
However, if you do need a pure bash solution, I see that your current solution has no way to print out the "no match" scenario. In addition, it may print out "match" multiple times.
To fix that, you can just store the fact that a match has been found, and use that to both:
exit the outer loop early as well as the inner loop; and
print the correct string at the end.
To do this, you can use something like:
#!/bin/bash
# Test data.
array1=(a b c 1 2 3)
array2=(b 1)
# Default to not-found state.
foundMatch=false
for a in "${array1[#]}" ; do
for b in "${array2[#]}" ; do
# Any match switches to found state and exits inner loop.
[[ "$a" == "$b" ]] && foundMatch=true && break
done
# If found, exit outer loop as well.
${foundMatch} && break
done
# Output appropriate message for found/not-found state.
$foundMatch && echo "Match" || echo "No match"
For array elements which does not contain newlines, the grep -qf with printf "%s\n" would be a good option. For comparing arrays with any elements, I ended with this:
cmp -s /dev/null <(comm -z12 <(printf "%s\0" "${array1[#]}" | sort -z) <(printf "%s\0" "${array2[#]}" | sort -z))
The printf "%s\0" "${array[#]}" | sort -z print a sorted list of zero terminated array elements. The comm -z12 then extracts common elements in both lists. The cmp -s /dev/null checks if the output of comm is empty, which will not be empty if any element is in both lists. You could use [ -z "$(comm -z ...)" ] to check if the output of comm would be empty, but bash will complain that the output of a command captured with $(..) contains a null byte, so it's better to cmp -s /dev/null.
I think | is faster then <(), so your if could be:
if ! printf "%s\0" "${array1[#]}" | sort -z |
comm -z12 - <(printf "%s\0" "${array2[#]}" | sort -z) |
cmp -s /dev/null -; then
echo "Some elements are in both array1 and array2"
fi
The following could work:
printf "%s\0" "${array1[#]}" | eval grep -qzFx "$(printf " -e %q" "${array2[#]}")"
But I believe I found a bug in grepv3.1 when matching a newline character with -x flag. If you don't use the newline character, the above line works.
Would you try the following:
array1=(a b c 1 2 3)
array2=(b 1)
declare -A seen # set marks of the elements of array1
for b in "${array2[#]}"; do
(( seen[$b]++ ))
done
for a in "${array1[#]}"; do
(( ${seen[$a]} )) && echo "match" && exit
done
echo "no match"
It may be efficient by avoiding the double loop, although the discussion of efficiency may be meaningless as long as using bash :)

How to replace a word from a file with elements in an array

I am trying to replace all the word "null" to elements in array. The problem is that after replacing one word of "null", I would like to replace the next "null" with next element in the array.
I am not very good with bash and I feel like this is quite a basic question.
Here is what I have so far:
for m in $(cat finalfile.csv)
do
if [ "$m" = "null" ]
then
m=cwearray[$counter]
let counter++
fi
done
This doesn't replace anything in the finalfile.csv.
For example if the file has:
"value1","value2","null","value3"\n
"value1","value2","null","value3"...
and the array has ["foo","bar"]
I would like it to be:
"value1","value2","foo","value3"\n
"value1","value2","bar","value3"...
can be done with bash, even with multiple nulls per line:
$ cat finalfile.csv
"value1","value2","null","null"
"value1","value2","null","value3"
$ cwearray=( foo bar baz )
$ idx=0
$ while read -r line; do
while [[ $line == *null* ]]; do
line=${line/null/${cwearray[idx++]}}
# ...............^^^^^^^^^^^^^^^^^^
# replace the _first_ "null" with the _next_ array element
done
echo "$line"
done < finalfile.csv > updatedfinalfile.csv
$ cat updatedfinalfile.csv
"value1","value2","foo","bar"
"value1","value2","baz","value3"
It's easier in Perl where you can increase the index directly in the replacement part of a substitution:
printf '%s\n' 1,2,3,null null,2,3,4 null,null,null,null \
| perl -pe 'BEGIN { #cwe = qw( A B C D E F ) }
s/(?:^|(?<=,))null(?=,|$)/$cwe[$i++]/g'
Update: It seems you've updated your question with a sample input. If nulls are double quoted, it gets even easier, as there's no need to check whether they're surrounded with commas or beginning/end of the line.
perl -pe 'BEGIN{ #cwe = qw( foo bar ) }
s/"null"/"$cwe[$i++]"/g'
An awk solution :
declare -a cwearray
cwearray=(foo bar)
awk -F, 'NR==FNR{repl[NR]=$0; next}{for(i=1;i<=NF;i++){if($i=="\"null\""){$i="\""repl[++counter]"\""}}}1' OFS="," <(for i in "${cwearray[#]}"; do echo "$i"; done) <file>
Read the file line by line. If a line contains null, then use sed to replace all occurrences of null with the corresponding value, retrieved via array index.
#!/bin/bash
file="finalfile.csv"
counter=0
array=(
"foo"
"bar"
)
while read -r line; do
item="${array[$counter]}"
echo "$line" | sed "s/null/$item/g"
((counter++))
done < "$file"

How do you unset all empty array elements in bash? [duplicate]

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

Bash: How to select elements in array

I have an array with subdirs and files:
~$ array=(elem1/a elem1/b elem2/a elem2/b)
I need to select all files (with its subdir) inside subdir elem1.
I've tried with a for loop and case:
for element in "${array[#]}"; do
case $element in
elem1*)
# action I need to perform on this element
echo "This is $element"
;;
esac
done
I'd like to get the list of elements I need using some glob, in a similar way as this (which substracts substrings from array elements):
~$ echo ${array[#]%/a}
elem1 elem1/b elem2 elem2/b
My main concern is that I shoudn't hardcode what strings I'm searching for, as I don't know its contents - in my final version of this script I'm getting them from another array.
My answer prints every element of the array in a different line, then filters the ones you are interested (grep) and put them into a new array
newarray=( $(printf "%s\n" "${array[#]}" | grep ^elem/) )
The only problem would be if the elements of your array (subdirs and file namews) might contain EOL or blank chars.
Bonus
If your elements contain EOL or blank chars, you should use the NUL (\000) char and grep -zZ to handle it; but then you need more code to turn out those NUL-char strings into something which the bash can work with -- the bash stops reading when it finds a NUL char.
In this example I use awk, which has no problem to handle NUL chars, and makes unnecessary the use of grep:
eval newarray=( $(printf "%s\000" "${A[#]}" |
awk 'BEGIN { FS="\000" }
/^elem\// {
gsub("\042","\\\042")
printf("[" NR-1 "]=\042%s\042 ",$0)
}'
)
)
You can filter your array with a variable query (prefix below) and store the results in a new array (result):
#!/bin/bash
array=(elem1/a elem1/b elem2/a elem2/b)
prefix=elem1
result=()
for element in "${array[#]}"; do
[[ $element == $prefix/* ]] && result+=("$element")
done
The result is:
$ declare -p result
declare -a result='([0]="elem1/a" [1]="elem1/b")'
here is one alternative
$ for e in "${array[#]}";
do case "${e%%/*}" in
elem1) echo "$e is in elem1" ;;
esac
done
elem1/a is in elem1
elem1/b is in elem1

Appending a character to every element except the last one of a bash array

I am looking to append each element of a bash array with | except the last one.
array=("element1" "element2" "element3")
My desired output would be
array=("element1"|"element2"|"element3")
What I have done
for i in ${!array[#]};
do
array1+=( "${array[$i]}|" )
done
Followed by
array=echo ${array1[#]}|sed 's/|$//'
Is there any other looping approach I can use that only appends the character until the last but one element?
For your special case, where only one single character has to be inserted between each array member, the simplest solution is probably to use array expansion (and changing the separation character IFS according to your need beforehand):
$ array=("element1" "element2" "element3")
$ array=( $( IFS="|" ; echo "${array[*]}") )
$ echo "\$array[0] is '${array[0]}'"
$array[0] is 'element1|element2|element3'
You can use:
# append | all except last element
read -ra array < <( printf "%s|" "${array[#]:0:$((${#array[#]} - 1))}"; echo "${array[#]: -1}"; )
# Check array content now
declare -p array
declare -a array='([0]="element1|element2|element3")'
"${array[#]:0:$((${#array[#]} - 1))}" will get all but last element of array
"${array[#]: -1}" will get the last element of array
printf "%s|" will append | in front of all the arguments
< <(...) is process substitution to read output of any command from stdin
read -ra will read the input in an array
$ array=("element1" "element2" "element3")
$ printf -v str "|%s" "${array[#]}"
$ array=("${str:1}")
$ declare -p array
declare -a array='([0]="element1|element2|element3")'
The printf statement creates a string str that contains |element1|element2|element3, i.e., one more | than we want (at the beginning).
The next statement uses substring parameter expansion, ${str:1}, to skip the first character and reassigns to array, which now consists of a single element.
A simple solution (if you don't mind changing IFS) is:
$ array=("element1" "element2" "element3")
$ IFS="|"; printf "%s\n" "${array[*]}"
And to re-assign to the variable array (which doesn't change IFS):
$ array=("$(IFS="|"; printf "%s\n" "${array[*]}")")
$ printf '%s\n' "${array[#]}"
element1|element2|element3
An alternative solution is:
$ array=($(printf '%s|' "${array[#]}")); array="${array%?}"
$ printf '%s\n' "${array}"
A more complex solution (script) is:
array=("element1" "element2" "element3")
delimiter='|'
unset newarr
for val in "${array[#]}"
do newarr=$newarr${newarr+"$delimiter"}$val
done
array=("$newarr")
echo "array=($array)"

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