Comparing two different arrays using shell script - shell

Question

How do we compare two arrays and display the result in a shell script?
Suppose we have two arrays as below:
list1=( 10 20 30 40 50 60 90 100 101 102 103 104)
list2=( 10 20 30 40 50 60 70 80 90 100 )
My requirement is to compare these two arrays in an order that it will only display the result as (101 102 103 104) from list1. It should not include the values 70 and 80 which are present in list2 but not in list1.
This does not help since it is including everything:
echo "${list1[#]}" "${list2[#]}" | tr ' ' '\n' | sort | uniq -u
I tried something like this below, but why is it not working?
list1=( 10 20 30 40 50 60 70 90 100 101 102 103 104)
list2=( 10 20 30 40 50 60 70 80 90 100 )
for (( i=0; i<${#list1[#]}; i++ )); do
for (( j=0; j<${#list2[#]}; j++ )); do
if [[ ${list1[#]} == ${list2[#] ]]; then
echo 0
break
if [[ ${#list2[#]} == ${#list1[#]-1} && ${list1[#]} != ${list2[#]} ]];then
echo ${list3[$i]}
fi
fi
done
done

Answer 1

You can use comm for this:
readarray -t unique < <( \
comm -23 \
<(printf '%s\n' "${list1[#]}" | sort) \
<(printf '%s\n' "${list2[#]}" | sort) \
)
resulting in
$ declare -p unique
declare -a unique=([0]="101" [1]="102" [2]="103" [3]="104")
or, to get your desired format,
$ printf '(%s)\n' "${unique[*]}"
(101 102 103 104)
comm -23 takes two sorted files (using sort here) and prints every line that is unique to the first one; process substitution is used to feed the lists into comm.
Then, readarray reads the output and puts each line into an element of the unique array. (Notice that this requires Bash.)
Your attempt failed, among other things, because you were trying to compare multiple elements in a single comparison:
[[ ${list1[#]} != ${list2[#]} ]]
expands to
[[ 10 20 30 40 50 60 90 100 101 102 103 104 != 10 20 30 40 50 60 70 80 90 100 ]]
and Bash complains about a binary operator expected instead of the second element, 20.

Answer 2

Could also use this kind of approach
#!/bin/ksh
list1=( 10 20 30 40 50 60 90 100 101 102 103 104 )
list2=( 10 20 30 40 50 60 70 80 90 100 )
# Creating a temp array with index being the same as the values in list1
for i in ${list1[*]}; do
list3[$i]=$i
done
# If value of list2 can be found in list3 forget this value
for j in ${list2[*]}; do
if [[ $j -eq ${list3[$j]} ]]; then
unset list3[$j]
fi
done
# Print the remaining values
print ${list3[*]}
Output is
101 102 103 104
Hope it could help
EDIT
In case of the 2 list are the same :
# Print the remaining values
if [[ ${#list3[*]} -eq 0 ]]; then
print "No differences between the list"
else
print ${list3[*]}
fi

Answer 3

ksh associative arrays are handy for this:
list1=( 10 20 30 40 50 60 90 100 101 102 103 104)
list2=( 10 20 30 40 50 60 70 80 90 100 )
typeset -a onlyList1
typeset -A inList2
for elem in "${list2[#]}"; do inList2["$elem"]=1; done
for elem in "${list1[#]}"; do [[ -v inList2["$elem"] ]] || onlyList1+=("$elem"); done
typeset -p onlyList1
typeset -a onlyList1=(101 102 103 104)
Or similarly, start with all of list1 and remove what's in list2:
typeset -A inList1
for elem in "${list1[#]}"; do inList1["$elem"]=1; done
for elem in "${list2[#]}"; do unset inList1["$elem"]; done
onlyList1=( "${!inList1[#]}" )