Replacing repeated elements in array

Question

I am new in bash scripting and I need help in my task... I have an array :

arr=( one two tree four five six seven one  four  nine one  two  one  ten )

I need to change it on the rule : If element repeated first time add 1 to the end, if twice - 2.

Expecting result:

arr=( one two tree four five six seven one1 four1 nine one2 two1 one3 ten )

My code :

for i in ${!arr[*]}
do
    k=1
    for j in ${!arr[*]}
        do
        if [[ ( ${arr[$i]} = ${arr[$j]} ) && ( $i > $j )  ]] ;then
            arr[$i]=$(echo ${arr[$j]}$k)
            ((j++))
            ((k++))
        fi
    done
    echo ${arr[$i]}
    ((i++))
done

Please, give me advice how to resolve this task...

Barmar · Accepted Answer · 2021-07-02 08:18:55Z

2

Use an associative array to hold the current count for each word.

declare -A count
for i in ${!arr[*]}
do
    if [[ -n "${count[${arr[$i]}]}" ]]
    then 
        ((count[${arr[$i]}]++))
        arr[$i]=${arr[$i]}${count[${arr[$i]}]}
    else
        count[${arr[$i]}]=0
    fi
done

edited Jul 2, 2021 at 8:18

answered Jul 1, 2021 at 18:57

Barmar

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8 Comments

QwertyBot Over a year ago

all count will be 1 in your solution

Barmar Over a year ago

Why? ((count[$arr[$i]]++)) increments the count each time it's used.

QwertyBot Over a year ago

when I make echo ${count[*]} it holds only 1 1 1 .....

Barmar Over a year ago

I fixed the problems.

KamilCuk Over a year ago

A shorter version: for i in ${!arr[*]}; do ((count[${arr[$i]}]++)) && arr[$i]+=$((count[${arr[$i]}]-1)); done

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