numbers as command-line arguments and prints the count

user_17266973 0
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I want to take numbers as command-line arguments and prints the count of numbers that end with 0, 1, 2, etc. up to 5.

Example:

bash test.sh 12 14 12 15 14

Expected Output:

Digit_ends_with   count
0                  0
1                  0
2                  2
3                  0
4                  2
5                  1

Mt Attempt:

read -a integers                                                                                                                                                                                                                                for i in ${integers[@]}                                                                                                 do                                                                                                                              if [[ grep -o '[0-9]' $i ]]                                                                                                     count=$(grep -c $i)                                                                                                     if [ "$count" -ge "0" ]                                                                                                 then 
echo "Digit_ends_with" $i                                                                                                                           echo -e "Count ""$count"                                                                                                fi                                                                                                              fi                                                                                                              done

But this is not working. How I can achieve this requirement?

shell bash
2 Answers
#!/bin/bash
echo "Digit_ends_with count"                                #print table header
for argument in "$@"                                        #loop over all given arguments
do 
    current=$(echo "$argument"  | tail -c 2 )               #get last digit
    if ((  "$current" <= 5 ))                               #check if lower than 6
    then
        echo "$current"                                     #echo if true
    fi
done | sort | uniq -c | sed -E 's/\s+//' | sed -E 's/([0-9]+).?([0-9]+)/\2\t\t\1/'  #sort, count, remove leading spaces and switch the fields

Example:

ā•°ā”€$ ./test.sh 188 182 182 12 13 14 18 15 16 17 18 19 10 0 0 0 0 0 0 0 0 0 0
Digit_ends_with count
0               11
2               3
3               1
4               1
5               1
user_150876220

Would you please try the following:

#!/bin/bash

for i in "$@"; do                               # loop over arguments
    (( count[i % 10]++ ))                       # index with the modulo 10
done

printf "%s %s\n" "Digit_ends_with" "count"      # header line
for (( i = 0; i < 6; i++ )); do                 # loop between 0 and 5
    printf "%d\t\t%d\n" "$i" "${count[$i]}"     # print the counts
done

Result of ./test.sh 12 14 12 15 14:

Digit_ends_with count
0               0
1               0
2               2
3               0
4               2
5               1
user_85723800
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