Bash: Count total number of keys in an associative array?

Question

Consider the associative array below:

declare -A shapingTimes
 shapingTimes=([0-start]=15 [0-stop]=21 [0-anotherkey]=foo)
shapingTimes+=([1-start]=4  [1-stop]=6  [1-anotherkey]=bar)
shapingTimes+=([2-start]=9  [2-stop]=11 [2-anotherkey]=blah)

Is there a way to find the total number of keys used per entry in an array? (Is this per 'index' in an array?)

For example, how to count: [start], [stop], [anotherkey] as = 3 keys?

At the moment I'm using the hardcoded value (3) from this code I found (as below) that does the job fine, but I'm wondering if this can be achieved dynamically?

totalshapingTimes=$((${#shapingTimes[*]} / 3))

I've found these variables that return various array aspects, but not the total number of keys.

echo "All of the items in the array:" ${shapingTimes[*]}
echo "All of the indexes in the array:" ${!shapingTimes[*]}
echo "Total number of items in the array:" ${#shapingTimes[*]}
echo "Total number of KEYS in each array entry:" #???

Desired output:

All of the items in the array: 21 6 11 blah 15 4 bar 9 foo
All of the indexes in the array: 0-stop 1-stop 2-stop 2-anotherkey 0-start 1-start 1-anotherkey 2-start 0-anotherkey
Total number of items in the array: 9
Total number of KEYS in each array entry: 3

Answer 1

declare -A shapingTimes
shapingTimes=([0-start]=15 [0-stop]=21 [0-anotherkey]=foo)
shapingTimes+=([1-start]=4  [1-stop]=6  [1-anotherkey]=bar)
shapingTimes+=([2-start]=9  [2-stop]=11 [2-anotherkey]=blah)

# output all keys
for i in "${!shapingTimes[@]}"; do echo $i; done

Output:

1-start
2-stop
1-stop
0-start
2-start
2-anotherkey
1-anotherkey
0-stop
0-anotherkey

---

# Leading numbers and "-" removed:
for i in "${!shapingTimes[@]}"; do echo ${i/#[0-9]*-/}; done

Output:

start
stop
stop
start
start
anotherkey
anotherkey
stop
anotherkey

---

# put shortend keys in new associative array
declare -A hash
for i in "${!shapingTimes[@]}"; do hash[${i/#[0-9]*-/}]=""; done
echo "${#hash[@]}"

Output:

3