arrays - Bash:计算关联数组中键的总数？

Question

考虑下面的关联数组:

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)

有没有办法找到数组中每个条目使用的键总数？ (这是数组中的每个“索引”吗？)
例如，如何将:[start], [stop], [anotherkey] 计算为= 3 个键？
目前，我正在使用此代码中的硬编码值 (3)，我发现(如下所示)可以很好地完成工作，但我想知道这是否可以动态实现？

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

我发现这些变量返回数组的各个方面，但不是键的总数。

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:" #???

所需输出:

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

输出:

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

输出:

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[@]}"

输出:
3