><> (Fish), 56 bytes

 i:0(?v:fg1+$fp
?\~1+a\~'!':fg:
.\{:o} 1-d1
)?;a1.\o:'~'

Try it

There is already a slightly shorter ><> answer but this one is mine and I'm proud of it.

><> (Fish), 53 50 bytes

l'~')?v0
v?(0:i<]r+1r[
~
r
>:?vao~l'!'(?;
/  \1-lo

Try it

This version actually completes before the end of the universe, which is nice. Uses the stack instead of the code to store how often each number appears.

><> (Fish), 53 50 46 bytes

l'~')?v0
v?(0:i<]r+1r[
\~rvol-1
^?:<;?('!'l~oa

Try it

Finally beat the current fish answer. Note the link pre-populates the stack to make it faster but this is unnecessary for the correctness of the program.

Each element of the stack stores how often the nth character appears in the input.

The top row pushes ord('~') 0s to the stack.

The second row reads the input. Use [r1+r] to get the nth element of the stack, add 1 to it, then put it back into place.

The left (red) half of the third row reverses the stack so the higher valued items are on the top. Now we know that the length of the stack is the character code of the count the top element represents. This allows very convenient printing.

The left (grey) half of the bottom row checks if the count of the nth character is 0. If so go to the right half of the third row.

The right (blue) half of the bottom row prints a line break ao. Then pops the top of the stack. If we now have less than ord('!') items on the stack only spaces remain and we exit (since we shouldn't print spaces)

The right half of the third row runs if there is more than one occurrence of the top character. If so, we subtract one occurrence, then print the length of the stack as a character.

><> (Fish), 56 bytes

 i:0(?v:fg1+$fp
?\~1+a\~'!':fg:
.\{:o} 1-d1
)?;a1.\o:'~'

Try it

There is already a slightly shorter ><> answer but this one is mine and I'm proud of it.

><> (Fish), 53 bytes

l'~')?v0
v?(0:i<]r+1r[
\~rl'!'(?;
>:  ?vao~22.
\ol-1/

Try it

This version actually completes before the end of the universe, which is nice. Uses the stack instead of the code to store how often each number appears.

><> (Fish), 56 bytes

 i:0(?v:fg1+$fp
?\~1+a\~'!':fg:
.\{:o} 1-d1
)?;a1.\o:'~'

Try it

There is already a slightly shorter ><> answer but this one is mine and I'm proud of it.

><> (Fish), 53 50 bytes

l'~')?v0
v?(0:i<]r+1r[
~
r
>:?vao~l'!'(?;
/  \1-lo

Try it

This version actually completes before the end of the universe, which is nice. Uses the stack instead of the code to store how often each number appears.

><> (Fish), 56 bytes

 i:0(?v:fg1+$fp
?\~1+a\~'!':fg:
.\{:o} 1-d1
)?;a1.\o:'~'

Try it

There is already a slightly shorter ><> answer but this one is mine and I'm proud of it.

><> (Fish), 56 bytes

 i:0(?v:fg1+$fp
?\~1+a\~'!':fg:
.\{:o} 1-d1
)?;a1.\o:'~'

Try it

There is already a slightly shorter ><> answer but this one is mine and I'm proud of it.

><> (Fish), 53 bytes

l'~')?v0
v?(0:i<]r+1r[
\~rl'!'(?;
>:  ?vao~22.
\ol-1/

Try it

This version actually completes before the end of the universe, which is nice. Uses the stack instead of the code to store how often each number appears.