Adding zeros to a string without generating a new variable - string

Question

I am trying to add zeros to a string variable in such a way that all levels of the variables have same number of digits (assume 3).
clear
input tina bina str4 pine
1 10 "99"
1 11 "99"
2 11 "99"
2 11 "99"
3 12 "."
4 12 "888"
5 14 "88"
6 15 "777"
7 16 "77"
8 17 "0"
8 18 "7"
end
I managed to do this by generating a new variable which stores the number of digits I need to add to each observation in order to reach 3:
generate pi=3-strlen(pine)
replace pine= ("0"*pi) + pine if strlen(pine)<3
I wonder if there is a way to obtain the same result but without generating the variable?
I tried the following but it does not work :
replace pine= ("0"*(`=3-strlen(pine)')) + pine if strlen(pine)<3
Probably I am not so clear about what happens when I evaluate expressions.

Answer 1

Your approach does not work because it evaluates the expression for the first observation only:
. display `= 3 - strlen(pine)'
1
The single quotes are not required:
replace pine = ("0" * (3-strlen(pine) ) ) + pine if strlen(pine) < 3
+--------------------+
| tina bina pine |
|--------------------|
1. | 1 10 099 |
2. | 1 11 099 |
3. | 2 11 099 |
4. | 2 11 099 |
5. | 3 12 00. |
|--------------------|
6. | 4 12 888 |
7. | 5 14 088 |
8. | 6 15 777 |
9. | 7 16 077 |
10. | 8 17 000 |
|--------------------|
11. | 8 18 007 |
+--------------------+

Answer 2

I know there is already an accepted answer, but I wanted to throw out my suggestion. This is maybe a little bit simpler than the other answer and is straightforward to explain. You just want to replace a string variable of real numbers with leading zeros and keep it as a string. You can easily do this by running:
replace pine = string(real(pine),"%03.0f")
Depending on your goal this is maybe better than the previous answer, because it maintains your missing value as missing and not add zeros to it. Hopefully this helpful.