Replace placeholders in string with replacements sequence

Question

I have a location string with placeholders, used as '#'. Another string which are replacements for the placeholders. I want to replace them sequentially, (like format specifiers). What is the way to do it in Python?

location = '/tmp/#/dir1/#/some_dirx/dir/var/2/#/dir3'
replacements = 'xyz'
result = '/tmp/x/dir1/y/some_dirx/dir/var/2/z/dir3'

• Prune: If you search in your browser for "Python string replace", you'll find references that can explain this much better than we can manage here.
• soap: I am not using the same replacement string for every replacement.
• Prune: Understood -- you replace each occurrence in turn. Where are you stuck? You've shown no attempt to solve the problem.
• Dani Mesejo: Where did somedir go? I mean the expected output does not match your the definition
• soap: My bad, corrected that.

Answer 1

You should use the replace method of a string as follows:

for replacement in replacements:
 location = location.replace('#', replacement, 1)

It is important you use the third argument, count, in order to replace that placeholder just once. Otherwise, it will replace every time you find your placeholder.

Answer 2

If your location string does not contains format specifiers ({}) you could do:

location = '/tmp/#/dir1/#/some_dirx/dir/var/2/#/dir3'
replacements='xyz'
print(location.replace("#", "{}").format(*replacements))

Output

/tmp/x/dir1/y/some_dirx/dir/var/2/z/dir3

As an alternative you could use the fact that repl in re.sub can be a function:

import re
from itertools import count
location = '/tmp/#/dir1/#/some_dirx/dir/var/2/#/dir3'
def repl(match, replacements='xyz', index=count()):
 return replacements[next(index)]
print(re.sub('#', repl, location))

Output

/tmp/x/dir1/y/some_dirx/dir/var/2/z/dir3