Regex for wrapping digits with curly braces?

I am trying to using Python's re.sub() to match a string with an e character and insert curly braces immediately after the e character and after the lastdigit. For example:

12.34e56 to 12.34e{56}
1e10 to 1e{10}

I can't seem to find the correct regex to insert the desired curly braces. For example, I can properly insert the left brace like this:

>>> import re
>>> x = '12.34e10'
>>> pattern = re.compile(r'(e)')
>>> sub = z = re.sub(pattern, "\1e{", x)
>>> print(sub)
    12.34e{10 # this is the correct placement for the left brace

My problem arises when using two back references.

>>> import re
>>> x = '12.34e10'
>>> pattern = re.compile(r'(e).+($)')
>>> sub = z = re.sub(pattern, "\1e{\2}", x)
>>> prin开发者_如何学编程t(sub)
    12.34e{} # this is not what I want, digits 10 have been removed

Can anyone point out my problem? Thanks for the help.

---

re.sub(r'e(\d+)', r'e{\1}', '12.34e56')

returns '12.34e{56}'

or, the same result but different logic (don't replace e with e):

re.sub(r'(?<=e)(\d+)', r'{\1}', '12.34e56')

---

Your brace placement is incorrect.

Here's a solution ensuring the that there's a number with optional decimal place before the e:

import re
samples = ['12.34e56','1e10']
for s in samples:
  print re.sub(r'(\d+(?:\.\d+)?)e([0-9]+)',"\g<1>e{\g<2>}",s)

Yields:

12.34e{56}
1e{10}