Regex currency Python 3.5 - regex

Question

I am trying to reformat the Euro currency in text data. The original format is like this: EUR 3.000.00 or also EUR 33.540.000.- .
I want to standardise the format to €3000.00 or €33540000.00.
I have reformatted EUR 2.500.- successfully using this code:
import re
format1 = "a piece of text with currency EUR 2.500.- and some other information"
regexObj = re.compile(r'EUR\s*\d{1,3}[.](\d{3}[.]-)')
text1 = regexObj.sub(lambda m:"\u20ac"+"{:0.2f}".format(float(re.compile('\d+(.\d+)?(\.\d+)?').search(m.group().replace('.','')).group())),format1)
Out: "a piece of text with currency €2500.00 and some other information"
This gives me €2500.00 which is correct. I've tried applying the same logic to the other formats to no avail.
format2 = "another piece of text EUR 3.000.00 and EUR 5.000.00. New sentence"
regexObj = re.compile('\d{1,3}[.](\d{3}[.])(\d{2})?')
text2 = regexObj.sub(lambda m:"\u20ac"+"{:0.2f}".format(float(re.compile('\d+(.\d+)?(\.\d+)?').search(m.group().replace('.','')).group())),format2)
Out: "another piece of text EUR €300000.00 and EUR €500000.00. New sentence"
and
format3 = "another piece of text EUR 33.540.000.- and more text"
regexObj = regexObj = re.compile(r'EUR\s*\d{1,3}[.](\d{3}[.])(\d{3}[.])(\d{3}[.]-)')
text3 = regexObj.sub(lambda m:"\u20ac"+"{:0.2f}".format(float(re.compile('\d+(.\d+)?(.\d+)?').search(m.group().replace('.','')).group())),format3)
Out: "another piece of text EUR 33.540.000.- and more text"
I think the problem might be with the regexObj.sub(), as the .format() part of it is confusing me. I've tried to change re.compile('\d+(.\d+)?(.\d+)?') within that, but I can't seem to generate the result I want. Any ideas much appreciated. Thanks!

Answer 1

Let's start with the regex. My propositions is:
EUR\s*(?:(\d{1,3}(?:\.\d{3})*)\.-|(\d{1,3}(?:\.\d{3})*)(\.\d{2}))
Details:
EUR\s* - The starting part.
(?: - Start of a non-capturing group - a container for alternatives.
( - Start of a capturing group #1 (Integer part with ".-" instead of
the decimal part).
\d{1,3} - Up to 3 digits.
(?:\.\d{3})* - ".ddd" part, 0 or more times.
) - End of group #1.
\.- - ".-" ending.
| - Alternative separator.
( - Start of a capturing group #2 (Integer part)
\d{1,3}(?:\.\d{3})* - Like in alternative 1.
) - End of group #2.
(\.\d{2}) - Capturing group #3 (dot and decimal part).
) - End of the non-capturing group.
Instead of a lambda function I used "ordinary" replicating function,
I called it repl. It contains 2 parts, for group 1 and group 2 + 3.
In both variants dots from the integer part are deleted, but the "final"
dot (after the integer part) is a part of group 3, so it is not deleted.
So the whole script can look like below:
import re
def repl(m):
g1 = m.group(1)
if g1: # Alternative 1: ".-" instead of decimal part
res = g1.replace('.','') + '.00'
else: # Alternative 2: integet part (group 2) + decimal part (group 3)
res = m.group(2).replace('.','') + m.group(3)
return "\u20ac" + res
# Source string
src = 'xxx EUR 33.540.000.- yyyy EUR 3.000.00 zzzz EUR 5.210.15 vvvv'
# Regex
pat = re.compile(r'EUR\s*(?:(\d{1,3}(?:\.\d{3})*)\.-|(\d{1,3}(?:\.\d{3})*)(\.\d{2}))')
# Replace
result = pat.sub(repl, src)
The result is:
xxx €33540000.00 yyyy €3000.00 zzzz €5210.15 vvvv
As you can see, no need to use float or format.