Rounding using regular expressions

Question

Is it possible to use a regular expression to perform rounding on a number? The problem is that I've got a lot of numbers to more than 2 decimal places in a file and I need to move them to 2 decimal places.

It's in source file so ideally I'd like to use Visual Studio's find and replace, but I'm not against the idea of writing a script to run on the file (treating it as plain text) if regex doesn't have the answer.

Answer 1

You can't do it with regexes alone, as Ryan pointed out, but you can use regexes to find decimals in your input and use a callback function on the replace operation.

In Python:

>>> import re
>>> text = "1.234 and 99.999; .555 12.345"
>>> simpledec = re.compile(r"\d*\.\d+")
>>> def mround(match):
...     return "{:.2f}".format(float(match.group()))
...
>>> re.sub(simpledec, mround, text)
'1.23 and 100.00; 0.56 12.35'

The regex to match decimals is very simplistic; it doesn't match exponential notation, for example, but it should give you an idea.

Answer 2

Well... at least you can search for (\d+\.\d\d\d)\d* and replace it with round($1, 2).

Yeah, perhaps this is not what you expected :)