I'm trying to replace all useless floats in a string (1.0, 2.0 etc.) by integers. So I'd turn a string like "15.0+abc-3" to "15+abc-3".
Do you know a way to do that?
I hope you understood my idea. If you didn't feel free to ask.
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2 Answers
You can use re.sub :
>>> s="15.0+abc-3"
>>>
>>> import re
>>> re.sub(r'\b(\d+)\.0+\b',r'\1',s)
'15+abc-3'
>>> s="15.0000+abc-333.0+er1102.05"
>>> re.sub(r'\b(\d+)\.0+\b',r'\1',s)
'15+abc-333+er1102.05'
\d+ will match any digit with length 1 or more and in sub function (\d+)\.0 will match the numbers with useless decimal zero.that will be replaced by the first group \1 that is your number (within capture group (\d+)).
And \b is word boundary that makes your regex doesn't match some numbers like 1102.05!
4 Comments
d0n.key
But it's only working with one 0 behind. Can you do one for infinite 0s?
Kasravnd
@Unknown Yeah, just add
+ after 0. fixed!d0n.key
I'm sorry I can't vote for both, but the other guy answered that faster :/ Thank you anyway.
Kasravnd
@Unknown It's OK buddy! don't care about this imaginary points!
(?<=\d)\.0+\b
You can simple use this and replace by empty string via re.sub.
See demo.
https://regex101.com/r/hI0qP0/22
import re
p = re.compile(r'(?<=\d)\.0+\b')
test_str = "15.0+abc-3"
subst = ""
result = re.sub(p, subst, test_str)
3 Comments
d0n.key
Thank you, but the other one was a bit faster :P
vks
@Unknown there is it capturing a group which we dont need.
d0n.key
But both are only working, with number with only one 0 behind. Do one for infinite 0s, and you're the winner ^^ EDIT: Got it my self. You'll get the vote anyway for giving the link :)