Regex to remove leading zero except for first

Question

I have the following cases where I'd like to remove the leading zeros

0 -> 0
0.1 -> 0.1
-0.1 -> -0.1
00.01 -> 0.01
001 -> 1

So whenever there are multiple zeros before the decimal or number, then we remove them. If the zero is by itself, we keep it. I have the following regex:

r'^[0]*'

but this removes all leading zeros. How can I fix this so that it does what I want it to do?

You could apply the regex, then after if the string has length 0 or if the first char is a `.` then you can append a 0 to the start of the string. — Erik McKelvey, Nov 23 '21 at 01:59

score 1 · Answer 1 · answered Nov 23 '21 at 02:03

1

You can use the Decimal class to convert the string to a number and back:

>>> from decimal import Decimal
>>> str(Decimal("0.01"))
'0.01'
>>> str(Decimal("000.01"))
'0.01'
>>> str(Decimal("-00.01"))
'-0.01'
>>> str(Decimal("1"))
'1'

answered Nov 23 '21 at 02:03

Selcuk

57,004
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score 1 · Answer 2 · answered Nov 23 '21 at 03:00

1

try this regex:

>>> import re
>>> text = '0\n0.1\n-0.1\n00.01\n001'
>>> print(re.sub(r'0+(\d+)(\.\d+)?', r'\1\2', text))
0
0.1
-0.1
0.01
1

answered Nov 23 '21 at 03:00

asiloisad

41
4

bobble bubble · Answer 3 · 2021-11-26T19:27:05.037

1

An idea with \B which matches a non word boundary.

^0+\B

See demo at regex101

edited Nov 26 '21 at 19:27

answered Nov 23 '21 at 11:57

bobble bubble

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Regex to remove leading zero except for first

3 Answers3