Python 正则表达式检查字符串

Question

im trying to set correct version on output. I have this possible strings:

• 0.0.4.1 # 1.
• 7.51.4.1 # 2.
• 0.1.4.1 # 3.

and i need to check, if the "0." is on the start (to set output without 0. or 0.0.)

Output 1. will have just "4.1" , 2. gonna stay the same and 3. will be 1.4.1

Im trying to set correct expresion like ^([0\.]{1})([.\d]{3})+$ but even if i use {1} quantifier (to just one "0.") it doesnt work correctly.

Anyone with some clue?

Answer 1

If there should be 4 digit parts with 3 dots, you can first assert the format using a positive lookahead.

In the replacement use an empty string.

^(?=\d+(?:\.\d+){3}$)0+\.(?:0+\.)*

• ^ Start of string
• (?=\d+(?:\.\d+){3}$) Positive lookahead, assert 4 digit parts with 3 dots
• 0+\. match 1+ zeroes and .
• (?:0+\.)* Optionally repeat matching 1+ zeroes and .

Regex demo | Python demo

Example

import re
strings = ["0.0.4.1", "7.51.4.1", "0.1.4.1", "001", "0.1", ".0.1", "0.0.0.0", "0.1.2.3.4.5"]
pattern = r"^(?=\d+(?:\.\d+){3}$)0+\.(?:0+\.)*"
for s in strings:
    print(re.sub(pattern, "", s))

Output

4.1
7.51.4.1
1.4.1
001
0.1
.0.1
0
0.1.2.3.4.5

---

Other options might be matching zeroes and a . at the start, and then repeatedly zeroes and a dot.

^0+\.(?:0+\.)*

Regex demo

Or a broader pattern to match all leading dots and zeroes:

^0+\.[.0]*

Regex demo

Answer 2

You can use

import re
texts = ['0.0.4.1','7.51.4.1','0.1.4.1']
for text in texts:
    m = re.fullmatch(r'(?:0+\.)*(\d+(?:\.\d+)*)', text)
    if m:
        print(m.group(1))
# => 4.1
#    7.51.4.1
#    1.4.1

See the Python demo . NOTE : re.fullmatch requires the full string match, so no need specifying the ^ and $ anchors.

NOTE 2 : If you do not want to match anything in 0.0.0.0 add the if m.group(1).replace('0',''): check before returning the value, see this Python demo .

See the regex demo meaning

• (?:0+\.)* - zero or more occurrences of one or more 0 s and a dot
• (\d+(?:\.\d+)*) - Group 1: one or more digits and then zero or more repetitions of a dot and one or more digits.

Answer 3

Use this simple regex: ^[0\.]+

https://regex101.com/r/zUKutI/1/