Get a substring for the main version from the full version

Question

I have a script in which I have to convert full version strings only to their main parts. For example, I should convert 1.2.3.4.5.6.7 to 1.2 .

I am currently using this: '.'.join(s.split('.', 2)[:-1])

 >>> s = '1.2.3.4.5.6.7' >>> '.'.join(s.split('.', 2)[:-1]) '1.2' 

Which works great. But it is very ugly, I hope there is a better way.

Edit:

• Performance is a problem, so answers that do not work well (although they may look nice) do not suit me.
• '.'.join(s.split('.', 2)[:-1]) can also be '.'.join(s.split('.', 2)[:2]) just as easy

Answer 1

Using regex is not redundant if you precompile the regular expression. Thus,

 import re pattern = re.compile(r'^[0-9]+\.[0-9]+') # ... later ... version = '1.2.3.4.5.6.7' def get_version(s): m = pattern.search(s) if m: return m.group() print get_version(version) 

It also ensures that your version matches the format.

Answer 2

Using regex :

 >>> s = '1.2.3.4.5.6.7' >>> re.search(r'(\d+\.\d+)', s).group() '1.2' 

Terms of comparison:

 >>> r = re.compile(r'^(\d+\.\d+)') >>> s = '100.21.3.4.5.6.7' >>> %timeit r.search(s).group() 100000 loops, best of 3: 1.43 us per loop >>> %timeit '.'.join(s.split('.')[:2]) 1000000 loops, best of 3: 2.32 us per loop >>> %timeit '.'.join(s.split('.', 2)[:-1]) 100000 loops, best of 3: 1.28 us per loop >>> s = '100.21.3.4.5.6.7'*100 >>> %timeit r.search(s).group() 1000000 loops, best of 3: 1.96 us per loop >>> %timeit '.'.join(s.split('.')[:2]) 10000 loops, best of 3: 40.4 us per loop >>> %timeit '.'.join(s.split('.', 2)[:-1]) 100000 loops, best of 3: 2.01 us per loop >>> s = '100.21.3.4.5.6.7'*1000 >>> %timeit r.search(s).group() 1000000 loops, best of 3: 1.94 us per loop >>> %timeit '.'.join(s.split('.')[:2]) 1000 loops, best of 3: 314 us per loop >>> %timeit '.'.join(s.split('.', 2)[:-1]) 100000 loops, best of 3: 6.76 us per loop >>> s = '100.21.3.4.5.6.7'*10000 >>> %timeit r.search(s).group() 100000 loops, best of 3: 1.42 us per loop >>> %timeit '.'.join(s.split('.')[:2]) 100 loops, best of 3: 5.3 ms per loop #millisecond >>> %timeit '.'.join(s.split('.', 2)[:-1]) 10000 loops, best of 3: 104 us per loop 

Based solution

re.findall will also be slow because it requires iterating over the entire string, and re.search stops at the first match.

Answer 3

Another way could be '.'.join(s.split('.')[:2]) , but it is very similar. I think there are no other effective features: you need to divide by dots, and then select only two elements.

This solution (or yours, the same) is perfect. Just avoid using regular expressions, it just overwhelms this task.

Answer 4

Another approach is to search for the second point and cut the line on it:

 s[:s.index('.', s.index('.')+1)] 

It should be about as fast as the split and regex versions .