[Tutor] Re: Question on Regular Expression Help

[alan.gauld@bt.com]

> But +42 is not an integer, right? This can be fixed of course.

 >>> def isnumber(str):
...     if str[0] in '+-':
...             start = 1
...     else:
...             start = 0

Or replace the if/else with:

      def isnumber(strng)
        start = strng[0] in '+-'
...     for c in strng[start:]:
...             if c not in '0123456789':
...                     return False
...     return True

But I prefer:
> Don Arnold wrote:
> >>> def isInt(aString):
>  try:
>   int(aString)
>   return True
>  except:
>   return False
>
>But note that int(3.14) => 3 ! Is that what you want? Is 3.14 an integer?

Doesn't matter since we only return a boolean, if the number can 
be converted to an int it is a number, thats all I care about here...

> import re
> def isInteger(s):
>     return re.match(r'[+-]?\d+$', s)

I note we've changed the test now to isInteger from isNumber...

>  From a performance point of view, the try/except version is
> the faster when the number is really and integer, 

It's faster regardless, it just doesn't guarantee the number 
is an integer.

> regular expression version is the slowest, but I think 
> regular expression might win big if applied to a bigger problem, 

Yes I agree, that's why I suggested it it to an earliuer poster 
as the best flexible solution for dealing with strange unformatted 
input.

> re.findall(r'\s([+-]?\d+)\s', s)

Just so.

Alan G.