Nice way to cast a homogeneous tuple

[Hexamorph]

wheres pythonmonks wrote:
> A new python convert is now looking for a replacement for another perl idiom.
> 
> In particular, since Perl is weakly typed, I used to be able to use
> unpack to unpack sequences from a string, that I could then use
> immediately as integers.
> 
> In python, I find that when I use struct.unpack I tend to get strings.
>  (Maybe I am using it wrong?)
> 
> def f(x,y,z): print x+y+z;
> 
> f( *struct.unpack('2s2s2s','123456'))
> 123456
> 
> (the plus concatenates the strings returned by unpack)
> 
> But what I want is:
> 
> f( *map(lambda x: int(x), struct.unpack('2s2s2s','123456')))
> 102
> 
> But this seems too complicated.
> 
> I see two resolutions:
> 
> 1.  There is a way using unpack to get out string-formatted ints?
> 
> 2.  There is something like map(lambda x: int(x).... without all the
> lambda function call overhead.  (e.g., cast tuple)?
>   [And yes: I know I can write my own "cast_tuple" function -- that's
> not my point.  My point is that I want a super-native python inline
> solution like (hopefully shorter than) my "map" version above.  I
> don't like defining trivial functions.]
> 
> W

In [31]: import re

In [32]: sum(int(x) for x in re.findall("..", s))
Out[32]: 102

To get a tuple instead of the sum, just do:

In [33]: tuple(int(x) for x in re.findall("..", s))
Out[33]: (12, 34, 56)