comparing tuples

Tim Chase python.list at
Sun Aug 22 20:12:00 CEST 2010

On 08/22/10 12:50, Baba wrote:
> level: beginners
> I was trying to write simple code that compares 2 tuples and returns
> any element in the second tuple that is not in the first tuple.
> def tuples(t1, t2):
>      result = []
>      for b in t2:
>          for a in t1:
>              if b == a:
>                  break
>          else:
>              result=result+[b,]
>      return result
> print tuples([0,5,6], [0,5,6,3,7])
> the code works but i was surprised by the following: my understanding
> was that an ELSE clause is part of an IF statement. Therefore it comes
> at the same indentation as the IF statement.

The ELSE clause can be used either with an IF (as you know) or 
with a FOR loop, which is interpreted as "if this loop reached 
the end naturally instead of exiting via a BREAK statement, 
execute this block of code".

If you reach the end of t1 without having found a value (and then 
issuing a "break"), then the current value of t2 (b) should be 
appended to the result.

That said, unless order matters, I'd just use sets:

   def tuples(t1, t2):
     return list(set(t2)-set(t1))

which should have better performance characteristics for large 


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