comparing tuples
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. However the above example only works if the ELSE clause is positioned under the second FOR loop. As if it was an ELSE clause without an IF statement!? Why/How does this work? tnx Baba -- http://mail.python.org/mailman/listinfo/python-list
Re: comparing tuples
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 inputs. -tkc -- http://mail.python.org/mailman/listinfo/python-list
Re: comparing tuples
On Aug 22, 7:12 pm, Tim Chase python.l...@tim.thechases.com wrote: 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 inputs. -tkc Thanks Tim! -- http://mail.python.org/mailman/listinfo/python-list
