Optimising list comparison

[Jim Dennis]

In article <kttk8.6969$OP.215047 at stones>, Graham Ashton wrote:

> A friend of mine is using Python in a research project where he needs to
> compare each element of two vectors, such that comparing [a, b, c] and 
> [d, e, f] would involve comparing a with d, b with e and c with f.

> The only possible values in either list are 0, 1 and -1. When comparing
> two values, if either value is set to -1 then both values are deemed to be
> equal. So we've got:

>    0 ==  0 
>    0 == -1
>    1 ==  1
>    1 == -1
>    1 !=  0
	

> What he wants to know is simply whether all the values in the two lists
> are "equal", in which case the comparison returns 1, otherwise it returns
> 0.

	If I'm understanding you correctly then you should bel 
	able to just use:

			map(lambda x,y: not (x+y == 1), a, b)

	Let me go back through your "truth" table and fill it out
	for all possible permutations:
		
	a     b		r   a+b		a+b == 1
    0 ==  0		1	 0			0
    0 == -1		1	-1			0
	0 !=  1  	0?	 1			1
    1 ==  1		1	 2			0
    1 == -1		1	 0			0
    1 !=  0		0	 1			1
   -1 ==  0 	1?	-1			0
   -1 == -1 	1?	-2			0
   -1 ==  1 	1?	 0			0

   ... the ? lines denote the ones that I've added, the r column is
   the (apparentely) desired result --- the only case where you want
   to return false (0) is if a or b are one AND the other is zero.

   I observe that in either of those cases the sum of a+b will be 
   1 (0+1 or 1+0).  More importantly I note that ALL other cases the
   sum of a+b must NOT be 1 (4th column in my table).  Finally I notice
   that this is the inverse of the (apparently) desired results in the
   third column.  Thus the expression lambda x,y: not (a+b == 1) seems
   to work.

   But perhaps I misunderstand. 

   I didn't do *any* benchmarking of this.  It's just the simplest
   bit of code I could see that seems to match the desired table in
   a functionesque way.

> ----cut----

> Cheers,
> Graham.