Re: comparing two lists

*To*: mathgroup at smc.vnet.net*Subject*: [mg54637] Re: comparing two lists*From*: Paul Abbott <paul at physics.uwa.edu.au>*Date*: Fri, 25 Feb 2005 01:18:36 -0500 (EST)*Organization*: The University of Western Australia*References*: <200502161936.OAA19139@smc.vnet.net> <cv3m5r$cb7$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

In article <cv3m5r$cb7$1 at smc.vnet.net>, Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote: > I would say the simplest way to solve your problem is: > > MyFunction1[mat_?MatrixQ, vec_?VectorQ] :=Position[mat - vec, _?(# > 0 > &)] > > For example with your data this gives: > > In[2]:= > mat={{0.183706,0.758693,0.462242,0.170041},{0.457054,0.349658,0.805455,0 > .\ > 127763}}; > > In[3]:= > vec={0.482259,0.314393}; > > In[4]:= > MyFunction1[mat,vec] > > Out[4]= > {{1,2},{2,1},{2,2},{2,3}} > > This tells you exactly the positions you wanted but not in the form you > asked for. Getting the output into that form will be the most > complicated part of the problem, so if you really do not need it I > would rather settle for the above. But of course there are lots of ways > to get exactly the answer you asked for. You could try to bring the > answer above into the form you want or you could use something like: > > MyFunction2[mat_?MatrixQ, vec_?VectorQ] := > Map[Flatten[Position[mat[[#]] - > vec[[#]], _?(# > 0 &)]] &, Range[Length[vec]]] A simpler solution, very close to the original proposal (Position[#1,_?(#>#2&)]&@@{mat,vec}) is Position[#, _?(# > 0 &)] & /@ (mat - vec) Cheers, Paul -- Paul Abbott Phone: +61 8 6488 2734 School of Physics, M013 Fax: +61 8 6488 1014 The University of Western Australia (CRICOS Provider No 00126G) 35 Stirling Highway Crawley WA 6009 mailto:paul at physics.uwa.edu.au AUSTRALIA http://physics.uwa.edu.au/~paul

**References**:**comparing two lists***From:*Curt Fischer <tentrillion@gmail.NOSPAM.com>