Re: An easier functional way to divide each Column of matrix by a row

*To*: mathgroup at smc.vnet.net*Subject*: [mg127153] Re: An easier functional way to divide each Column of matrix by a row*From*: Dana DeLouis <dana01 at me.com>*Date*: Mon, 2 Jul 2012 22:17:44 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com

> mat = {{a1,a2},{b1,b2}}; > v = {v1,v2}; > Inner[Divide,mat,v,List] Hi. The ideas given by others in the similar thread (about multiplying) could apply here. Just divide instead. For example: mat={{a1,a2},{b1,b2}}; v={v1,v2}; // Yours Inner[Divide,mat,v,List] {{a1/v1,a2/v2},{b1/v1,b2/v2}} Map[#/v &,mat] {{a1/v1,a2/v2},{b1/v1,b2/v2}} %% == % True // Your other example: mat = {{1,2},{3,4}}; v={5,10}; Inner[Divide,mat,v,List] {{1/5,1/5},{3/5,2/5}} Map[#/v &,mat] {{1/5,1/5},{3/5,2/5}} %% == % True If one inverts the vector at the start, one has the same solution as the other tread using Times. mat={{a1,a2},{b1,b2}}; v=1 / {v1,v2}; Map[# * v &, mat] {{a1/v1,a2/v2},{b1/v1,b2/v2}} = = = = = = = = = = HTH :>) Dana DeLouis Mac & Math 8 = = = = = = = = = = On Jun 28, 4:03 am, "Nasser M. Abbasi" <n... at 12000.org> wrote: > hello; > > I found a better solution. > > (after a strong coffee and staring on it for sometime) > > method(5) > ----------- > mat = {{a1,a2},{b1,b2}}; > v = {v1,v2}; > Inner[Divide,mat,v,List] > > Out[61]= { {a1/v1, a2/v2}, {b1/v1,b2/v2} } > > But I can't say though it was easy and intuitive to find > for me but at least the above solution is a functional and > I think the right Mathematica way of doing it. So I am > happy. Was good practice though. > > --Nasser > > On 6/27/2012 8:13 PM, Nasser M. Abbasi wrote: > > > > > I have a list like this 'mat' and 'v' like this > > > mat = { {a1,a2},{b1,b2} } > > v = {v1,v2} > > > I want to generate > > > mat={ {a1/v1, a2/v2}, { b1/v1, b2/v2 } } > > > I can't just do mat/v since this does > > > mat={ {a1/v1, a2/v1}, { b1/v2, b2/v2 } } > > > I solved this 2 ways, but I am still not happy. > > I think there should be an easier way. > > > method 1 (not too natural) > > ------------------------------- > > Clear["Global`*"] > > mat={{a1,a2},{b1,b2}}; > > v={v1,v2}; > > Transpose[Transpose[mat]/v] > > > Out[93]= { {a1/v1, a2/v2}, {b1/v1, b2/v2} } > > > method 2 (too complicated) > > --------------------------- > > In[94]:= MapIndexed[Divide[#1,v[[#2[[2]]]]]&,mat,{2}] > > > Out[94]= { {a1/v1, a2/v2}, {b1/v1, b2/v2}} > > > method 3 (using a Table, oh no !) > > ------------------------------------ > > In[96]:= Table[mat[[i,j]]/v[[j]],{i,2},{j,2}] > > > Out[96]= {{a1/v1, a2/v2} , {b1/v1, b2/v2} } > > > method 4 (a not good way to do it ) > > ---------------------------------------- > > In[108]:= mat.v/.Plus->List/.Times->Divide > > > Out[108]= {{a1/v1, a2/v2}, {b1/v1, b2/v2}} > > > I looked at real Mathematical tricks using Inenr and Outer and > > something like this, but I do see a way so far. (I also did not > > have my morning coffee yet), so I wanted to ask if > > someone can see one of those elegant super functional > > correct ways to do this. > > > ps. fyi, in that other system (starts with O and ends with VE) > > I can do this like this: > > > mat=[1 2;3 4] > > v=[5 10] > > bsxfun(@rdivide,mat,v) > > > 0.20000 0.20000 > > 0.60000 0.40000 > > > thanks, > > --Nasser