Re: Trouble with tables

*To*: mathgroup at smc.vnet.net*Subject*: [mg64841] Re: Trouble with tables*From*: Peter Pein <petsie at dordos.net>*Date*: Sun, 5 Mar 2006 03:19:01 -0500 (EST)*References*: <dubjlr$gua$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Luiz Melo schrieb: > Hi there, > I'm having a little trouble with the function "Table". Using: > > t1 = Table[{x[1, 1], x[1, 2], y}, {x[1, 1], 1, 3}, {x[1, 2], 1, 3}] > > the variables x[1,1] and x[1,2] assume the values 1, 2 and 3, that's fine. > Now I want to automatically generate the argument of the function Table above > and then evaluate it. I tried the following program: > > ni = 1; nj = 2; > a = Flatten[Table[x[i, j], {i, 1, ni}, {j, 1, nj}]]; d = Dimensions[a][[1]]; > b = {}; Do[b = Append[b, {a[[i]], 1, 3}], {i, 1, d}]; > t2 = Table[Join[{Append[a, y]}, b]]; > > and I got a result (t2) which is different from t1. Why? > > Thanks, > Luiz > > > Hello Luiz, Join[{Append[a, y]}, b] gives you {{x[1,1],x[1,2],y},{x[1,1],1,3},{x[1,2],1,3}}. You've got the parameters to the call of Table[] as a list. You have to apply the function Table to that list and must not take the list as only parameter. t2 = Table @@ Join[{Append[a, y]}, b] gives the desired result: --> {{{1,1,y},{1,2,y},{1,3,y}},{{2,1,y},{2,2,y},{2,3,y}},{{3,1,y},{3,2,y},{3,3, y}}} Peter b.t.w. Why do you use Dimensions[a][[1]] instead of Length[a]? and why don't you use the simple Array[{#1, #2, y} &, {3, 3}] ?