Re: How to use the multiple varible Interpolation?

*To*: mathgroup at smc.vnet.net*Subject*: [mg26298] Re: [mg26169] How to use the multiple varible Interpolation?*From*: Tomas Garza <tgarza01 at prodigy.net.mx>*Date*: Sun, 10 Dec 2000 00:20:08 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

I notice you haven't had an answer from the group yet, so I thought I might venture a possible approach. First, when you say you have "many" points, I expect they are many more than those in your example "tab". The trouble with "tab" is that you have a very incomplete grid in the x-y plane. The thing is, for y = 0 you have the x- values {-5, -4, -3, 0, 3, 4, 5} , whereas for y = 2 you have only the x-values {-3, -2, 0, 2, 3}. This makes it very awkward to work with. Anyway, my approach would be to get an interpolation function for each y value and using it, complete your observations for all values of x = {-4, -3, -2, -1, 0, 1, 2, 3, 4} (I omit x = -5 and 5 in order to avoid too many negative z-values). You would then have 27 points on a regular grid. Then you may use ListInterpolation to estimate the value at any x,y point. To illustrate the point, I obtain an "augmented tab", say tabAug , with two-dimensional interpolated values at all the original missing points: In[61}:= tabAug Out[61]= {{{-4, 0, 1.}, {-3, 0, 2.}, {-2, 0, 3.}, {-1, 0, 4.}, {0, 0, 5.}, {1, 0, 4.}, {2, 0, 3.}, {3, 0, 2.}, {4, 0, 1.}}, {{-4, 1, 0.}, {-3, 1, 1.}, {-2, 1, 2.19048}, {-1, 1, 3.28571}, {0, 1, 4.}, {1, 1, 3.28571}, {2, 1, 2.19048}, {3, 1, 1.}, {4, 1, 0.}}, {{-4, 2, -0.6}, {-3, 2, 0.}, {-2, 2, 0.8}, {-1, 2, 1.8}, {0, 2, 3.}, {1, 2, 1.8}, {2, 2, 0.8}, {3, 2, 0.}, {4, 2, -0.6}}} which you may compare with your original tab. Then I use ListInterpolation on tabAug: In[62]:= vals = Transpose[Partition[Flatten[tabAug], 3] Out[62]= {1., 2., 3., 4., 5., 4., 3., 2., 1., 0., 1., 2.19048, 3.28571, 4., 3.28571, 2.19048, 1., 0., -0.6, 0., 0.8, 1.8, 3., 1.8, 0.8, 0., -0.6}; In[63}:= h = ListInterpolation[Partition[vals, 9], {{0, 2}, {-4, 4}}] ListInterpolation::"inhr": "Requested order is too high; order has been reduced to {2, 3}" Out[63]= InterpolatingFunction[{{0., 2.}, {-4., 4.}}, "<>"] We check the quality of the 3-dimensional interpolation: In[64]:= Table[h[i, j], {i, 0, 2}, {j, -4, 4}] Out[64]= {{1., 2., 3., 4., 5., 4., 3., 2., 1.}, {0., 1., 2.19048, 3.28571, 4., 3.28571, 2.19048, 1., 0.}, {-0.6, 0., 0.8, 1.8, 3., 1.8, 0.8, 0., -0.6}} which is obviously very good in comparison with "vals" above. In[65]:= h[1.5, 3.5] Out[65]= 0.034375 I'm sure that a more decent set of observations "tab" will get you very reasonable results. Tomas Garza Mexico City "liwen liwen" <gzgear at yahoo.com> wrote: > Now I have many points which are on a 3D > surface. > for > example,tab={{-5,0,0},{-4,0,1},{-3,0,2},{0,0,5},{3,0,2}, > {4,0,1},{5,0,0}, > {-4,1,0},{-3,1,1},{0,1,4},{3,1,1},{4,1,0}, > {-3,2,0},{-2,2,0.8},{0,2,3},{2,2,0.8},{3,2,0}}; > I want to find the z value of (x=3.5,y=1.5) by the > method > of Interpolation,what should I do? > Is there any other method to solve those problem like > this?