Re: ListPlot vs ListPlot3D

*To*: mathgroup at smc.vnet.net*Subject*: [mg28906] Re: ListPlot vs ListPlot3D*From*: "Allan Hayes" <hay at haystack.demon.co.uk>*Date*: Fri, 18 May 2001 01:13:09 -0400 (EDT)*References*: <9e02ft$2vu@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Otto, The options MeshRange works if the array is evenly spaced. data= Table[{x,y,x^2+y^2},{x,-4,4},{y,-5,7}]; Extract array of z values dataz =data/.{__,z_?NumberQ}->z Get the x and y ranges mr={Min[#],Max[#]}&/@{data[[All,All,1]],data[[All,All,2]]}; Use ListPlot3D with MeshRange ->mr; ListPlot3D[dataz, MeshRange->mr]; If we have an uneven rectangular grid we can make an interpolation function for data and plot that: for example fn=Interpolation[Flatten[data,1],InterpolationOrder\[Rule]1] InterpolatingFunction[{{-4,4},{-5,7}},<>] Plot3D[fn[x,y],{x,-4,4},{y,-5,7}] If the points are not on a rectangular grid then the function TriangularSurfacePlot from the standard add-on <<DiscreteMath`ComputationalGeometry` can be used. -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay at haystack.demon.co.uk Voice: +44 (0)116 271 4198 Fax: +44 (0)870 164 0565 "Otto Linsuain" <linsuain+ at andrew.cmu.edu> wrote in message news:9e02ft$2vu at smc.vnet.net... > > Dear Mathematica experts, I was wondering if there is a way of plotting > a 3D graph using a rectangular array of z-values USING THE CORRESPONDING > VALUES OF X AND Y. > > Here is what I mean: > > with a 2D plot, one can do: > > ListPlot[ {y1,y2,y3,.........yn} ] and this will give a graph of the points > > {y1,1} ,{y2,2}, {y3,3},....{yn,n}, i.e. it will use the integers > 1,2,3,...n as the x values. > > One can, however, write > > ListPlot[ { {x1,y1}, {x2,y2}, {x3,y3}, ... {xn,yn} }] > > and get a graph with user-specified values for x and y. > > There doesn't seem to be an analogous thing for ListPlot3D. The values > of x and y seem to be always {1,1}, {1,2},{2,1},{2,2},.......{n,n} > > Any ideas? Thanks in advance. Otto Linsuain. >