Message Boards Message Boards

1
|
6254 Views
|
3 Replies
|
1 Total Likes
View groups...
Share
Share this post:

ListPointPlot3D & Interpolation: alternative way to enter the arguments?

Gernot H
Posted 10 years ago

Hi,

I am using ListPointPlot3D a lot recently. The arguments seem to have been given in the pattern:

ListPointPlot3D[{...,{xi,yj,zij},...}]

Also, I use Interpolation[ ] to interpolate the data. And this command wants the arguments in the form:

Interpolation[{...,{{xi,yj},zij},...}]

I usually have my data in the form:

x={x1,x2,...}; (* 1D array with the x-values *)
y={y1,y2,...}; (* 1D array with the y-values *)
z={{z11,z12,...},{z21,z22,...},...}; (* 2D array with the z-values *)

And it is a bit cumbersome to having to rearrange the data every time, especially when entering the data for Interpolation.

Therefore my question: Is there a way to either give my data to ListPointPlot3D and Interpolation in this way, or alternatively, is there a simple way to rearrange my data in the way so I can feed it to ListPointPlot3D?

Cheers! Gernot
POSTED BY: Gernot H
3 Replies
Sort By:

Try

Transpose[{x,y,#}]&/@z

Here's a (simple) symbolic example:

In[10]:= x = {x1, x2};(*1D array with the x-values*)

In[11]:= y = {y1, y2};(*1D array with the y-values*)

In[12]:= z = {{z11, z12}, {z21, z22}};(*2D array with the z-values*)

In[13]:= Transpose[{x, y, #}] & /@ z

Out[13]= {{{x1, y1, z11}, {x2, y2, z12}}, {{x1, y1, z21}, {x2, y2, z22}}}
POSTED BY: David Reiss
Gernot H
Gernot H
Posted 10 years ago

Hi, thanks for the help.

Thats not quite what I need, since zij should correspond to xi and yj.

I did it this way so far:

Table[{x[[i]],y[[j]],z[[i,j]]},{i,Length[x]},{j,Length[y]}]

That works. I was just wondering if there was a smarter way to do it. But anyway, since it works, maybe i shouldn't bother too much.

Edit: Now I remember again why I found it so annoying:

In certain cases I want to interpolate over the data, and the Interpolation command only accepts the arguments in the form:

{...,{{xi,yj},zij},...}

And THIS is really the cumbersome part. Sorry for my confusion, I will edit my start post to address this question.
POSTED BY: Gernot H

Hi Gernot,

Try the following:

x = Range[10]
y = Range[12, 14]
z = Table[i^2 j^2, {j, Length[y]}, {i, Length[x]}]
Transpose[{Tuples[{x, y}], Flatten[z, 1]}]
POSTED BY: Sander Huisman
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Tag limit exceeded
Note: Only the first five people you tag will receive an email notification; the other tagged names will appear as links to their profiles.
Reply Preview
Attachments
Remove
or Discard

Group Abstract Group Abstract