Mathematical large-panel periodic interpolation

Question

I have a very large table in Mathematica ((dimcub-1) ^ 3 elements) coming from the inverse FFT. I need to use periodic interpolation in this table. Since periodic interpolation requires the first elements and the last elements to be equal, I manually create a new element table dim ^ 3 and use it in my interpolation. It works, but it is ugly / slow, and because of my superfluous staging table, I rather hit memory. Can someone tell me how to turn my old table into a periodic one by adding elements or using my non-periodic table to create periodic interpolation? Here is my current code snippet:

mr 1 - new table:

 mr1 = Table[ 0. , {i, 1, dimcub}, {j, 1, dimcub}, {k, 1, dimcub}]; Do[Do[ Do[ mr1[[m, n, k]] = oldtable[[m, n, k]] ; , {m, 1, dimcub - 1}]; , {n, 1, dimcub - 1}]; , {k, 1, dimcub - 1}]; Do[Do[ mr1[[m, n, dimcub]] = mr1[[m, n, 1]]; mr1[[m, dimcub, n]] = mr1[[m, 1, n]]; mr1[[dimcub, m, n]] = mr1[[1, m, n]]; , {m, 1, dimcub - 1}]; mr1[[n, dimcub, dimcub]] = mr1[[n, 1, 1]]; mr1[[dimcub, n, dimcub]] = mr1[[1, n, 1]]; mr1[[dimcub, dimcub, n]] = mr1[[1, 1, n]]; , {n, 1, dimcub - 1}]; mr1[[dimcub, dimcub, dimcub]] = mr1[[1, 1, 1]]; Remove[oldtable]; myinterpolatingfunction = ListInterpolation[mr1, {{1, dimcub}, {1, dimcub}, {1, dimcub}}, InterpolationOrder -> 1, PeriodicInterpolation -> True]; Remove[mr1]; 

myinterpolatingfunction takes up much less memory and works great after deleting old tables. Any help would be greatly appreciated.

Answer 1

Thanks for all the answers. I tried the Leonid sentence, but when I print my antique table, it was still (dimcub -1) ^ 3 dimensional. New elements have been defined, and I see them individually, but they do not appear as part of the old table when printing the entire table. So, I got something similar that does exactly what I need:

 oldtable= PadRight[oldtable, {dimcub, dimcub, dimcub}]; oldtable[[All, All, dimcub]] = oldtable[[All, All, 1]]; oldtable[[All, dimcub, All]] = oldtable[[All, 1, All]]; oldtable[[dimcub, All, All]] = oldtable[[1, All, All]]; oldtable[[All, dimcub, dimcub]] = oldtable[[All, 1, 1]]; oldtable[[dimcub, All, dimcub]] = oldtable[[1, All, 1]]; oldtable[[dimcub, dimcub, All]] = oldtable[[1, 1, All]]; oldtable[[dimcub, dimcub, dimcub]] = oldtable[[1, 1, 1]]; 

The Wizard answer is too advanced for my math level ..

Answer 2

The answers of Leonid and Mr. Weiser do too much work. In the case of Leonid, only the first three lines are needed. To show this, I changed the last 4 Set to Equal s:

 In[65]:= len = 4; oldtable = Partition[Partition[Range[len^3], len], len] Out[65]= {{{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}, {13, 14, 15, 16}}, {{17, 18, 19, 20}, {21, 22, 23, 24}, {25, 26, 27, 28}, {29, 30, 31, 32}}, {{33, 34, 35, 36}, {37, 38, 39, 40}, {41, 42, 43, 44}, {45, 46, 47, 48}}, {{49, 50, 51, 52}, {53, 54, 55, 56}, {57, 58, 59, 60}, {61, 62, 63, 64}}} In[66]:= oldtable[[All, All, -1]] = oldtable[[All, All, 1]]; oldtable[[All, -1, All]] = oldtable[[All, 1, All]]; oldtable[[-1, All, All]] = oldtable[[1, All, All]]; oldtable[[All, -1, -1]] == oldtable[[All, 1, 1]] oldtable[[-1, All, -1]] == oldtable[[1, All, 1]] oldtable[[-1, -1, All]] == oldtable[[1, 1, All]] oldtable[[-1, -1, -1]] == oldtable[[1, 1, 1]] Out[69]= True Out[70]= True Out[71]= True Out[72]= True 

What Leonid does is illustrated in the figures below. Lines 4-6 of his code do something as shown in the left panel: copying a line (darker color) of an already copied plane (light colors). Line 7 is shown in the right pane. This is a copy of a cell to a cell with diagonally opposite positions, and its action is not included in one of the first three copy actions separately, but is the result of their sequential operation.

Answer 3

You can get it much faster and more efficiently with memory by modifying the original table as follows:

 oldtable[[All, All, -1]] = oldtable[[All, All, 1]]; oldtable[[All, -1, All]] = oldtable[[All, 1, All]]; oldtable[[-1, All, All]] = oldtable[[1, All, All]]; oldtable[[All, -1, -1]] = oldtable[[All, 1, 1]]; oldtable[[-1, All, -1]] = oldtable[[1, All, 1]]; oldtable[[-1, -1, All]] = oldtable[[1, 1, All]]; oldtable[[-1, -1, -1]] = oldtable[[1, 1, 1]]; 

These assignments replace nested loops and work much faster, plus you do not need memory to store a copy. This is based on the extended vectorized functionality of the Part command (indexing an array and general expression), in particular vectorized assignments. It is also important to have a numerical array in the form of a Packed array , which is often the case.

Answer 4

Just because I'm a little versed, Leonid’s solution can be written as:

 a = {0, 1}~Tuples~3~SortBy~Tr // Rest; MapThread[ (oldtable[[Sequence @@ #]] = oldtable[[Sequence @@ #2]]) &, {-a, a} /. 0 -> All ];