Re: Summary:Ways to get Odd Columns&Rows of Matrix
- To: mathgroup at christensen.cybernetics.net
- Subject: [mg667] Re: Summary:Ways to get Odd Columns&Rows of Matrix
- From: villegas (Robert Villegas)
- Date: Sat, 8 Apr 1995 03:31:30 -0500
> (* using Fold and Drop *)
> (*From: bob Hanlon <hanlon at pafosu2.hq.af.mil>*)
> (*works on matrices with even number of terms.
> Not a good example of coding.
> The use of Fold, Transpose and Drop are too complex.
> Slow as well.
[bob's original code is here:]
> Fold[ Transpose[ Fold[ Drop, #,
> Table[ {-k}, {k, Length[ u ]/2} ] ] ]&,
> u, {1, 2} ] // MatrixForm
> (* Select odd rows and columns from square matrix *)
> Fold[ Transpose[ Fold[ Drop, #,
> Table[ {-k-1}, {k, Length[ u ]/2} ] ] ]&,
> u, {1, 2} ] // MatrixForm
Actually, I thought Bob's solution was rather clever, even if not as
practical for this application. The cumulative deletion of an
element, moving one further from the end each time, is kind of
neat, and I liked the Transpose for switching focus to the columns
after doing the rows, too. I don't think this would have occurred to
me.
I think some of the weaknesses in terms of generality can be fixed,
preserving the original ideas. I'm not sure it will be as general
as the others, and not as fast, but it seems like a worthwhile programming
and conceptual exercise.
The code works from the end instead of the beginning, and would
have to be changed if you wanted to apply it to an (odd x odd) matrix
instead of (even x even). Perhaps
Fold[Delete, #, 1 + Range[1/2 Length[#]] ]&
This deletes element 2, then 3, then 4, et cetera, to the end or
the (end - 1). I think Delete makes it just a tad easier because
you don't have to wrap the position in a list.
Also, the outer Fold needn't be a Fold, because it uses only one variable,
#1, which is the previous expression in the sequence. You are just
applying the same transformation several times (well, twice for a
2-dim matrix), so Nest will work.
In[47]:= Nest[Transpose @ Fold[Delete, #, 1 + Range[1/2 Length[#]] ]&,
Array[u, {6, 6}], 2]
Out[47]= {{u[1, 1], u[1, 3], u[1, 5]}, {u[3, 1], u[3, 3], u[3, 5]},
> {u[5, 1], u[5, 3], u[5, 5]}}
I think you can generalize it to an n-dimensional tensor, as well.
Use the Transpose idea, giving a rotated list of dimensions as the
second argument to Transpose in this case.
An example for a 3D matrix:
In[41]:= Nest[Transpose[Part[#, Range[1, Length[#], 2]],
RotateRight @ Range[3]]&, Array[a, {5, 6, 7}], 3] //MatrixForm
Out[41]//MatrixForm= a[1, 1, 1] a[1, 3, 1] a[1, 5, 1]
a[1, 1, 3] a[1, 3, 3] a[1, 5, 3]
a[1, 1, 5] a[1, 3, 5] a[1, 5, 5]
a[1, 1, 7] a[1, 3, 7] a[1, 5, 7]
a[3, 1, 1] a[3, 3, 1] a[3, 5, 1]
a[3, 1, 3] a[3, 3, 3] a[3, 5, 3]
a[3, 1, 5] a[3, 3, 5] a[3, 5, 5]
a[3, 1, 7] a[3, 3, 7] a[3, 5, 7]
a[5, 1, 1] a[5, 3, 1] a[5, 5, 1]
a[5, 1, 3] a[5, 3, 3] a[5, 5, 3]
a[5, 1, 5] a[5, 3, 5] a[5, 5, 5]
a[5, 1, 7] a[5, 3, 7] a[5, 5, 7]
I think this general coding will work for higher tensors. The levels of
the tensor get rotated around a full cycle, and end up where they were
originally:
everyother[tensor_List] :=
With[{n = TensorRank[tensor]},
Nest[Transpose[Part[#, Range[1, Length[#], 2]], RotateRight @ Range[n]]&,
tensor, n]
]
A test, using a matrix whose entries are just the indices concatenated
together, for compactness of display (I always use this sort of thing
when I'm having difficulty visualizing what's going on with higher
dimensional matrices):
In[29]:= everyother[ Array[SequenceForm, {3, 4, 5, 6}] ] //MatrixForm
Out[29]//MatrixForm= 1111 1113 1115 1311 1313 1315
1131 1133 1135 1331 1333 1335
1151 1153 1155 1351 1353 1355
3111 3113 3115 3311 3313 3315
3131 3133 3135 3331 3333 3335
3151 3153 3155 3351 3353 3355
Robby Villegas