Comparison Operation for Nested Matrices

Question

I have a nested matrix n as bellow

n = {{a, b}, {c, d}}
a = {{0, t, q, dh}, {0, 1, 0, th}, {1, 0, 0, sh}}
b = {{0, t, q, dh}, {0, 0, 0, th}, {0, 0, 1, sh}}
c = {{0, t, q, dh}, {1, 0, 1, th}, {0, 0, 0, sh}}
d = {{0, t, q, dh}, {0, 1, 0, th}, {1, 0, 0, sh}}

enter image description here

containing letters and numbers. I am going to do a particular operation in each row yielding the result shown bellow:

enter image description here

 {{{{0, t, q, dh}, {0, 1, 0, th}, {1, 0, 1, sh}}}, 
  {{{0, t, q, dh}, {1, 1, 1, th}, {1, 0, 0, sh}}}}M

Please post code instead of images so we have something to work with. — mfvonh, Jun 25 '14 at 18:53
Also, how exactly is this supposed to work? Are the differences always going to be simply nonzero vs zero? — mfvonh, Jun 25 '14 at 18:55
in sub matrices in each row, and for corresponding elemenets in each of these sub matrices, the comparison between 0 and 0 is 0, between 0 and 1, yields 1. and between 1 and 1 results 1. and letters such as t, q, dh and so on must be repeated. — Unbelievable, Jun 25 '14 at 19:00
in 'n', as n = {{a, b}, {c, d}}, which is written above a11 must be comprised with b11, a12 (that is 't' will must be comprise just for b12 (that is same as a12), the correspoding elements: a12,b12.........a13,b13........a31,b31. and so on. — Unbelievable, Jun 25 '14 at 19:12
Actually the corresponded elements that contains letter are same. and will be repeated in the result matrix (m). — Unbelievable, Jun 25 '14 at 19:13
Please edit your comments above into the question. Without these the question does not make any sense at all as you are not specifying clearly what the operation is to involve, and nor does it actually ask a question. — Oleksandr R., Jun 26 '14 at 07:24

score 7 · Accepted Answer · edited Jun 26 '14 at 05:47

7

I don't know what would be a general pattern but for this case you can use:

{MapThread[Max, #, 2]} & /@ n

{{
   {{0, t, q, dh}, 
    {0, 1, 0, th}, 
    {1, 0, 1, sh}}},
   {{{0, t, q, dh}, 
    {1, 1, 1, th}, 
    {1, 0, 0, sh}}}}

Alternatively:

List /@ MapThread[Max, n, 3] // MatrixForm

$\left( \begin{array}{c} \left( \begin{array}{cccc} 0 & t & q & \text{dh} \\ 1 & 1 & 1 & \text{th} \\ 1 & 0 & 0 & \text{sh} \end{array} \right) \\ \left( \begin{array}{cccc} 0 & t & q & \text{dh} \\ 0 & 1 & 0 & \text{th} \\ 1 & 0 & 1 & \text{sh} \end{array} \right) \end{array} \right)$

edited Jun 26 '14 at 05:47

Mr.Wizard

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answered Jun 25 '14 at 19:23

Kuba

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Your result is : {{{0, t, q, dh}, {0, 1, 0, th}, {1, 0, 1, sh}}, {{0, t, q, dh}, {1, 1, 1, th}, {1, 0, 0, sh}}} ,while my desire is: {{{{0, t, q, dh}, {0, 1, 0, th}, {1, 0, 1, sh}}}, {{{0, t, q, dh}, {1, 1, 1, th}, {1, 0, 0, sh}}}} – Unbelievable Jun 25 '14 at 19:40
It can't as far as I know -- I don't think that is actually happening with these particular matrices. But you could use Block[{Max},Unprotect@Max;Max[s_Symbol,0]:=s; ... – mfvonh Jun 25 '14 at 19:55
1

@mfvonh It does not matter since there will not be a case of two different symbols. Or a case of symbol and 0. That's why I've asked many questions before answering. – Kuba Jun 25 '14 at 19:59
@Kuba Gotcha, I was just trying to clarify for mostafa , not to correct your answer :) – mfvonh Jun 25 '14 at 20:03
@mfvonh right, I probably should've explained that earlier so thanks :) – Kuba Jun 25 '14 at 20:05
1

I independently arrived at almost exactly the same solution so I appended it to this answer. – Mr.Wizard Jun 26 '14 at 05:48
@Kuba, No why?! I am practicing this one, again, for not forgetting which I have learn of you. – Unbelievable Jul 01 '14 at 11:46
I am worried if other users who had tried to answer my question be unhappy if I accept just one answer while their answers are completely correct too. – Unbelievable Jul 01 '14 at 12:00
@mostafa no, of course not, you should accept the answer that fits your need the best. which you are using or you just want :) Upvotes are indicators of correctness. – Kuba Jul 01 '14 at 12:01
@mostafa take a look at meta there are couple of related topics. p.s. could you please delete old not relevant comments? – Kuba Jul 01 '14 at 12:02
I deleted that were related to me. Ok I am going to take a look. – Unbelievable Jul 01 '14 at 12:14

score 4 · Answer 2 · answered Jun 26 '14 at 05:54

If only Max were Listable:

listMax = Function[, Max[##], Listable];

Then:

List /@ listMax @@@ n // MatrixForm

$\left( \begin{array}{c} \left( \begin{array}{cccc} 0 & t & q & \text{dh} \\ 1 & 1 & 1 & \text{th} \\ 1 & 0 & 0 & \text{sh} \end{array} \right) \\ \left( \begin{array}{cccc} 0 & t & q & \text{dh} \\ 0 & 1 & 0 & \text{th} \\ 1 & 0 & 1 & \text{sh} \end{array} \right) \end{array} \right)$

Comparison Operation for Nested Matrices

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