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I have a $N$ dimensional array $M$, and a function $f(\{M_{i}\})$ in terms of the array elements, where each matrix element $M_{i}$ can be 0 or 1. I'd like to construct a table 

Table[f(\{M_{ij}\}), {M_{1},0,1},{M_{2},0,1},...,{M_{N},0,1}]

For a given N, I can write down the code to construct the table, but for general N, it there a convenient way to write the above code? Namely, I'd like to keep N as an input variable.

As an example, let us use 

n=3;
Mat=Table[M[i],{i,1,n}];

and define f as the sum of all the elements in Mat. so the table I want to construct is 

Table[Sum[M[i],{i,1,n}], {M[1],0,1},{M[2],0,1}, {M[3],0,1}]

The outcome is 

{{{0, 1}, {1, 2}}, {{1, 2}, {2, 3}}}

Of course, when I change n to other values, I need to rewrite the code for the table. So there should be a way for us to construct the table without modifying the code.

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    $\begingroup$ It seems to me that you want to Map your function $f$ on each element of the matrix $M$. Would Map[f, M, {2}] do what you want? $\endgroup$
    – MarcoB
    May 23, 2020 at 18:03
  • $\begingroup$ $f$ depends on every element of $M$, not just one particular element. So I think we need some more complicated form. $\endgroup$
    – user34104
    May 23, 2020 at 18:15
  • $\begingroup$ Can you give an example input with the desired example output? That would help. $\endgroup$
    – march
    May 23, 2020 at 18:20
  • $\begingroup$ @user34104 Yes, so you will have defined f[mi_] := <do something with the mi value> so for each different value in $M$, $f$ will do whatever you want with it. I second @march’s request though. If you show us an example of input and desired output, this will be much easier. $\endgroup$
    – MarcoB
    May 23, 2020 at 18:33
  • $\begingroup$ Thanks! I've updated with an example. $\endgroup$
    – user34104
    May 23, 2020 at 18:37

2 Answers 2

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From one example it is not clear what you want for other than n = 3, perhaps

Clear["Global`*"]

Mat[n_] := 
 Table[Sum[M[i], {i, 1, n}], 
  Evaluate[Sequence @@ ({M[#], 0, 1} & /@ Range[n])]]

Mat[3]

(* {{{0, 1}, {1, 2}}, {{1, 2}, {2, 3}}} *)

Mat[5]

(* {{{{{0, 1}, {1, 2}}, {{1, 2}, {2, 3}}}, {{{1, 2}, {2, 3}}, {{2, 
     3}, {3, 4}}}}, {{{{1, 2}, {2, 3}}, {{2, 3}, {3, 4}}}, {{{2, 
     3}, {3, 4}}, {{3, 4}, {4, 5}}}}} *)
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Working from your example:

ClearAll[Mat]
Mat[n_Integer] := Nest[Partition[#, 2, 1] &, Range[0, n], n - 1]

Mat[3]

(* Out: {{{0, 1}, {1, 2}}, {{1, 2}, {2, 3}}} *)

Checking with your desired output:

% == {{{0, 1}, {1, 2}}, {{1, 2}, {2, 3}}} (* Out: True *)

Also: 

Mat[5]

(* Out: 
{{{{{0, 1}, {1, 2}}, {{1, 2}, {2, 3}}}, {{{1, 2}, {2, 3}}, {{2, 3}, {3, 4}}}}, {{{{1, 2}, {2, 3}}, {{2, 3}, {3, 4}}}, {{{2, 3}, {3, 4}}, {{3, 4}, {4, 5}}}}}
*)
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