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I have a matrix where I would like to look for all elements around element x (in this example -2) and replace all the ones that are 0 with the number 1. My matrix (matrix0) looks as follows:

\begin{array}{cccccccccc} -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -2 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ \end{array}

matrix0 = {{-1, -1, -1, -1, -1, -1, -1, -1, -1, -1},
           {-1, 0, 0, 0, 0, 0, 0, 0, 0, -1},
           {-1, 0, 0, 0, 0, 0, 0, 0, 0, -1},
           {-1, 0, 0, 0, 0, 0, 0, 0, 0, -1},
           {-1, 0, 0, 0, 0, 0, 0, 0, 0, -2},
           {-1, 0, 0, 0, 0, 0, 0, 0, 0, -1},
           {-1, 0, 0, 0, 0, 0, 0, 0, 0, -1},
           {-1, 0, 0, 0, 0, 0, 0, 0, 0, -1},
           {-1, 0, 0, 0, 0, 0, 0, 0, 0, -1},
           {-1, -1, -1, -1, -1, -1, -1, -1, -1, -1}};

I wrote a module, which gives me the positions of the elements around my x like this:

Clear[around] (* The following module gives a list of all positions next to that of x0
in matrix mat0 *)

around[x0_, mat0_] :=
 Module[{x, mat, pos},
  x = x0;
  mat = mat0;
  pos[x_] := Flatten[Position[mat, x]];
  Cases[Tuples[Range[Dimensions[mat][[1]]], 
    2], {a_, 
     b_} /; (a == pos[x][[1]] && 
       b == pos[x][[2]] - 1) || (a == pos[x][[1]] - 1 && 
       b == pos[x][[2]]) || (a == pos[x][[1]] - 1 && 
       b == pos[x][[2]] - 1) || (a == pos[x][[1]] + 1 && 
       b == pos[x][[2]] - 1) || (a == pos[x][[1]] - 1 && 
       b == pos[x][[2]] + 1) || (a == pos[x][[1]] && 
       b == pos[x][[2]] + 1) || (a == pos[x][[1]] + 1 && 
       b == pos[x][[2]]) || (a == pos[x][[1]] + 1 && 
       b == pos[x][[2]] + 1)]
  ]

For -2 this yields the result:

around[-2, matrix0]
{{4,9},{4,10},{5,9},{6,9},{6,10}}

I now tried to let all of these elements which are 0 be replaced by 1, however Mathematica is only giving me a Null output. Any ideas why this doesn't work?

For[i=1,i<=Length[around[-2,matrix0]],i++,
If[Part[matrix0,around[-2,matrix0][[i,1]],around[-2,matrix0][[i,2]]]==0,
ReplacePart[matrix0,around[-2,matrix0][[i]]->1],Null]]

EDIT: These answers are already helping me learn new functions thank you. However my question mainly was why my version does not work seeing as I can't find an illogical step in it. The reason my version is so unnecessarily complicated is partly because I am a Mathematica beginner but also because I want to generalise this method to build up my matrix. After having changed the 0s around -2 to 1s, I would like to change the 0s around all 1s to 2s and the 0s around all 2s to 3s and so on. Any suggestion what I should look into to solve this on my own? Or is my approach just fundamentally doomed to failure?

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  • $\begingroup$ Try {m, n} = Dimensions[mat0]; ReplacePart[mat0, Cases[First[Position[mat0, x0]] + # & /@ DeleteCases[Tuples[{-1, 0, 1}, 2], {0, 0}], {j_, k_} /; 1 <= j <= m && 1 <= k <= n] -> 1]. $\endgroup$
    – J. M.'s persistent exhaustion ♦
    Jul 31, 2017 at 15:28
  • 1
    $\begingroup$ It would be very helpful if you could share your matrix as Mathematica code rather than as TeX. $\endgroup$
    – MarcoB
    Jul 31, 2017 at 16:14
  • $\begingroup$ Somewhat related: (140099) $\endgroup$
    – Mr.Wizard
    Jul 31, 2017 at 21:26
  • $\begingroup$ As noted, please include data that can be easily copied the next time. $\endgroup$
    – J. M.'s persistent exhaustion ♦
    Aug 1, 2017 at 2:34

5 Answers 5

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Update: Re: why my version does not work

Documentation >> ReplacePart:

  • ReplacePart[expr, i -> new] yields an expression in which the i'th part of expr is replaced by new.

that is, it does not replace expr with the expression it yields.

So, with a small change, namely assigning the value of ReplacePart[...] to matrix0 in each step of the For loop, your version also works:

For[i = 1, i <= Length[around[-2, matrix0]], i++, 
 If[Part[matrix0, around[-2, matrix0][[i, 1]], around[-2, matrix0][[i, 2]]] == 0, 
 matrix0 = ReplacePart[matrix0, around[-2, matrix0][[i]] -> 1], Null]]

matrix0 // TeXForm

$\left( \begin{array}{cccccccccc} -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -2 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ \end{array} \right)$

By the way, since

  • ReplacePart[expr, {i, j, ...} -> new] replaces the part at position {i, j, ...}

you can use much simpler

 matrix0 = ReplacePart[matrix0, around[-2, matrix0] -> 1]

instead of using a For loop.

Original answer - updated:

ClearAll[aroundF, replaceF]
aroundF[m_, t_] := DeleteDuplicates[Join @@ Function[{k}, 
 Clip[#, {1, #2}] & @@@ Transpose[{k + #, Dimensions[m]}] & /@ 
 Tuples[{-1, 0, 1}, {2}]] /@ Position[m, t]]



replaceF[old_: 0, new_: 1][m_, t_] := MapAt[If[#===old, new, #] &, m, aroundF[m,t]]

replaceF[][matrix0, -2]  // TeXForm

$\left( \begin{array}{cccccccccc} -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -2 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ \end{array} \right)$

replaceF[0, aa][matrix0, -2]  // TeXForm

$ \left( \begin{array}{cccccccccc} -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{aa} & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{aa} & -2 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{aa} & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ \end{array} \right) $

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  • $\begingroup$ To the RE: Using only ReplacePart would be a good trick but that does not include that I only want parts that are 0 to be replaced does it? $\endgroup$
    – TheSeparatrix
    Aug 2, 2017 at 8:11
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Here is a version that is pretty fast:

Clear[setNeighbors]

setNeighbors[m_, n_, foo_Rule:(0->1)] := Module[{nmatrix,omatrix,old,new},
    {old,new} = List @@ foo;
    nmatrix=Unitize @ ListCorrelate[
        {{1,1,1}, {1,0,1}, {1,1,1}},
        Unitize @ Clip[m, {n, n}, {0, 0}],
        {2, -2},
        0
    ];
    omatrix = 1 - Unitize[m - old];
    m + omatrix nmatrix (new-old)
]

For your example:

m = {{-1, -1, -1, -1, -1, -1, -1, -1, -1, -1},
{-1, 0, 0, 0, 0, 0, 0, 0, 0, -1},
{-1, 0, 0, 0, 0, 0, 0, 0, 0, -1},
{-1, 0, 0, 0, 0, 0, 0, 0, 0, -1},
{-1, 0, 0, 0, 0, 0, 0, 0, 0, -2},
{-1, 0, 0, 0, 0, 0, 0, 0, 0, -1},
{-1, 0, 0, 0, 0, 0, 0, 0, 0, -1},
{-1, 0, 0, 0, 0, 0, 0, 0, 0, -1},
{-1, 0, 0, 0, 0, 0, 0, 0, 0, -1},
{-1, -1, -1, -1, -1, -1, -1, -1, -1, -1}};

setNeighbors[m, -2] //TeXForm

$\left( \begin{array}{cccccccccc} -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -2 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ \end{array} \right)$

Here is the timing for a much larger matrix:

data = RandomInteger[100, {1000, 1000}];

setNeighbors[data, 3, 1->3]; //AbsoluteTiming

{0.051501, Null}

Finally, it seems that @klgr's solution might have a bug. Compare:

SeedRandom[2];
m = RandomInteger[10, {5, 5}];
m //TeXForm

$\left( \begin{array}{ccccc} 8 & 4 & 5 & 4 & 7 \\ 4 & 0 & 1 & 0 & 4 \\ 3 & 7 & 3 & 0 & 2 \\ 7 & 8 & 7 & 9 & 3 \\ 6 & 2 & 3 & 8 & 9 \\ \end{array} \right)$

r1 = setNeighbors[m, 3, 1->3];
r1 //TeXForm

$\left( \begin{array}{ccccc} 8 & 4 & 5 & 4 & 7 \\ 4 & 0 & 3 & 0 & 4 \\ 3 & 7 & 3 & 0 & 2 \\ 7 & 8 & 7 & 9 & 3 \\ 6 & 2 & 3 & 8 & 9 \\ \end{array} \right)$

r2 = setNeighborValues[1,3][m,3];
r2 //TeXForm

$\left( \begin{array}{ccccc} 8 & 4 & 5 & 4 & 7 \\ 4 & 0 & 1 & 0 & 4 \\ 3 & 7 & 3 & 0 & 2 \\ 7 & 8 & 7 & 9 & 3 \\ 6 & 2 & 3 & 8 & 9 \\ \end{array} \right)$

r1 === r2

False

I think the $m_{2, 3}$ entry should be a 3.

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  • $\begingroup$ Very nice! I wanted to apply ListCorrelate too but I ran out of time yesterday so I just left a link in the comments. $\endgroup$
    – Mr.Wizard
    Aug 1, 2017 at 10:12
  • $\begingroup$ @Carl, thank you. The bug is fixed now. $\endgroup$
    – kglr
    Aug 1, 2017 at 11:57
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Here is a cleaner approach using Developer`PartitionMap.

(m as defined in my first answer.)

fn[b_] := If[# != 0, #, Total[1 - Unitize[b + 2], 2]] & @ b[[2, 2]]

Developer`PartitionMap[fn, m, {3, 3}, 1, 2] // MatrixForm

$\left( \begin{array}{cccccccccc} -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -2 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ \end{array} \right)$

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A related post:

I'll try to apply the same method here.

Your data

I'll start by assigning your data to m:

m = {{-1, -1, -1, -1, -1, -1, -1, -1, -1, -1},
   {-1, 0, 0, 0, 0, 0, 0, 0, 0, -1},
   {-1, 0, 0, 0, 0, 0, 0, 0, 0, -1},
   {-1, 0, 0, 0, 0, 0, 0, 0, 0, -1},
   {-1, 0, 0, 0, 0, 0, 0, 0, 0, -2},
   {-1, 0, 0, 0, 0, 0, 0, 0, 0, -1},
   {-1, 0, 0, 0, 0, 0, 0, 0, 0, -1},
   {-1, 0, 0, 0, 0, 0, 0, 0, 0, -1},
   {-1, 0, 0, 0, 0, 0, 0, 0, 0, -1},
   {-1, -1, -1, -1, -1, -1, -1, -1, -1, -1}};

Two-dimensional replacement rules

Using CellularAutomaton:

p = Partition[#, 3] & /@ Permutations[Table[_, {8}]~Append~-2];
p[[All, 2, 2]] = 0;

rules = Append[Thread[(Delete[p, 5]) -> 1], ArrayPad[{{x_}}, 1, _] :> x];

CellularAutomaton[rules, m, 1][[2]] // MatrixForm

$\left( \begin{array}{cccccccccc} -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -2 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ \end{array} \right)$

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$\begingroup$ 
m = {{-1, -1, -1, -1, -1, -1, -1, -1, -1, -1},
   {-1, 0, 0, 0, 0, 0, 0, 0, 0, -1},
   {-1, 0, 0, 0, 0, 0, 0, 0, 0, -1},
   {-1, 0, 0, 0, 0, 0, 0, 0, 0, -1},
   {-1, 0, 0, 0, 0, 0, 0, 0, 0, -2},
   {-1, 0, 0, 0, 0, 0, 0, 0, 0, -1},
   {-1, 0, 0, 0, 0, 0, 0, 0, 0, -1},
   {-1, 0, 0, 0, 0, 0, 0, 0, 0, -1},
   {-1, 0, 0, 0, 0, 0, 0, 0, 0, -1},
   {-1, -1, -1, -1, -1, -1, -1, -1, -1, -1}};

rep[m_, v_] :=
 Module[{a, r, c, p},
  p = {-1, 0, 1};
  a = ArrayPad[m, 1];
  {r, c} = FirstPosition[a, v];
  a[[r + p, c + p]] = a[[r + p, c + p]] /. (0) -> 1;
  a[[2 ;; -2, 2 ;; -2]]]

rep[m, -2] // MatrixForm

enter image description here

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