Passing a matrix as a function argument

Question

I am trying to do the following:

q = Table[i + j, {i, 1, 15}, {j, 1, 10}];
normalize[mat_] := (temp = Transpose[mat[[2 ;;, 5 ;;]]];
  temp1 = Transpose[N[Map[Normalize[#] &, temp]]];
  mat[[2 ;;, 5 ;;]] = temp1; Return[mat])
q1=normalize[q];

I get the following error:

Set::setps: 
{{2,3,4,5,6,7,8,9,10,11},
{3,4,5,6,7,8,9,10,11,12},
{4,5,6,7,8,9,10,11,12,13},
{5,6,7,8,9,10,11,12,13,14},
{6,7,8,9,10,11,12,13,14,15},<<5>>,
{12,13,14,15,16,17,18,19,20,21},
{13,14,15,16,17,18,19,20,21,22},
{14,15,16,17,18,19,20,21,22,23},
{15,16,17,18,19,20,21,22,23,24},
{16,17,18,19,20,21,22,23,24,25}} 
in the part assignment is not a symbol. >>

Can you please help me in finding out what is wrong? I am new to Mathematica.

Answer 1

To complement the other answer presented, I would like to suggest that it is often cleaner not to carry around additional symbols such as temp and temp1 in your code.

If we want to modify a copy of the matrix we still need a "proxy" symbol which I will create with Module. (This is safer than Block as the latter will replace any symbols seen in the course of evaluation, even ones that are in the array; for example if x were in the array and I used Block it would conflict.) You also do not need explicit Return except in special circumstances.

normalize[mat_] :=
 Module[{x = mat},
   x[[2 ;;, 5 ;;]] = 
     N @ Map[Normalize, x[[2 ;;, 5 ;;]]\[Transpose]]\[Transpose];
   x
 ]

Or, to modify the array in-place no proxy is needed, but HoldFirst is:

SetAttributes[normalizeInPlace, HoldFirst]

normalizeInPlace[mat_] :=
 (
   mat[[2 ;;, 5 ;;]] = 
     N @ Map[Normalize, mat[[2 ;;, 5 ;;]]\[Transpose]]\[Transpose];
   mat
 )

Answer 2

Answer

You can use the HoldAll attribute. Suppose I want a function that takes some list or matrix as input and sets its 4th element to 42. This can be done like so:

ClearAll[f];
SetAttributes[f, HoldAll]
f[m_] := (m[[4]]= 42; m)

then

t = RandomInteger[{0, 10}, {5, 2}]
f[t]

(*
{{6, 2}, {7, 4}, {0, 4}, {10, 6}, {8, 7}}
{{6, 2}, {7, 4}, {0, 4}, 42, {8, 7}}
*)

In your case, it would be

ClearAll[normalize, q, q1];
q = Table[i + j, {i, 1, 15}, {j, 1, 10}];
SetAttributes[normalize, HoldAll];
normalize[mat_] := (temp = Transpose[mat[[2 ;;, 5 ;;]]];
  temp1 = Transpose[N[Map[Normalize[#] &, temp]]];
  mat[[2 ;;, 5 ;;]] = temp1; Return[mat])
q1 = normalize[q];

Explanation

The HoldAll attribute tells the evaluator not to evaluate the argument of your function before trying to do operations on it. Otherwise, look at what happens (using Trace which shows what is being evaluated):

ClearAll[f, g];
SetAttributes[f, HoldAll]
f[m_] := (m[[4]] = 42; m)

g[m_] := (m[[4]] = 42; m)

t = RandomInteger[{0, 10}, {5}]
f[t] // Trace

(*
{7, 6, 3, 7, 6}

{f[t],t[[4]]=42;t,{t[[4]]=42,42},{t,{7,6,3,42,6}},{7,6,3,42,6}}
*)

while without HoldAll,

g[t] // Trace

issues an error and dumps some stuff on the screen, most relevant of which (and immediately before the error) is this:

{7, 6, 3, 42, 6}[[4]] = 42

which is what causes the error. In the previous case, at this point there was a t[[4]]=42 which is fine.

You can think of this as passing by reference, versus passing by value if the HoldAll attribute is not set (although this is not entirely accurate).

Warning

Note that in this approach, you are manipulating the t that is in the global context. If you are writing more complicated code, you probably want to do what JM described.

Answer 3

You need another temporary variable to be able to do what you want:

normalize[mat_] := Block[{matTemp = mat, temp, temp1},
  temp = Transpose[matTemp[[2 ;;, 5 ;;]]]; 
  temp1 = Transpose[N[Map[Normalize, temp]]];
  matTemp[[2 ;;, 5 ;;]] = temp1;
  matTemp]

Note that in the line mat[[2 ;;, 5 ;;]] = temp1 of the original code, the mat is replaced by the matrix you have given as argument, and since you can't assign to a matrix of a numbers, you get that error you saw.