p = {{1, 0, 0}, {0, 1, 0}, {0, 1, 0}};
p[[All, 3]][[1 ;; 2]] = {1, 1};
p
I want to replace the first two elements of the third column of a 3x3 identity matrix.
The above code does not work. I get a depth-error message.
Set::partd: Part specification is longer than depth of object
I am confused because the following code is functional.
p[[All, 3]][[1 ;; 2]]
The way Set works in setting parts of an expression is this:
symb[[..<part specification>..]] = values;
The component symb must be a symbol (i.e. with head Symbol).
In the OP's code,
p[[All,3]][[1;;2]] = {1,1};
The symb component is p[[All,3]], which is not a symbol.
Fix as @kglr suggests,
p[[1 ;; 2, 3]] = {1, 1};
As for evaluating p[[All,3]][[1;;2]], you can see the procedure in Trace[p[[All, 3]][[1 ;; 2]]]. First p[[All, 3]] is evaluated. The expression then becomes
{0, 0, 0}[[1 ;; 2]]
(If p is meant to be the identity matrix, then the third row is wrong.)
p = {{1, 0, 0}, {0, 1, 0}, {0, 1, 0}};
f = MapAt[1 &, {1 ;; 2, 3}];
f @ p
{{1, 0, 1}, {0, 1, 1}, {0, 1, 0}}
If we want to change p inline we can use ApplyTo (new in 12.2)
p //= f;
p
{{1, 0, 1}, {0, 1, 1}, {0, 1, 0}}
Using SubsetMap:
p = {{1, 0, 0}, {0, 1, 0}, {0, 1, 0}};
SubsetMap[{1, 1} &, p, {{1, 3}, {2, 3}}]
({{1, 0, 1}, {0, 1, 1}, {0, 1, 0}})
p[[1 ;; 2, 3]] = {1, 1}? – kglr Jun 15 '17 at 01:18