Part Assignment of variables with argument

Question

I do not really understand why this:

a = {1, 2, 3};
a[[1]] = 3;
a

{3,2,3}

works as desired, but this does not:

b[1] = {1, 2, 3};
b[1][[1]] = 3;
b[1]

Set::setps: b[1] in the part assignment is not a symbol.

Can I not treat variables with arguments like 'normal' variables?

$\begingroup$ closely related: 148387 $\endgroup$
– Kuba ♦
Oct 22, 2019 at 9:20 — Kuba, Oct 22, 2019 at 9:20

kglr · Accepted Answer · 2019-10-22 09:20:10Z

1

ClearAll[b]
b[1] = {1, 2, 3};
b[1] = ReplacePart[ b[1], 1 -> 3];
b[1]

{3, 2, 3}

Alternatively, MapAt:

ClearAll[b]
b[1] = {1, 2, 3};
b[1] = MapAt[3 &, b[1], {1}];
b[1]

{3, 2, 3}

answered Oct 22, 2019 at 9:20

kglr

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$\begingroup$ Interesting thank you. What would I do If I wanted to set a column in a two Dimensional array? Like this: a = ConstantArray[1, {10, 10}] a[[;; , 3]] = ConstantArray[2, 10]; a $\endgroup$
– Mr Puh
Oct 22, 2019 at 9:33
$\begingroup$ @MrPuh, the code in your comment works as expected. $\endgroup$
– kglr
Oct 22, 2019 at 9:35
$\begingroup$ yeah, I mean in the sense of the question, I would want something like this now: a[1] = ConstantArray[1, {10, 10}] a[1][[;; , 3]] = ConstantArray[2, 10]; a[1] $\endgroup$
– Mr Puh
Oct 22, 2019 at 9:43
1

$\begingroup$ @MrPuh, you can do ClearAll[a];a[1] = ConstantArray[1, {10, 10}] ;a[1]=MapAt[ 2&,a[1], {All,3}];a[1] or ClearAll[a];a[1] = ConstantArray[1, {10, 10}] ; a[1]=ReplacePart[ a[1], {_,3}->2];a[1] $\endgroup$
– kglr
Oct 22, 2019 at 9:53
$\begingroup$ ... you can also use SubsetMap: ClearAll[a];a[1] = ConstantArray[1, {10, 10}] ; a[1]=SubsetMap[ConstantArray[2,10]&,a[1], {All,3}];a[1] $\endgroup$
– kglr
Oct 22, 2019 at 9:55

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