Comma-separated assignment to a table variable

The code below tests assignment by comma-separated expansion to a cell array in 3 different scenarios. In the specific case when the cell array X is the variable of a table, it fails. Why is that?
X=cell(1,3);
S.X=X;
T=table(X);
[X{:}]=deal(1,2,3) %Case (1) : works fine
X = 1×3 cell array
{[1]} {[2]} {[3]}
[S.X{:}]=deal(1,2,3) %Case (2) : works fine
S = struct with fields:
X: {[1] [2] [3]}
[T.X{:}]=deal(1,2,3) %Case (3) : fails
Error using deal (line 37)
The number of outputs should match the number of inputs.

13 Commenti

Table overloads both subsref and subsasgn. Its indexing is nothing like any other class.
Paul
Paul il 19 Giu 2026 alle 20:44
Spostato: Paul il 19 Giu 2026 alle 21:23
Why does the second call to deal result in three copies of the same displayed output?
X = cell(1,3);
T = table(X);
[X{:}] = deal(1,2,3)
X = 1×3 cell array
{[1]} {[2]} {[3]}
[T.X{1},T.X{2},T.X{3}] = deal(1,2,3)
T = table
X _______________________ {[1]} {[2]} {[3]}
T = table
X _______________________ {[1]} {[2]} {[3]}
T = table
X _______________________ {[1]} {[2]} {[3]}
Matt J
Matt J il 19 Giu 2026 alle 21:42
Table overloads both subsref and subsasgn.
Does that explain why comma-separated assignment would not be supported?
Matt J
Matt J il 19 Giu 2026 alle 22:40
Modificato: Matt J il 19 Giu 2026 alle 22:41
Why does the second call to deal result in three copies of the same displayed output?
@Paul It is normal for unsuppressed output of CSL assignment to make nargout displays if it thinks the outputs are separate and unrelated variables, e.g.,
[a,b]=deal(1,2)
a = 1
b = 2
When you assign to separately indexed regions of the same variable, it apparently cannot tell they are related,
[X{1},X{2}] = deal(4,5)
X = 1×2 cell array
{[4]} {[5]}
X = 1×2 cell array
{[4]} {[5]}
as distinct from,
[X{1:2}]=deal(6,7)
X = 1×2 cell array
{[6]} {[7]}
Stephen23
Stephen23 il 20 Giu 2026 alle 12:14
Modificato: Stephen23 il 20 Giu 2026 alle 13:33
"Does that explain why comma-separated assignment would not be supported?"
No, it explains why comparisons to other data types are not particularly useful.
Matt J
Matt J il 20 Giu 2026 alle 12:57
No, it explains why comparisons to other data types are not particualrly useful.
I don't know about that. This is a built-in Matlab data type meant as a general kind of container. All other containers support the full range of indexing.
dpb
dpb il 20 Giu 2026 alle 13:47
" All other containers support the full range of indexing."
"All"?
Matt J
Matt J il 20 Giu 2026 alle 14:16
Modificato: Matt J il 20 Giu 2026 alle 14:19
All containers within the set of Fundamental Classes. Regardless, the main thing was that there doesn't seem to be a good reason for tables not to allow this.
dpb
dpb il 20 Giu 2026 alle 14:25
Modificato: dpb il 20 Giu 2026 alle 14:29
I note there that the table/timetable are notably missing the "Use in comma separated lists" line item in the "Intended Use" column....guess it also shows how long it's been since I did more than look up individual functions in the doc -- I had forgotten about that overview page existing.
While I'd never attempted the addressing to a table in that fashion and wouldn't have thought of doing so in all likelihood anyway, I'll confess I don't know enough about the internals for it to be apparent to me why it couldn't be done; I would presume adding the additional complexity on top of existing logic could be a performance issue in adding yet another layer of complexity.
Paul
Paul il 21 Giu 2026 alle 16:03
Modificato: Paul il 22 Giu 2026 alle 12:01
AFAICT, numArgumentsFromSubscript alway returns 1 for a table for assignment, at least for the various cases that I tried, which is why deal fails for the example at hand. But that also means that deal works for assigning to a scalar cell in the table variable
X = cell(1,1);
T = table(X);
[T.X{:}] = deal(pi)
T = table
X __________ {[3.1416]}
or to one element of vector cell
X = cell(1,3);
T = table(X)
T = table
X ____________________________________________ {0×0 double} {0×0 double} {0×0 double}
[T.X{2}] = deal(pi)
T = table
X __________________________________________ {0×0 double} {[3.1416]} {0×0 double}
And deal can expand the table variable
[T.X{4}] = deal(8)
T = table
X ___________________________________________________ {0×0 double} {[3.1416]} {0×0 double} {[8]}
[T.X{2,3}] = deal(100)
T = 2×1 table
X ____________________________________________________________ {0×0 double} {[ 3.1416]} {0×0 double} {[ 8]} {0×0 double} {0×0 double} {[ 100]} {0×0 double}
Paul, I can't tell what you are trying to demonstrate. Putting square brackets around things like X{2} and then using deal makes no sense to me. T.X{2} is one thing, i.e. "the contents of" the 1x3 cell array t.X. You don't need square brackets or deal for this any more than you need them for X{2} = pi.
>> X = cell(1,3);
>> T = table(X)
T =
table
X
____________________________________________
{0×0 double} {0×0 double} {0×0 double}
>> T.X{2} = pi
T =
table
X
__________________________________________
{0×0 double} {[3.1416]} {0×0 double}
Even assigning two elements:
>> T.X(1,2:3) = {100}
T =
table
X
__________________________________
{0×0 double} {[100]} {[100]}
I will also say that
>> table(cell(3,1))
ans =
3×1 table
Var1
____________
{0×0 double}
{0×0 double}
{0×0 double}
is probably what you want, not
>> table(cell(1,3))
ans =
table
Var1
____________________________________________
{0×0 double} {0×0 double} {0×0 double}
but I can't say for sure.
This is just to point out that the issue affects table variables that are struct-valued as well.
[Sarr(1:3,1).X]=deal(0,0,0)
Sarr = 3×1 struct array with fields:
X
S.Sarr=Sarr; %Contain in struct
T=table(Sarr); %Contain in table
[S.Sarr.X]=deal(1,2,3); S.Sarr.X
ans = 1
ans = 2
ans = 3
[T.Sarr.X]=deal(4,5,6); T.Sarr.X
Error using deal (line 37)
The number of outputs should match the number of inputs.
I was just trying to show that deal works when assigning scalar to scalar when using the general form
[CSL] = deal(CSL)
even though deal doesn't work for other cases as shown. That seemed like an inconsistency to me, even if one would never need to use deal for those scalar to scalar assignments.
I used table(cell(1,3)) because that was the example used in the Question that started this thread.

Accedi per commentare.

 Risposta accettata

Matt J
Matt J il 22 Giu 2026 alle 14:57
According to Tech support, it is not a bug, just something that wasn't implemented for tables. However, they see it as a legitimate enhancement, and have filed it as an enhancement request.

2 Commenti

Hi Matt,
Yes, we're aware of this indexing inconsistency and consider it a valid enhancement. Do you have any examples of workflows or use-cases where comma-separated list assignment comes up/is more suitable than its alternatives?
Like @dpb pointed out, there are other ways to do this kind of assignment. More commonly, users have a cell array and not a comma-separated list, so these alternatives are usually sufficient.
Matt J
Matt J il 10 Lug 2026 alle 17:59
Modificato: Matt J circa 18 ore fa
Hi @Jack,
More commonly, users have a cell array and not a comma-separated list, so these alternatives are usually sufficient.
I assume you meant to say "more commonly users have a cell array and not a struct array"? Comma-separated-list assignment is not a distinct thing from cell arrays. It is an operation that both struct and cell arrays support.
The alternative assignment methods you mention appear to apply only to cells and not to structs. If struct assignment were uncommon compared to cell assignment, I guess that limitation wouldn't matter as much, but I don't know how anyone could know which is more common.
Do you have any examples of workflows or use-cases where comma-separated list assignment comes up/is more suitable than its alternatives?
The use cases are those where you want to populate a container with varargout. The advantage of varargout in conjunction with comma separated lists is that you don't have to write the function with any prior knowledge of whether the output is going to be assigned to brace-indexed or to dot-indexed containers. As an example, the function dealfun below is a utility I use to invoke a function on a series of inputs and send the results to a container. It works for at least 3 different container forms, but not when the container is a table variable.
X=cell(1,3);
S.X=X;
Sarr(3).X=[];
T=table(X);
[X{:}]=dealfun(@(x) x.^2, 2,3,4)
X = 1×3 cell array
{[4]} {[9]} {[16]}
[Sarr(1:3).X] = dealfun(@(x) x.^2, 2,3,4); Sarr.X
ans = 4
ans = 9
ans = 16
[S.X{:}]=dealfun(@(x) x.^2, 2,3,4)
S = struct with fields:
X: {[4] [9] [16]}
[T.X{:}]=dealfun(@(x) x.^2, 2,3,4)
Error using . (line 514)
Assigning to 3 elements using a simple assignment statement is not supported. Consider using comma-separated list assignment.
function [varargout]=dealfun(fun,varargin)
%Apply function to all arguments and deal() the results
varargin=cellfun(fun,varargin,'uni',0);
varargout=varargin(1:nargout);
end

Accedi per commentare.

Più risposte (1)

dpb
dpb il 19 Giu 2026 alle 20:38
Modificato: dpb il 19 Giu 2026 alle 20:39
As @Stephen23 notes, table addressing is its own animal...
The table is a one-row cell array in one vairable so it can be assigned directly as a cell can be...
X=cell(1,3);
T=table(X)
T = table
X ____________________________________________ {0×0 double} {0×0 double} {0×0 double}
T.X={1 2 3}
T = table
X _______________________ {[1]} {[2]} {[3]}
No need for deal() here as the dot-X notation returns the cell array in its entirety. Of course, in a table cell, it can contain anything any other cell can
T.X(1:2)={1 2};
T.X(3)={'Fred Flintstone'}
T = table
X _____________________________________ {[1]} {[2]} {'Fred Flintstone'}
T.X
ans = 1×3 cell array
{[1]} {[2]} {'Fred Flintstone'}

2 Commenti

Ah, I see that dpb has beat me to it. I will just add: If you find yourself relying on CSLs and/or deal anywhere in MATLAB, there is almost certainly a better way to do it.
In this case, the better way is, leave the cell array (whether it be X or T.X) as a cell array. Don't CSL expand it. I'm skeptical (but of course might be wrong) that deal(1,2,3) represents a real use case where there are three separate things in separate workspacce vars. But even if it is, just wrap them in braces and assign to the cell array:
>> X = cell(3,1);
>> T = table(X)
T =
3×1 table
X
____________
{0×0 double}
{0×0 double}
{0×0 double}
>> T.X = {1;2;3}
T =
3×1 table
X
_____
{[1]}
{[2]}
{[3]}
>> T.X(:) = {11;22;33}
T =
3×1 table
X
_____
{[11]}
{[22]}
{[33]}
>> T.X(2:3) = {222;333}
T =
3×1 table
X
______
{[ 11]}
{[222]}
{[333]}
Matt J
Matt J il 10 Lug 2026 alle 18:15
Modificato: Matt J circa 3 ore fa
The table is a one-row cell array in one vairable so it can be assigned directly as a cell can be...
That assumes the intended contents of the cell array are available as variables in the current workspace. But that is not always the case. It might be that the contents are the output arguments of a function call. In that case, you need comma separated list assignment to draw the required number of outputs from the function.

Accedi per commentare.

Prodotti

Release

R2024b

Richiesto:

il 19 Giu 2026 alle 19:41

Modificato:

circa 23 ore fa

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by