Comma-separated assignment to a table variable
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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
[S.X{:}]=deal(1,2,3) %Case (2) : works fine
[T.X{:}]=deal(1,2,3) %Case (3) : fails
11 个评论
Stephen23
2026-6-19，20:00
Table overloads both subsref and subsasgn. Its indexing is nothing like any other class.
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)
[T.X{1},T.X{2},T.X{3}] = deal(1,2,3)
Matt J
2026-6-19，21:42
Matt J
2026-6-20，12:57
dpb
2026-6-20，13:47
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.
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)
or to one element of vector cell
X = cell(1,3);
T = table(X)
[T.X{2}] = deal(pi)
And deal can expand the table variable
[T.X{4}] = deal(8)
[T.X{2,3}] = deal(100)
Peter Perkins
about 4 hours 前
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.
采纳的回答
Matt J
2026-6-22，14:57
1 个投票
2 个评论
Jack
about 5 hours 前
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.
更多回答（1 个）
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.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.X
2 个评论
Peter Perkins
about 4 hours 前
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]}
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