randperm function for a table - keep rows together

Question

Alessandro Livi 2024-7-10

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2136318-randperm-function-for-a-table-keep-rows-together

编辑： Stephen23 2024-7-10

采纳的回答： Alessandro Livi

在 MATLAB Online 中打开

I looked in help but everone seems to be randomizing a vector.

The closest ask I can find is

https://www.mathworks.com/matlabcentral/answers/155207-matrix-with-different-randperm-rows?s_tid=sug_su

I have a table (including a category column) I asked (Q )about using Shuffle but if that doesn't handle table shuffling like I need i can take the performance hit and use randperm()

Here's what I'm doing:

app.TrialsTable = table('Size',[2*C 4],'VariableTypes',{'uint8','uint8','categorical','int16'});
... % fill the table from given input table, expanding rows as needed for number of repeats
    app.TrialsTable(index, 1) = ... %and so on
    ...
    app.TrialsTable = Shuffle(app.TrialsTable, ???);  % if anyone knows how to do it this way
& or
app.TrialsTable(randperm(height(app.TrialsTable(:,:))));   % How do I make this line work??
% and then finally
set(app.RandTrialsTable,'Data',app.TrialsTable(:,:)); % save it to the UITable for display in the Figure

Testing at the command prompt for various options of randperm(...)

app.TrialsTable(randperm(height(app.TrialsTable(:,:))))

Error using ()

Subscripting into a table using one subscript (as in t(i)) is not supported. Specify

a row subscript and a variable subscript, as in t(rows,vars). To select variables,

use t(:,i) or for one variable t.(i). To select rows, use t(i,:).

I didn't use 1 subscript so the Error message doesn't help

Another try:

app.TrialsTable(:,:)(randperm(height(app.TrialsTable(:,:))))

Error: Invalid array indexing.

Test

(randperm(height(app.TrialsTable(:,:))))

ans =

Columns 1 through 14

17 18 13 8 3 9 19 27 14 6 25 15 2 5

Columns 15 through 28

7 23 24 12 22 20 28 21 26 4 16 1 10 11

I only have 4 columns (28 rows)

Try

(randperm(height(app.TrialsTable)))

ans =

Columns 1 through 14

27 14 22 4 10 2 21 3 5 15 18 28 23 7

Columns 15 through 28

24 16 25 9 6 13 26 17 19 20 1 11 8 12

No sign of the category column ('L' or 'R') sorting along with it.

OH! maybe I don't need to tell the height (as the Shuffle guy said)

(randperm(app.TrialsTable))

Error using randperm

Conversion to double from table is not possible.

try

(randperm(app.TrialsTable(:,:)))

Error using randperm

Conversion to double from table is not possible.

randperm(1:size(app.TrialsTable.Row))

ans =

1×0 empty double row vector

I tried on my own, time to ask for help!

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Stephen23 2024-7-10

编辑：Stephen23 2024-7-10

在 MATLAB Online 中打开

"I didn't use 1 subscript so the Error message doesn't help"

Yes you did, exactly on the line where the error occured:

app.TrialsTable(randperm(height(app.TrialsTable(:,:))))

% ^ indexing using exactly one subscript ^

Lets split the code onto two lines to make it clearer:

X = randperm(height(app.TrialsTable(:,:)));

app.TrialsTable(X) % <- that is exactly one subscript

"How do I make this line work??"

You were so close: if you had paid attention to the error message then you might have gotten there yourself. Just generate the desired indices and then apply them to the rows:

T = array2table(magic(4))
T = 4x4 table
    Var1    Var2    Var3    Var4
    ____    ____    ____    ____

     16       2       3      13 
      5      11      10       8 
      9       7       6      12 
      4      14      15       1 
X = randperm(height(T))
X = 1x4
     3     4     2     1
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
T(X,:) % two subscripts, not one
ans = 4x4 table
    Var1    Var2    Var3    Var4
    ____    ____    ____    ____

      9       7       6      12 
      4      14      15       1 
      5      11      10       8 
     16       2       3      13 