How to efficiently replace value at the n last column?

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Dear all, The objective was to replace the value at the end of a column. I have make a simple case but wonder if there is another way to make it more compact and efficient?
Thanks in advance
f_sbj=2
sA=23;
A =1:sA;
appndOut=cell(sA,1);
for f_x=1:sA
Out = nchoosek(A, f_x);
[rows, columns] = find(Out==f_sbj);
Out(rows, :) = [];
[m,n] = size(Out );
newNan=nan(m,(sA-n));
appndOut{f_x}=[newNan Out];
end
  2 Comments
Walter Roberson
Walter Roberson on 4 Sep 2018
That code appears to insert at the beginning of columns, not replace at the end of columns.
The code can be made more compact, but not necessarily more efficient.
balandong
balandong on 4 Sep 2018
Thanks for the quick response, Appreciate if you can some idea to make it more compact

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Accepted Answer

Walter Roberson
Walter Roberson on 4 Sep 2018
lastNcol = @(M, N) M(:,end-N+1:end);
NanPrePad = @(M, N) lastNcol( [nan(size(M,1), N), M], N);
select_and_prepad = @(M, N) NanPrePad( M(all(M~=f_sbj, 2),:) );
appndOut = arrayfun(@(f_x) select_and_prepad(nchoosek(A, f_x), sA), (1:sA).', 'uniform', 0);
You can use shorter names for the anonymous functions if you find the above to not be compact enough.
Though I wonder why you do not do something like
A = setdiff(1:sA, f_sbj);
That would eliminate the possibility of f_sbj being generated by the nchoosek, so it would not be necessary to delete any rows containing f_sbj.
  3 Comments
Walter Roberson
Walter Roberson on 4 Sep 2018
f_sbj=2
sA=23;
A =1:sA;
lastNcol = @(M, N) M(:,end-N+1:end);
NanPrePad = @(M, N) lastNcol( [nan(size(M,1), N), M], N);
select_and_prepad = @(M, N) NanPrePad( M(all(M~=f_sbj, 2),:), N );
appndOut = arrayfun(@(f_x) select_and_prepad(nchoosek(A, f_x), sA), (1:sA).', 'uniform', 0);
This takes a moment to execute.

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