Suppose I have K variables, each of which can take the values 1 through N. I want to create a table such that each column corresponds to the Kth variable, and each row corresponds to a particular combination. This table will be N^K entries long and K entries wide. One way to do this would be, for example
[V1 V2 ... VK] = ngrid(1:N,1:N,...,1:N)
V = [V1 V2 ... VK]
However, this clearly does not generalize nicely to variable N. Is there a simple way to extend this code so it works for different K?

댓글 수: 8

James Tursa
James Tursa 2020년 12월 18일
How do you have your K variables stored? Hopefully not as named-numbered variables. In a cell array or as columns of a matrix?
David Cyncynates
David Cyncynates 2020년 12월 18일
편집: David Cyncynates 2020년 12월 18일
I called them variables just because it makes sense to think of them that way to me, but ultimately I just want to have a N^K by K array with each combination stored in a given row. I suppose I could do this the old-fashioned way, where I create the matrix row by row, but I feel like there aught to be a better way.
James Tursa
James Tursa 2020년 12월 18일
That doesn't answer my question. How do you have them currently stored?
David Cyncynates
David Cyncynates 2020년 12월 18일
편집: David Cyncynates 2020년 12월 18일
They're not stored. I would like it to be stored in an array. To be clear - this is my point of confusion! It feels very cumbersome to make this object in matlab, so I think I am missing something.
James Tursa
James Tursa 2020년 12월 18일
OK, I will ask it another way. Are the inputs always the exact values 1:N?
David Cyncynates
David Cyncynates 2020년 12월 18일
Yes, the only informatino you need to create this array is the values N and K: Basically, each row stores a number written in N symbols.
Here's an example of what I was trying to write. Perhaps I'll just stick with this:
N = 4;
K = 3;
V = zeros(K,N^K);
ii = 0 : N^K - 1;
for jj = 1 : K
V(jj,:) = floor((mod(ii,N^jj))/N^(jj-1));
end
James Tursa
James Tursa 2020년 12월 18일
편집: James Tursa 2020년 12월 18일
OK, got it. Looks like you are basically counting in base N with K digits. How large can K and N be? This could easily eat all your memory if they are too large.

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답변 (2개)

Stephen23
Stephen23 2020년 12월 18일
편집: Stephen23 2025년 7월 27일

0 개 추천

N = 4;
K = 3;
C = cell(1,K);
[C{:}] = ndgrid(1:N);
M = reshape(cat(K+1,C{:}),[],K)
M = 64×3
1 1 1 2 1 1 3 1 1 4 1 1 1 2 1 2 2 1 3 2 1 4 2 1 1 3 1 2 3 1 3 3 1 4 3 1 1 4 1 2 4 1 3 4 1
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How it works:
As James Tursa pointed out, you could quickly run out of memory for large values of K.

댓글 수: 4

Hi Stephen,
I thought this would work, but it throws an error:
N = 4;
K = 3;
try
T = combinations(struct('a',repmat({1:N},1,K)).a)
catch ME
ME.message
end
ans = 'Argument number is not valid.'
But this works:
C = repmat({1:N},1,K);
T = combinations(C{:});
Is it not the case that both calls to combinations use the same comma-separated-list on input?
It appears the issue is with a call to inputname in combinations
try
inputnametest(struct('a',repmat({1:N},1,K)).a)
catch ME
ME.message
end
nargin =
ans = 3
varargin = 1×3 cell array
{[1 2 3 4]} {[1 2 3 4]} {[1 2 3 4]}
ans = 'Argument number is not valid.'
Why is 2 not a valid input argument number when nargin == 3?
But this works fine:
inputnametest(C{:})
nargin =
ans = 3
varargin = 1×3 cell array
{[1 2 3 4]} {[1 2 3 4]} {[1 2 3 4]}
ans = 0×0 empty char array
function inputnametest(varargin)
disp('nargin = ');nargin
varargin
inputname(2)
end
Stephen23
Stephen23 2025년 7월 27일
편집: Stephen23 2025년 7월 28일
@Paul: that appears to be a bug in how the fun(struct(..).fieldname) syntax is parsed when the function FUN just happens to call INPUTNAME. It seems to not depend on COMBINATIONS in particular as it also occurs with other functions that call INPUTNAME, e.g. TABLE:
S = struct('a',{1,2});
T = table(S.a) % this works (two lines)...
T = 1×2 table
Var1 Var2 ____ ____ 1 2
[X,Y] = ndgrid(struct('a',{1,2}).a) % and this too (no INPUTNAME)...
X = 1
Y = 2
T = table(struct('a',{1,2}).a) % yet this does not.
Error using inputname
Argument number is not valid.

Error in table (line 159)
for i = 1:numVars, varnames{i} = inputname(i); end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
I reported the bug. My guess is that it has something to do with how the number of input names are identified from a static code analysis, and perhaps has never been updated for this (still relatively new) syntax.
Paul
Paul 2025년 7월 27일
Would you mind posting back here the outcome of the bug report? Thx.
Stephen23
Stephen23 2025년 8월 1일
@Paul: TMW confirmed that it is a bug, which apparently arose due to changes made in R2020b. It will get fixed in a future release.

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Bruno Luong
Bruno Luong 2020년 12월 19일

0 개 추천

N=3
K=5
[~,A] = ismember(dec2base(0:N^K-1,N),['0':'9' 'A':'Z']);
A = A-1

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