Splitting an array in every possible combination
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Hello,
I am attempting to split an array in all possible combination. I have illustrated it in the figure given below.
Depending on the number of rows, I wish to split the given array in all possible combinations.
One of the condition is Number of rows <= Number of elements in given array
After toiling for 3 days, I am posting this here.
Kindly guide me on how to accomplish this task.
Thanks for your time and valuable guidance.
12 个评论
Selva Kumar
2022-7-14
编辑：Selva Kumar
2022-7-14
maybe it helps you to see how such a matrix is generated. You can run this script multiple times to see different variants
you did still not tell us how many columns are allowed
numElements=randi([2 10]); % number of elements shall be random between 2 and 10
elements=1:numElements; % define elements
numOfRows=randi([2 numElements]); % random number of rows
elInRowNrX=zeros(numOfRows,1); % vector containing the number of elements per row
createdMatrix=[];
for rowNr=1:numOfRows
if rowNr<numOfRows
elInRowNrX(rowNr)=randi(numElements-sum(elInRowNrX)-(numOfRows-rowNr)); % we cannot enter more values than total number of values minus already given values minus 1 for each remaining row
createdMatrix(rowNr,1:elInRowNrX(rowNr))=elements(1:elInRowNrX(rowNr));
elements(1:elInRowNrX(rowNr))=[];
else
elInRowNrX(rowNr)=numElements-sum(elInRowNrX); % last row has to contian the remaining number of elements
createdMatrix(rowNr,1:elInRowNrX(rowNr))=elements;
end
end
disp(createdMatrix)
Steven Lord
2022-7-14
Those aren't "every possible combination". What about:
[x6 0 0 0 0 0;
x1 x2 x3 x4 x5 0]
What about:
[ 0 0 0 0 0 x1;
x2 x3 x4 x5 x6 0]
or:
[x1 0 0 0 0;
x2 x3 x4 x5 x6]
or
[x1 x2 x3;
x4 x5 x6]
What rules haven't you stated that disallow those options?
@Steven Lord to my understanding we have to fill the rows from the left and put in the elements one by one
your last case depends on a definition of number of columns, which was also not given. but the requested output can be generated by my script
numElements=6; % number of elements shall be random between 2 and 10
elements=1:numElements; % define elements
numOfRows=2; % random number of rows
elInRowNrX=zeros(numOfRows,1); % vector containing the number of elements per row
createdMatrix=[];
for rowNr=1:numOfRows
if rowNr<numOfRows
elInRowNrX(rowNr)=randi(numElements-sum(elInRowNrX)-(numOfRows-rowNr)); % we cannot enter more values than total number of values minus already given values minus 1 for each remaining row
createdMatrix(rowNr,1:elInRowNrX(rowNr))=elements(1:elInRowNrX(rowNr));
elements(1:elInRowNrX(rowNr))=[];
else
elInRowNrX(rowNr)=numElements-sum(elInRowNrX); % last row has to contian the remaining number of elements
createdMatrix(rowNr,1:elInRowNrX(rowNr))=elements;
end
end
disp(createdMatrix)
Selva Kumar
2022-7-14
编辑：Selva Kumar
2022-7-14
Dyuman Joshi
2022-7-14
Is the number of rows allowed only 2 and 3?
Selva Kumar
2022-7-14
Selva Kumar
2022-7-14
Dyuman Joshi
2022-7-14
This looks like a herculean task. I'll give it a try and update you.
Jonas
2022-7-14
@Selva Kumar, i know that it gives only one solution, i also wrote that it gives one at a time. but all of the solutions are at least valid. it should just be a help how the problem could finally be solved, that's why the code is a comment and not in the answer section
Jonas
2022-7-14
one 'solution' could be to run the script multiple times and collect unique solutions. but we still do not know restrictions on the number of columns
采纳的回答
I believe this will do the required "splitting" of the array:
format compact
x = 1:6;
disp(split_x(x,1));
disp(split_x(x,2));
disp(split_x(x,3)); % scroll down to see them all
disp(split_x(x,4)); % scroll down to see them all
disp(split_x(x,5)); % scroll down to see them all
disp(split_x(x,6));
function x_split = split_x(x,n_rows)
N = numel(x);
n_cols = N-n_rows+1;
split_idx = nchoosek(1:N-1,n_rows-1);
n_combos = size(split_idx,1);
split_idx = [zeros(n_combos,1) split_idx N*ones(n_combos,1)];
n_split = diff(split_idx,1,2);
x_split = zeros(n_rows,n_cols,n_combos);
for ii = 1:n_combos
for jj = 1:n_rows
x_split(jj,1:n_split(ii,jj),ii) = x(split_idx(ii,jj)+(1:n_split(ii,jj)));
end
end
end
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