sparse2matrix random error

function [matrix]= sparse2matrix(incell);
X=size(incell);
q=X(2)-2;
msize=incell{1};
mdef=incell{2};
matrix=repmat(mdef,msize);
while q > 0
    msize= size(matrix);
    matrix(incell{q+2}(1), incell{q+2}(2)) = incell{q+2}(3);
    q = q-1 ;
end

getting right answer but

Variable solution has an incorrect value.

sparse2matrix( { [ 6 11 ], 9, [ 4 11 4 ], [ 2 8 0 ], [ 4 9 7 ], [ 2 7 8 ], [ 5 5 -7 ], [ 3 10 6 ], [ 2 11 8 ],

[ 2 5 -7 ], [ 2 8 -4 ], [ 5 7 3 ] } )

failed...

[Abhishek singh]

please if you have any idea or correct code .I'm sick of this problem

[dpb]

Not enough context...what's the problem?

[Abhishek singh]

getting error for this input

matrix = sparse ( { [ 6 11 ], 9, [ 4 11 4 ], [ 2 8 0 ], [ 4 9 7 ], [ 2 7 8 ], [ 5 5 -7 ], [ 3 10 6 ], [ 2 11 8 ],

[ 2 5 -7 ], [ 2 8 -4 ], [ 5 7 3 ] } )

Answer 1

[Prathiksha RD]

%This function is a bit bigger but it is just to help you guys to understand better
%This function is to get a sparse matrix
function matrix = sparse2matrix(cellvec)
row = cellvec{1}(1);
col = cellvec{1}(2);
a = cellvec{2};
matrix = a * ones(row,col);   %This is to obtain the initial matrix with the desired number
for ii = 3: length(cellvec)
    r = cellvec{ii}(1,1);
    c = cellvec{ii}(1,2);
    n = cellvec{ii}(1,3);
    matrix(r,c) = n;
end
end

[Michelle Zheng]

Thank you for this answer! It was actually really helpful to for you to break it down step by step, especially since I didn't really understand the question at first.

Answer 2

[Wilver Sánchez]

% 
function matrix=sparse2matrix(cell)
M=cell{2}*ones(cell{1}(1,1),cell{1}(1,2));
for ii=1:length(cell)-2
    M(cell{2+ii}(1,1),cell{2+ii}(1,2))=cell{2+ii}(1,3);
end
matrix=M;
end

[Wilver Sánchez]

I've used this code to solve the problem. I hope it could help to someone! Greetings!

[saiyad junaid]

Great it worked!!!

but what this line means

M=cell{2}*ones(cell{1}(1,1),cell{1}(1,2));

and please explain some further lines too. Grateful for this kardes!!!

[TANUJA GUPTA]

M = cell{2}* ones(cell{1}(1,1), cell{1}(1,2));

This statement means creating a new vector; M, which is a product of the second element of the cell vector and an identity matrix of size row = 1st element's 1st value and column = 1st elements 2nd value.

Answer 3

[James Tursa]

You have two entries for the (2,8) spot, namely [2 8 0] and [2 8 -4]. The way you have your algorithm coded, the first one in the list will prevail. If you want different behavior you will have to change your code, but you haven't told us the rules for this situation. Do you want the last one to prevail? Do you want to add the values for these cases? Or ...?

Answer 4

[KOUSHIK NANDY]

please anyone give the solution

[Walter Roberson]

The solution has been posted in other questions on the same topic. However since this is an assignment then using the solution could risk plagiarism.

Answer 5

[TANUJA GUPTA]

% This function is to get a sparse matrix 
% The first element's first value represent the size of the sparse matrix, whereas the first element's 
% second value represent the column of the sparse matrix. The second element of the vector represent the 
% default value of the sparse matrix.
% When we come to the third element of vector matrix, the first value of the element represent the rows 
% and column of that matrix; whereas the third element represent the actual value of that location.
% The representation of next vector element is similar to that of the previous one
 
function new_matrix = sparse2matrix(vector) % function definition
    sparse_row = vector{1}(1);     % assignment of no of rows to the vectors 1'st elements first value.                               
    sparse_column = vector{1}(2);   % assignment of no of columns to the vectors 1'st elements second value
    new_matrix = vector{2}*ones(sparse_row,sparse_column);%This will create a new vector, which is obtained
                            % by multiplying the value of first vector to the identity matrix of size
                            % i.e. of sparse matrix. Thus generating a new vector which is of same size
                            % as sparse matrix. in other words we are preallocating the elements to the
                            %  new vector,  with default value as 0.
    for i= 3:length(vector) % This helps in reaching to the last element of the vector form the 3rd element
        new_matrix(vector{i}(1),vector{i}(2))= vector{i}(3); % giving the values to the new vector
                                                             %  which were intitially 0.

    end
end

Answer 6

[Aakash Kotia]

function [matrix]= sparse2matrix(cell)
s=cell{2}(1,1)*ones(cell{1}(1,1),cell{1}(1,2));
for i=3:length(cell)
    s(cell{i}(1,1),cell{i}(1,2))=cell{i}(1,3);
end
matrix=s;
end

Answer 7

[Arifuzzaman Joy]

function matrix = sparse2matrix(cellvec)
a=cellvec{1,1};
z=zeros(a(1,1),a(1,2))++cellvec{1,2};
for ii=3:(numel(cellvec))
a=cellvec{1,ii};
z(a(1,1),a(1,2))=a(1,3);
end
matrix=z;

Answer 8

[Nguyen Bui]

function maxtrix = sparse2matrix(cellvec)
a = cellvec{1};
maxtrix = cellvec{2}*ones(a(1,1),a(1,2));
for ii = 3:length(cellvec)
    maxtrix(cellvec{ii}(1,1),cellvec{ii}(1,2)) = cellvec{ii}(1,3);
end
end

Answer 9

[Milad Mehrnia]

function matrix = sparse2matrix(c)  % c = cellvec
matrix = ones(c{1}).*c{2};      
for i = 3:length(c)
    p = c{i}(1:2);      
    matrix(p(1),p(2)) = c{i}(end);      
end