sparse2matrix random error

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Abhishek singh
Abhishek singh 2019-4-25
编辑： Milad Mehrnia 2021-10-24
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...
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dpb
dpb 2019-4-25
Not enough context...what's the problem?
Abhishek singh
Abhishek singh 2019-4-25
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 ] } )

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回答（9 个）

Prathiksha RD
Prathiksha RD 2020-9-25
%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
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Michelle Zheng
Michelle Zheng 2020-12-4
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.

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Wilver Sánchez
Wilver Sánchez 2020-2-11
%
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
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saiyad junaid
saiyad junaid 2020-5-18
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
TANUJA GUPTA 2020-7-31
编辑：TANUJA GUPTA 2020-7-31
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.

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James Tursa
James Tursa 2019-4-25
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 ...?
KOUSHIK NANDY
KOUSHIK NANDY 2019-5-3
please anyone give the solution
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Walter Roberson
Walter Roberson 2019-5-3
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.

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TANUJA GUPTA
TANUJA GUPTA 2020-7-31
% 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
Aakash Kotia
Aakash Kotia 2020-8-8
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
Arifuzzaman Joy
Arifuzzaman Joy 2021-2-14
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;
Nguyen Bui
Nguyen Bui 2021-6-7
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
Milad Mehrnia
Milad Mehrnia 2021-10-24
编辑：Milad Mehrnia 2021-10-24
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

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