sparse2matrix random error

7 vues (au cours des 30 derniers jours)
Abhishek singh
Abhishek singh le 25 Avr 2019
Modifié(e) : Milad Mehrnia le 24 Oct 2021
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...
  3 commentaires
dpb
dpb le 25 Avr 2019
Not enough context...what's the problem?
Abhishek singh
Abhishek singh le 25 Avr 2019
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 ] } )

Connectez-vous pour commenter.

Réponses (9)

Prathiksha RD
Prathiksha RD le 25 Sep 2020
%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
  1 commentaire
Michelle Zheng
Michelle Zheng le 4 Déc 2020
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.

Connectez-vous pour commenter.


Wilver Sánchez
Wilver Sánchez le 11 Fév 2020
%
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
  3 commentaires
saiyad junaid
saiyad junaid le 18 Mai 2020
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 le 31 Juil 2020
Modifié(e) : TANUJA GUPTA le 31 Juil 2020
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.

Connectez-vous pour commenter.


James Tursa
James Tursa le 25 Avr 2019
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 le 3 Mai 2019
please anyone give the solution
  1 commentaire
Walter Roberson
Walter Roberson le 3 Mai 2019
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.

Connectez-vous pour commenter.


TANUJA GUPTA
TANUJA GUPTA le 31 Juil 2020
% 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 le 8 Août 2020
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 le 14 Fév 2021
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 le 7 Juin 2021
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 le 24 Oct 2021
Modifié(e) : Milad Mehrnia le 24 Oct 2021
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

Catégories

En savoir plus sur Logical dans Help Center et File Exchange

Tags

Produits


Version

R2015b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by