Info

This question is closed. Reopen it to edit or answer.

# Write a function called sparse2matrix that takes a single input of a cell vector as defined above and returns the output argument called matrix, the matrix in its traditional form

1 view (last 30 days)
Abhishek singh on 21 Apr 2019
Closed: Rik on 23 Jun 2020
cellvec = {[2 3], 0, [1 2 3], [2 2 -3]};
matrix = sparse2matrix(cellvec)
matrix =
0 3 0
0 -3 0
function [matrix]=sparse2matrix(incell)
msize = incell{1};
mdef = incell{2};
matrix = repmat(mdef,msize);
for n = 3:numel(incell)
RCV = incell{n};
end
matrix = sparse2matrix({[2 3], 0, [1 2 3], [2 2 -3]})
matrix =
0 0 0
0 0 0
Assessment result: incorrectA few simple cases
Variable solution has an incorrect value.
sparse2matrix( { [ 3 4 ], 0, [ 2 2 -3 ], [ 1 3 3 ] } ) failed...
Assessment result: incorrectRandom cases
Variable solution has an incorrect value.
sparse2matrix( { [ 10 10 ], 3, [ 9 9 0 ], [ 9 8 8 ], [ 8 6 -7 ], [ 7 7 4 ], [ 1 1 0 ], [ 4 8 7 ], [ 1 4 1 ], [ 4 8 -1 ], [ 8 7 6 ] } ) failed..
##### 6 CommentsShowHide 5 older comments
Rik on 14 Jun 2020
My reason for closing this question: people are just posting their answers to this homework question, without substantial discussion happening. Once there is an option to disallow new answers while allowing comments that option should be considered for this thread.

### Answers (10)

Arafat Roney on 11 May 2020
function matrix=sparse2matrix(p)
m=p{1}(1,1);
n=p{1}(1,2);
o=p{2}(1,1);
s=o.*ones(m,n);
for i=3:length(p)
r=p{i}(1,1);
c=p{i}(1,2);
v=p{i}(1,3);
s(r,c)=v;
end
matrix=s;
end
##### 0 CommentsShowHide -1 older comments

Emine Ertugrul on 17 Apr 2020
function matrix = sparse2matrix(cell)
matrix = cell{2}*ones(cell{1}(1),cell{1}(2))
for ii = 3:length(cell)
matrix(cell{ii}(1),cell{ii}(2)) = cell{ii}(3);
end
##### 1 CommentShowHide None
Rik on 17 Apr 2020
There are several issues with this answer. Firstly it is not formatted correctly, making it less readable. Secondly it seems intended to be a fully working solution to a homework question, encouraging cheating.
But more importantly, it is using cell as a variable name while using the cell data type. This will very likely lead to confusion. Also, since the question definition is not entirely clear, it is possibly not the correct answer to the question.

Saibalaji Kokate on 23 Apr 2020
function ans=sparse2matrix(m)
x=m{1,1};
class(x)
val=m{1,2};
mat=repmat(val,x);
len=length(m);
for y=3:len
a=m{y};
row=a(1,1);
col=a(1,2);
mat(row,col)=a(1,3);
end
ans=mat;
end
##### 0 CommentsShowHide -1 older comments

Soham Khadilkar on 26 Apr 2020
Edited: Soham Khadilkar on 26 Apr 2020
function matrix = sparse2matrix(cellvec)
r = cellvec{1,1}(1,1);
c = cellvec{1,1}(1,2);
[e,l] = size(cellvec)
matrix = ones(r,c)*cellvec{1,2};
for i= 3:l
r1 = cellvec{1,i}(1,1);
c1 = cellvec{1,i}(1,2);
matrix(r1,c1) = cellvec{1,i}(1,3);
i=i+1;
end
%This works for any length of the cellvec
%the code is probably a little long so suggest some stuff to make it short.
##### 2 CommentsShowHide 1 older comment
Rik on 23 Jun 2020
They are created on the third line of this function. Have you read the documentation for the size function?

Ved Prakash on 15 May 2020
function matrix=sparse2matrix(P)
r=P{1}(1);
c=P{1}(2);
D=P{2}*ones(r,c);
for i=3:length(P)
D(P{i}(1),P{i}(2))=P{i}(3);
end
matrix=D;
##### 0 CommentsShowHide -1 older comments

Syed Zuhair Ali Razvi on 22 May 2020
function mat=sparse2matrix(cellvec)
m1=zeros(cellvec{1});
m2=cellvec{2}+m1;
if length(cellvec)<=2
mat=m2;
else
for i=cellvec(3:end)
for n=1:i{end}
a1=i{1,n}(1,1);
a2=i{1,n}(1,2);
m2(a1,a2)=i{n}(1,3);
mat=m2;
n=n+1;
break
end
end
end
end
##### 1 CommentShowHide None
Rik on 22 May 2020
If you are going to post a complete solution to a homework question, at least write proper documentation and comments for your function, and make sure to properly allign the code.

Tahsin Oishee on 3 Jun 2020
function matrix=sparse2matrix(x)
matrix=zeros(x{1})
matrix(1:end)=x{2}
a=length(x);
i=3;
for i=3:a
matrix(x{1,i}(1),x{1,i}(2))=x{1,i}(1,3)
i=i+1
end
end
##### 1 CommentShowHide None
Walter Roberson on 8 Jun 2020
What benefit do you see for incrementing i within the for i loop?

Raymond He on 8 Jun 2020
function matrix = sparse2matrix(a)
matrix = a{2}(1,1) * ones(a{1}(1,1),a{1}(1,2));
for i = 3:length(a)
matrix(a{i}(1,1),a{i}(1,2)) = a{i}(1,3);
end
end
##### 0 CommentsShowHide -1 older comments

UJJWAL Padha on 10 Jun 2020
Edited: UJJWAL Padha on 10 Jun 2020
function matrix = sparse2matrix(a)
sparse_matrix = 0; % assigning variable sparse_matrix = 0
for i = 1:a{1,1}(1,1) %running for loop from 1st to the first element of vector(i.e rows of matrix) assigned to first cell of a
for j = 1:a{1,1}(1,2) %running for loop from 2nd to the first element of vector(i.e columns of matrix) assigned to first cell of a
sparse_matrix(i,j) = a{1,2} ; %here all the elements of sparse_matrix will become default i.e 2nd cell of a
end
end
for l= 3:length(a) % running for loop from 3 to length of a
sparse_matrix(a{1,l}(1,1) , a{1,l}(1,2)) = a{1,l}(1,3); %as the loop runs from 3 to length of a the elements, the non-default values will be assigned to respective elements of sparsematrix
end
matrix = sparse_matrix;
end
##### 4 CommentsShowHide 3 older comments
UJJWAL Padha on 11 Jun 2020
i understand now.....
Thankyou for sharing this piece of knowledge....
Appreciate it :)

Md Nazmus Sakib on 19 Jun 2020
function y = sparse2matrix(p)
row = p{1,1}(1,1);
col = p{1,1}(1,2);
default_val = p{1,2};
matrix = [];
%assigning default value to all positions
for i = 1 : row
for j = 1 : col
matrix(i,j) = default_val;
end
end
row_mat = [];
col_mat = [];
val_mat = [];
%fetching all rows info and creating a vector
for m = 1 : length(p)
if ((m == 1) || (m == 2))
continue
else
row_mat(m) = p{1,m}(1,1);
end
end
%fetching all columns info and creating a vector
for n = 1 : length(p)
if ((n == 1) || (n == 2))
continue
else
col_mat(n) = p{1,n}(1,2);
end
end
%fetching all values and creating a vector
for no = 1 : length(p)
if ((no == 1) || (no == 2))
continue
else
val_mat(no) = p{1,no}(1,3);
end
end
%rewriting the values to corresponding positions
for x = 3 : length(row_mat)
matrix(row_mat(x),col_mat(x)) = val_mat(x);
end
y = matrix;
end

R2016b

### Community Treasure Hunt

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

Start Hunting!