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# 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

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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 Comments

Walter Roberson
on 21 Apr 2019

Rik
on 14 Jun 2020

### 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 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 Comment

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 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 Comments

Rik
on 23 Jun 2020

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 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 Comment

Rik
on 22 May 2020

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 Comment

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 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 Comments

Walter Roberson
on 10 Jun 2020

Suppose the data indicates that the matrix is to be 3 x 4. Then you would do

for i = 1 : 3; for j = 1 : 4; sparse_matrix(i,j) = a{1,2}; end; end

The first iteration, sparse_matrix is a 1 x 1 scalar, 0, and a 1 x 1 scalar is the same as indexing at (1,1), so assigning to sparse_matrix(1,1) is re-using the storage that had the 0 and that is fine.

The second iteration, sparse_matrix is a 1 x 1 scalar, and you assign to sparse_matrix(1,2) but that is larger than the existing matrix. MATLAB deals with that by creating a new temporary matrix that is large enough to hold the existing data and new data (so 1 x 2) and then copies the existing data into it (so (1,1) would be copied into the temporary matrix); then releases the old matrix and replaces the pointers to it with the new 1 x 2 matrix, and then it does the assignment of the value into the now-larger matrix putting it into location (1,2) .

The third iteration, sparse_matrix is 1 x 2, and you assign to (1,3) but that is larger than the existing matrix. A new larger tempory matrix that is 1 x 3 has to be built, the existing 1 x 2 is copied into it, the 1 x 2 is released, the new value is copied into (1,3).

The fourth iteration, sparse_matrix is 1 x 3, you assign to (1,4), new tempory 1 x 4 is created, existing 1 x 3 is copied into it, old one is released, (1,4) is assigned into.

That is the last j iteration; then you move on to the second i iteration, so i=2, j=1 . sparse_matrix(2,1) is being assigned into, but existing is 1 x 4, so a complete second row needs to be allocated making a 2 x 4 , the existing 1 x 4 copied into it, the 1 x 4 released, then (2,1) is updated.

Next is i=2, j=2. This time the array is already 2 x 4, so no growing is needed, so (2,2) gets updated directly. Same for (2,3) and (2,4).

Then at i=3, j=1, once more you have to allocate a complete new row, copying the existing 2 x 4 into it, release old one, assign into the new (3,1). Then for i=3, j=2 to 4 no new allocation is needed.

So when you assign into a matrix in this way, the matrix ends up getting grown with all the existing items having to be copied, once for each column (for the first row), and then once again for each new row added. That requires copying 1 + 2 + 3 + ... cols = cols * (cols+1) / 2 the first time around. Then each additional row requires copying (row_number-1)*cols items, and by the end of rows that would be rows * (rows-1)/2 * cols items copied around. For 50 rows and 30 columns that would add up to having to copy around about 37215 items. For 100 rows and 30 columns it would be about 148965 data items copied.

Now on the other hand if you were to initialize sparse_array to its known final size, then no copying around would be needed at all.

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

##### 0 Comments

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