# I get an error, what's wrong? on Sparse matrix logic and answer

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Write the function for

A sparse matrix is a large matrix with almost all elements of the same value (typically zero). The normal representation of a sparse matrix takes up lots of memory when the useful information can be captured with much less. A possible way to represent a sparse matrix is with a cell vector whose first element is a 2-element vector representing the size of the sparse matrix. The second element is a scalar specifying the default value of the sparse matrix. Each successive element of the cell vector is a 3-element vector representing one element of the sparse matrix that has a value other than the default. The three elements are the row index, the column index and the actual value. 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. Consider the following run:

cellvec = {[2 3], 0, [1 2 3], [2 2 -3]};

matrix = sparse2matrix(cellvec)

matrix =

0 3 0

0 -3 0

##### 23 Comments

Walter Roberson
on 9 Nov 2023

Locking threads is something that Mathworks is working on, but it isn't available yet.

It will likely take the form of requiring a minimum reputation to add to soft-locked Questions.

DGM
on 9 Nov 2023

Edited: DGM
on 9 Nov 2023

All we can really do is post a notice clarifying that duplicates will be subject to deletion, and then manually follow up on it whenever new activity occurs. That's what I mean when I say "soft lock". Sort of like:

It helps to have the thread cleaned and cataloged enough to be able to summarize the forms that have already been exhausted. Rik's comment on that question makes it easier for the reader to grasp the scope of the whole thread, and it also makes it easier to moderate it when the rules are clearly stated and a summary is available.

Of course, that would require cleaning up and summarizing all these sprawling threads, in many of which the repeated patterns aren't as easy to distinguish. It makes it all the more time consuming when the volume of content make the page laggy. When the answers are large, it makes comparison all the more tedious. One thread I don't even want to deal with is this one:

I've tried a number of ideas to steamline the cleanup task, but if I'm honest, I don't really have a good way. After an hour or three staring at piles of slightly embellished variations with vague timestamps, I tend to start losing perspective and patience and it's hard for me to feel confident that I'm being fair.

Even without a full cleanup and formal notice, there's no reason we can't check new posts as they arrive on duplicate-prone types of threads. I tend to just keep an eye out for old threads that get bumped. It helps catch traffic on these sorts of homework//onramp/coursera threads, but it also catches the cases where people try burying their questions where they don't belong.

### Answers (30)

stanleo
on 7 Jul 2019

%simple version

function matrix = sparse2matrix(cellvec)

matrix = cellvec{2}*ones(cellvec{1});

for m=3:size(cellvec,2)

matrix(cellvec{m}(1),cellvec{m}(2))=cellvec{m}(3);

end

##### 6 Comments

Rik
on 29 Aug 2020

Walter Roberson
on 29 Aug 2020

Bhoomika:

We do not have any idea what your level of experience in programming is. We would have to start from the basics of mathematics and computer science to explain the entire code in a way that we could relatively sure you would understand. That would take at least two textbooks of explanation. None of us has time to write all that.

We suggest you ask more specific questions that can be more easily answered.

AYUSH GURTU
on 28 May 2019

function [matrix]=sparse2matrix(incell)

S=size(incell);

q=S(2)-2;

msize=incell{1};

mdef=incell{2};

matrix=repmat(mdef,msize);

while q>0

matrix(incell{q+2}(1), incell{q+2}(2)) = incell{q+2}(3);

q = q-1;

end

##### 4 Comments

Kaushik Hariharan
on 13 Feb 2024 at 19:35

Edited: Kaushik Hariharan
on 13 Feb 2024 at 19:41

Could anyone explain why q=S(2)-2 = 2? I would have thought it was zero since S is size of incell which is 1*4. And s(2) could either be the 2nd index value. I am not sure how it is 4-2.

for the while loop, why are we calling q+2? and then doing q=q-1? in Incell isn;t there two values of q+2 = 4 for the first test case and if we do that we would skip the first 3-element vector. When I tried this code, I saw that it chaged 1,2 first to value of 3 then when down to 2,2 then changed the value of -3. I am trying to understand the process matlab code is taking, perhaps understanding the when s(2) = 4 would probably explain more.

DGM
on 14 Feb 2024 at 11:02

There's an offset of 2 because the first two elements of the cell array store the size of the original numeric array and the fill value.

The reason that q is decremented from S(2)-2 to zero is because whoever wrote this chose to use a while loop and count backwards instead of just using a for loop.

Pavel Radko
on 13 Aug 2020

Edited: Pavel Radko
on 13 Aug 2020

Passed all tests solution. May be not the best one (because I have no idea how to biuld default matrix in easier way), but works 100%.

% Build a matrix called "matrix" using instrictions of input "cellvec"

function matrix = sparse2matrix(cellvec)

% first we build a default matrix with size ii*jj

% we use 1st element of "cellvec" to get the size of matrix

for ii = 1:cellvec{1}(1,1)

for jj = 1:cellvec{1}(1,2)

% all elements of matrix equals to the 2nd element of "cellvec"

matrix(ii,jj) = cellvec{2};

end

end

% now we need to change elements of our default matrix

% instructions for place and value of this elements comes in "cellvec"

% from 3rd element till the end of "cellvec"

for zz = 3:length(cellvec)

% we call "matrix" elements and assign values to them from every 3rd element of subarrays of "cellvec"

matrix(cellvec{zz}(1,1),cellvec{zz}(1,2)) = cellvec{zz}(1,3);

end

end

##### 3 Comments

THIERNO AMADOU MOUCTAR BALDE
on 29 Dec 2020

working for the example given in the problem but others no thanks for sharing

manish Singh
on 18 Jun 2021

You wrote down the complex code into very simple manner and it do work for any problem. And I understand it very well

Thanks man,

Abhishek singh
on 24 Apr 2019

function [matrix]= sparse2matrix(incell);

X=size(incell);

q=X(2)-2;

msize=incell{1};

mdef=incell{2};

matrix=repmat(mdef,msize);

while q > 0

matrix(incell{q+2}(1), incell{q+2}(2)) = incell{q+2}(3);

break

end

output

matrix =

0 0 0

0 -3 0

required output

matrix =

0 3 0

0 -3 0

##### 2 Comments

Abhishek singh
on 24 Apr 2019

# added q+1

function [matrix]= sparse2matrix(incell);

X=size(incell);

q=X(2)-2;

msize=incell{1};

mdef=incell{2};

matrix=repmat(mdef,msize);

while q > 0

matrix(incell{q+1}(1), incell{q+1}(2)) = incell{q+1}(3);

matrix(incell{q+2}(1), incell{q+2}(2)) = incell{q+2}(3);

break

end

matrix =

0 3 0

0 -3 0

but failed for

ariable solution has an incorrect value.

sparse2matrix( { [ 9 12 ], 3, [ 6 2 6 ], [ 7 1 -6 ], [ 1 10 -7 ], [ 2 2 -3 ], [ 1 4 -8 ], [ 1 11 -8 ], [ 9 11 -8 ], [ 7 8 5 ], [ 9 8 4 ], [ 9 11 7 ], [ 5 9 -4 ], [ 8 12 8 ], [ 3 6 5 ] } ) failed...

Walter Roberson
on 24 Apr 2019

Why are you using break after one iteration of the loop ? If you are only going to do a set of instructions once, do not bother to put it in a loop.

I suggest that you read about for loops.

Jaimin Motavar
on 30 Jun 2019

Edited: Jaimin Motavar
on 30 Jun 2019

can you tell me what is wrong in this answer?

function matrix = sparse2matrix(a)

e=length(a);

b=rand(a{1,1});

[m,n]=size(b);

c=a{1,3};

d=a{1,4};

for i=1:m

for j=1:n

b(i,j)=a{1,2};

end

end

for g=3:e

for f=(g-2):(e-2)

p(1,f)=a{1,g}(1,1);

end

end

for g=3:e

for f=(g-2):(e-2)

q(1,f)=a{1,g}(1,2);

end

end

for g=3:e

for f=(g-2):(e-2)

r(1,f)=a{1,g}(1,3);

end

end

for o=1:(e-2)

b(p(o),q(o))=r(o);

end

matrix=b;

end

##### 0 Comments

Litesh Ghute
on 20 Mar 2020

What's wrong with my code ?

function matrix= sparse2matrix(v)

mat=zeros([v{1}(1),v{1}(2)]);

r=size(mat);

m=3;

while m <= 4

i=v{m}(1);

j=v{m}(2);

mat(v{m}(i,j))=v{m}(3);

m=m+1;

end

matrix=mat;

end

##### 1 Comment

ABINAND PANDIYAN
on 23 Apr 2020

Edited: ABINAND PANDIYAN
on 23 Apr 2020

%All the test cases are right. try this

function matrix= sparse2matrix(cellvec)

a= cellvec{1};

row=a(1);

column=a(2);

main_value= cellvec{2};

sparse_matrix= main_value * ones(row, column);

len= length(cellvec);

for i= 3:length(cellvec)

change = cellvec{i};

r=change(1);

c=change(2);

m=change(3);

sparse_matrix(r,c)=m;

end

matrix=sparse_matrix;

end

##### 4 Comments

THIERNO AMADOU MOUCTAR BALDE
on 29 Dec 2020

thank you so much!

it is working just one suggestion for using the variable len

len = length(cellvec);

for i = i= 3:len

...

......

end

SAMARTH MAHESHKUMAR GEMLAWALA
on 15 May 2020

% Compteled all the test cases successfully.

function matrix = sparse2matrix(a)

cellvec = a

p= size(cellvec)

z = cellvec{1,1}

x = cellvec{1,2}

matrix = zeros(z(1,1),z(1,2));

for i=1:z(1,1)

for j= 1:z(1,2)

matrix(i,j) = x;

end

end

for j= 3: p(1,2)

y = cellvec{1,j}

matrix(y(1,1),y(1,2)) = y(1,3);

end

##### 1 Comment

Priyansh Kushwaha
on 16 May 2020

Edited: Priyansh Kushwaha
on 17 May 2020

function matrix=sparse2matrix(a)

b=a{1,1}

b1=b(1,1);

b2=b(1,2);

e=ones(b1,b2);

b=a{1,2}

e=b.*(e);

for i=3:length(a)

c=a{1,i};

d1=c(1,1);

d2=c(1,2);

d3=c(1,3);

e(d1,d2)=d3;

matrix=e;

end

matrix=e;

end

##### 3 Comments

Priyansh Kushwaha
on 17 May 2020

utkarsh singh
on 21 May 2020

Edited: utkarsh singh
on 21 May 2020

function matrix=sparse2matrix(a)

row=a{1,1}(1,1);

col=a{1,1}(1,2);

default=a{1,2}; % or simply default=a{1,2}

matrix=ones(row,col)*default; % matrix=ones(a{1})*deafult.....no need of finding row and col

for m=3:length(a)

matrix(a{m}(1,1),a{m}(1,2))=a{m}(1,3);

end

##### 0 Comments

Taif Ahmed BIpul
on 24 May 2020

function matrix=sparse2matrix(cellvec)

m=ones(cellvec{1}(1),cellvec{1}(2));

m=m.*cellvec{2};

for i=3:length(cellvec)

m(cellvec{i}(1),cellvec{i}(2))=cellvec{i}(3);

end

matrix=m;

Ahmed Mamdouh
on 7 Jun 2020

function matrix = sparse2matrix(ce)

matri=ones(ce{1,1}(1,1),ce{1,1}(1,2))*ce{1,2};

siz=size(ce);

i=siz(1,2);

for ii=3:i

matri( ce{1,ii}(1,1),ce{1,ii}(1,2))=ce{1,ii}(1,3);

end

matrix=matri;

##### 0 Comments

Shikha Thapa
on 13 Jun 2020

function matrix=sparse2matrix(cellvec)

matrix=cellvec{1,2}*ones([cellvec{1}(1),cellvec{1}(2)]);

for a=3:length(cellvec)

matrix(cellvec{1,a}(1,1), cellvec{1,a}(1,2))=cellvec{1,a}(1,3);

end

##### 1 Comment

Kumar Shubham
on 12 Jul 2020

Edited: Kumar Shubham
on 12 Jul 2020

function matrix = sparse2matrix(cellvec)

%creates matrix of req. size.

matrix=zeros(cellvec{1});

%allots assigned scalar value to all elements.

matrix(:)=deal(cellvec{2});

%used loop to maipulate matrix for result.

%use breakpoints to see approach to result step by step .

for ii = 3:length(cellvec)

matrix(cellvec{ii}(1,1),cellvec{ii}(1,2))=cellvec{ii}(1,3);

end

##### 1 Comment

Walter Roberson
on 12 Jul 2020

Why are you using deal? Are you expecting that cellvec{2} will expand to multiple comma-separated elements? That is not going to happen with a scalar index like {2}

If you are wanting to copy the one value expectd in cellvec{2} to all elements on the left, then you do not need deal() .

Ishani Uthpala
on 1 Aug 2020

function matrix=sparse2matrix(v)

matrix=v{1,2}*ones(v{1,1}(1,1),v{1,1}(1,2));

y=length(v);

for x=3:y

matrix(v{1,x}(1,1),v{1,x}(1,2))=v{1,x}(1,3);

x=x+1;

end

matrix;

end

##### 2 Comments

Walter Roberson
on 1 Aug 2020

What is the purpose of your line

x=x+1;

??

What is the purpose of your line

matrix;

??

sushmanth pulavarthi
on 3 Aug 2020

function matrix=sparse2matrix(v)

rows=v{1,1}(1);columns=v{1,1}(2); %extracting total no.of rows and columns for sprase matrix

magnitude=v{2}; %extracting the default value

m=magnitude*ones(rows,columns);

for i=3:length(v) %creating the loop foor changing the values other than default

r=v{i}(1);

c=v{i}(2);

m(r,c)=v{i}(3);

end

matrix=m;

end

%this works for any no.of elements

##### 0 Comments

A.H.M.Shahidul Islam
on 6 Aug 2020

% 100% accurate

function matrix=sparse2matrix(m)

m=cell(m);

r=m{1}(1);c=m{1}(2);dv=m{2};

ss=size(m);

matrix=sparse(r,c)+dv;

q=ss(1,2);

for ii=3:q

matrix(m{ii}(1),m{ii}(2))=m{ii}(3);

end

##### 1 Comment

Rik
on 6 Aug 2020

Thanks, now I can cheat on my homework without having to bother understanding the problem or the solution.

On a slightly more serious note: you forgot the closing end. Although you don't need it, it has become a lot more common, especially since it is possible to put functions in script files.

Ali Raza
on 9 Sep 2020

function matrix = sparse2matrix(x)

M = x{1};

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

[~,len] = size(x);

if len == 3

i = 3;

m(x{i}(1),x{i}(2)) = x{i}(3);

else

for i = 3:len

m(x{i}(1),x{i}(2)) = x{i}(3);

end

end

matrix = m;

end

##### 1 Comment

Abdul Quadir Khan
on 6 Nov 2020

function matrix = sparse2matrix (cellvec)

m = cellvec{1}(1,1);

n = cellvec{1}(1,2);

defult = ones(m,n) .* cellvec{1,2};

for i= 3:length(cellvec)

r1 = cellvec{i}(1,1);

c1 = cellvec{i}(1,2);

defult(r1,c1) = cellvec{i}(1,3);

end

matrix = defult;

end

##### 0 Comments

zehra ülgen
on 12 Nov 2020

Here is another solution..

function m = sparse2matrix(a)

[t c] = size(a);

m = zeros(a{1,1});

[x y] = size(m);

for ii = 1:x;

for jj = 1:y;

m(ii,jj) = a{1,2};

end

end

for i = 3:c;

v = a(1,i);

m(v{1,1}(1,1),v{1,1}(1,2)) = v{1,1}(1,3);

end

##### 0 Comments

Alberto Gil
on 29 Dec 2020

Edited: Alberto Gil
on 29 Dec 2020

Hello people,

What do you think about this code?

function matrix=sparse2matrix(cll)

if iscell(cll)==1

% Declare values, cs=size of the array; cdn=the nominal value; N=greatest value;

cs=cll{1,1}; cdn=cll{1,2}; N=size(cll,2);

% Create the matrix with the nominal value and the size.

cm=ones(cs)*cdn;

for n=3:N;

cxdn=cll{1,n};

% Select the values of the input values.

cm(cxdn(1,1), cxdn(1,2))=cxdn(1,3);

end

matrix= cm;

else

error('The input must be a cell class');

end

end

##### 1 Comment

Walter Roberson
on 30 Dec 2020

The question does not seem to require that you verify that the input is a cell.

xin yi leow
on 19 Jan 2021

function matrix=sparse2matrix(cellx)

matrix=zeros(cellx{1});

matrix(:,:)=cellx{2};

for ii=3:length(cellx)

num=cellx{ii};

matrix(num(1),num(2))=num(3);

end

end

##### 1 Comment

Minh Nguyen
on 27 Mar 2021

Edited: Minh Nguyen
on 27 Mar 2021

my idea about this

function matrix = sparse2matrix(ABC)

r = ABC{1}(1);

c = ABC{1}(2);

B = zeros(r,c); %make a zero matrix

B(1:end) = ABC{2}; % the sparse matrix with the second element

for i = 3:length(ABC) % and adding

a1 = ABC{i}(1,1);

a2 = ABC{i}(1,2);

B(a1,a2) = ABC{i}(1,3);

end

matrix = B;

end

##### 0 Comments

Blaze Shah
on 13 Sep 2021

Edited: Walter Roberson
on 13 Sep 2021

function matrix = sparse2matrix(cellvec)

jj = cell2mat(cellvec);

m = jj(3)*ones(jj(1,[1,2]));

n = 4;

while n<=length(jj)

m(jj(n),jj(n+1)) = jj(n+2);

n = n+3;

end

matrix = m;

##### 0 Comments

Sumanth Bayya
on 19 Oct 2021

Edited: Sumanth Bayya
on 19 Oct 2021

function M = sparse2matrix(cellvec)

sz = cellvec{1};

val = cellvec{2};

M = val*ones(sz);

for i = 3:length(cellvec)

el = cellvec{i};

M(el(1), el(2)) = el(3);

end

end

##### 0 Comments

Ujwal Dhakal
on 7 Jan 2022

function matrix = sparse2matrix (cellvec)

[a b] = size(cellvec);% stores the number of cell elements in b whereas a is always 1 as cellvec is a vector

a = cellvec{1,1};% is size of the matrix that loads into a vector a

default_element=cellvec{1,2};

%preallocating the matrix to be of size m*n with all elements default

for i=1:a(1)%a(1) is the no of rows in the matrix

for j=1:a(2) %a(2) is the no of columns in the matrix

matrix(i,j)=cellvec{1,2};

end

end%matrix is generated with all elements set to default value

for ii=3:b%this loop runs from 3 to no of elements in cell vec

matrix(cellvec{1,ii}(1,1),cellvec{1,ii}(1,2))=cellvec{1,ii}(1,3);

end

##### 0 Comments

昱安 朱
on 11 Mar 2023

function matrix=sparse2matrix(cellvec)

matrix=cellvec{2}*ones(cellvec{1,1});

for ii=3:length(cellvec)

matrix(cellvec{ii}(1),cellvec{ii}(2))=cellvec{ii}(3);

end

##### 0 Comments

abdul kabeer
on 14 Jun 2023

I solve it this way but dont know if this can be any shorter:

function [matrix]=sparse2matrix(v)

matrix = zeros(v{1}(1),v{1}(2))+v{2};

for i = 3:length(v)

matrix(v{i}(1),v{i}(2)) = v{i}(3);

end

##### 0 Comments

Muhammad Saleh
on 2 Nov 2023

function matrix = sparse2matrix(sparce_rep)

matrix=ones(sparce_rep{1});

matrix=matrix*sparce_rep{2};

for ii=3:length(sparce_rep)

matrix(sparce_rep{ii}(1),sparce_rep{ii}(2))= sparce_rep{ii}(3);

end

end

##### 1 Comment

DGM
on 2 Nov 2023

How many copies of the same answer does anybody need?

Is it an improvement to use meaningless variable names?

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