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

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Suraj Tawde on 20 Apr 2019
Locked: DGM on 4 Jul 2024 at 0:22
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
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.

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
Rik on 29 Aug 2020
There are several complete solutions on this page. What have you tried so far to piece together what every part means?
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
Kaushik Hariharan on 13 Feb 2024
Edited: Kaushik Hariharan on 13 Feb 2024
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
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 CommentsShow 1 older commentHide 1 older comment
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
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+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.

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

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
Walter Roberson on 21 Mar 2020
You did not use the default value for the matrix.

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
Priya Dwivedi on 18 May 2020
Bt ans coming is incorrect
thank you so much!
it is working just one suggestion for using the variable len
len = length(cellvec);
for i = i= 3:len
...
......
end

% 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
Amith Anoop Kumar on 26 Jun 2020
can you just expalin me the for loop 1:z(1,1)

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 CommentsShow 1 older commentHide 1 older comment
Priyansh Kushwaha on 17 May 2020
When there are less element in the 'a' (less than 3), So ''matrix=e'' assignment is helpful to display the output/ assigning value because in this condition(length(a)<3) the loop does not initiate.
Walter Roberson on 17 May 2020
but there is semicolon so it is not going to display anything.

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

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;
Chandrashekar Upadhya b r on 28 Jun 2020

Thank u It was very simple

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;

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
Shikha Thapa on 13 Jun 2020
You can get help from the answer and code your own logic accordingly!!

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
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
Ishani Uthpala on 1 Aug 2020
I think this will helpfull for you
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

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
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
Walter Roberson on 9 Sep 2020
What is your reason for treating len == 3 differently ?

Mukul Rai on 22 Oct 2020
Edited: Mukul Rai on 22 Oct 2020
function matrix = sparse2matrix(a)
asize=length(a);
r = a{1}(1,1);
c = a{1}(1,2);
z = zeros(r,c)
z(:)= a{2};
if asize<=2
matrix =z
return
end
for jj=3:asize
r1 = a{jj}(1,1);
c1 = a{jj}(1,2);
n1 = a{jj}(1,3);
z(r1,c1)=n1
end
matrix =z

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

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

Alberto Gil on 29 Dec 2020
Edited: Alberto Gil on 29 Dec 2020
Hello people,
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
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
Rik on 19 Jan 2021
What does this answer add? What does it teach? Why should it not be deleted?

Minh Nguyen on 27 Mar 2021
Edited: Minh Nguyen on 27 Mar 2021
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

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;

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

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

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

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

Chaohua on 3 Jul 2024 at 21:09
function matrix = sparse2matrix(A)
matrix_m = ones(A{1});
matrix_m = A{2} * matrix_m;
for i = 3:length(A)
r = A{i}(1);
c = A{i}(2);
matrix_m(r,c) = A{i}(3);
end
matrix = matrix_m;

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