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There are several commands that provide high-level information about the nonzero elements of a sparse matrix:

To try some of these, load the supplied sparse matrix `west0479`

,
one of the Harwell-Boeing collection.

```
load west0479
whos
```

Name Size Bytes Class Attributes west0479 479x479 34032 double sparse

This matrix models an eight-stage chemical distillation column.

Try these commands.

nnz(west0479)

ans = 1887

format short e west0479

west0479 = (25,1) 1.0000e+00 (31,1) -3.7648e-02 (87,1) -3.4424e-01 (26,2) 1.0000e+00 (31,2) -2.4523e-02 (88,2) -3.7371e-01 (27,3) 1.0000e+00 (31,3) -3.6613e-02 (89,3) -8.3694e-01 (28,4) 1.3000e+02 . . .

nonzeros(west0479)

ans = 1.0000e+00 -3.7648e-02 -3.4424e-01 1.0000e+00 -2.4523e-02 -3.7371e-01 1.0000e+00 -3.6613e-02 -8.3694e-01 1.3000e+02 . . .

**Note**

Use **Ctrl+C** to stop the `nonzeros`

listing
at any time.

Note that initially `nnz`

has the same value
as `nzmax`

by default. That is, the number of nonzero
elements is equivalent to the number of storage locations allocated
for nonzeros. However, MATLAB^{®} does not dynamically release memory
if you zero out additional array elements. Changing the value of some
matrix elements to zero changes the value of `nnz`

,
but not that of `nzmax`

.

However, you can add as many nonzero elements to the matrix
as desired. You are not constrained by the original value of `nzmax`

.

For any matrix, full or sparse, the `find`

function
returns the indices and values of nonzero elements. Its syntax is

[i,j,s] = find(S);

`find`

returns the row indices of nonzero values
in vector `i`

, the column indices in vector `j`

,
and the nonzero values themselves in the vector `s`

.
The example below uses `find`

to locate the indices
and values of the nonzeros in a sparse matrix. The `sparse`

function
uses the `find`

output, together with the size of
the matrix, to recreate the matrix.

S1 = west0479; [i,j,s] = find(S1); [m,n] = size(S1); S2 = sparse(i,j,s,m,n);

Because sparse matrices are stored in compressed sparse column format, there are different costs associated with indexing into a sparse matrix than there are with indexing into a full matrix. Such costs are negligible when you need to change only a few elements in a sparse matrix, so in those cases it’s normal to use regular array indexing to reassign values:

B = speye(4); [i,j,s] = find(B); [i,j,s]

ans = 1 1 1 2 2 1 3 3 1 4 4 1

B(3,1) = 42; [i,j,s] = find(B); [i,j,s]

ans = 1 1 1 3 1 42 2 2 1 3 3 1 4 4 1

`42`

at `(3,1)`

, MATLAB inserts
an additional row into the nonzero values vector and subscript vectors,
then shifts all matrix values after `(3,1)`

.Using linear indexing to access or assign an element in a large
sparse matrix will fail if the linear index exceeds `2^48-1`

,
which is the current upper bound for the number of elements allowed
in a matrix.

S = spalloc(2^30,2^30,2); S(end) = 1

Maximum variable size allowed by the program is exceeded.

To access an element whose linear index is greater than `intmax`

,
use array indexing:

S(2^30,2^30) = 1

S = (1073741824,1073741824) 1

While the cost of indexing into a sparse matrix to change a single element is negligible, it is compounded in the context of a loop and can become quite slow for large matrices. For that reason, in cases where many sparse matrix elements need to be changed, it is best to vectorize the operation instead of using a loop. For example, consider a sparse identity matrix:

n = 10000; A = 4*speye(n);

`A`

within a loop is slower than a similar vectorized
operation:tic A(1:n-1,n) = -1; A(n,1:n-1) = -1; toc

Elapsed time is 0.003344 seconds.

tic for k = 1:n-1 C(k,n) = -1; C(n,k) = -1; end toc

Elapsed time is 0.448069 seconds.

`A`

during each pass through the loop.Preallocating the memory for a sparse matrix and then filling it in an element-wise manner similarly causes a significant amount of overhead in indexing into the sparse array:

S1 = spalloc(1000,1000,100000); tic; for n = 1:100000 i = ceil(1000*rand(1,1)); j = ceil(1000*rand(1,1)); S1(i,j) = rand(1,1); end toc

Elapsed time is 2.577527 seconds.

Constructing the vectors of indices and values eliminates the need to index into the sparse array, and thus is significantly faster:

i = ceil(1000*rand(100000,1)); j = ceil(1000*rand(100000,1)); v = zeros(size(i)); for n = 1:100000 v(n) = rand(1,1); end tic; S2 = sparse(i,j,v,1000,1000); toc

Elapsed time is 0.017676 seconds.

For that reason, it’s best to construct sparse matrices
all at once using a construction function, like the `sparse`

or `spdiags`

functions.

For example, suppose you wanted the sparse form of the coordinate
matrix `C`

:

$$\text{C}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\begin{array}{cccc}4& 0& 0& \begin{array}{cc}0& -1\end{array}\\ 0& 4& 0& \begin{array}{cc}0& -1\end{array}\\ 0& 0& 4& \begin{array}{cc}0& -1\end{array}\\ \begin{array}{c}0\\ 1\end{array}& \begin{array}{c}0\\ 1\end{array}& \begin{array}{c}0\\ 1\end{array}& \begin{array}{cc}\begin{array}{c}4\\ 1\end{array}& \begin{array}{c}-1\\ 4\end{array}\end{array}\end{array}\right)$$

Construct the five-column matrix directly with the `sparse`

function using the triplet pairs for the row subscripts, column
subscripts, and
values:

i = [1 5 2 5 3 5 4 5 1 2 3 4 5]'; j = [1 1 2 2 3 3 4 4 5 5 5 5 5]'; s = [4 1 4 1 4 1 4 1 -1 -1 -1 -1 4]'; C = sparse(i,j,s)

C = (1,1) 4 (5,1) 1 (2,2) 4 (5,2) 1 (3,3) 4 (5,3) 1 (4,4) 4 (5,4) 1 (1,5) -1 (2,5) -1 (3,5) -1 (4,5) -1 (5,5) 4

It is often useful to use a graphical format to view the distribution of the nonzero elements within a sparse matrix. The MATLAB `spy`

function produces a template view of the sparsity structure, where each point on the graph represents the location of a nonzero array element.

For example:

Load the supplied sparse matrix `west0479`

, one of the Harwell-Boeing collection.

`load west0479`

View the sparsity structure.

spy(west0479)